Mahdi Abdeljaouad and Jeffrey Oaks

Al-Hawārī’s Essential Commentary

Series Editors
Ian T. Baldwin, Gerd Graßhoff, Jürgen Renn, Dagmar Schäfer, Robert Schlögl, Bernard F. Schutz

Edition Open Access Development Team
Lindy Divarci, Sylvia Szenti, Klaus Thoden

Editor-in-chief
Matteo Valleriani
Max Planck Institute for History of Science, Berlin

Other Editors
Stephen P. Weldon
Department of History of Science, University of Oklahoma
Esther Chen
Library of the Max Planck Institute for the History of Science, Berlin
Kerry V. Magruder
History of Science Collections, University of Oklahoma Libraries
Anne-Laurence Caudano
History Faculty, The University of Winnipeg
Massimiliano Badino
Facultat de Ciéncies, Universitat Autònoma de Barcelona
Robert G. Morrison
Department of Religion, Bowdoin College

e-ISBN [EPUB] 978-3-945561-64-5

DOI 10.34663/9783945561638-00

First published 2021 by Max-Planck-Gesellschaft zur Förderung der Wissenschaften
Max Planck Institute for the History of Science
https://edition-open-sources.org
Published under Creative Commons Attribution-ShareAlike 4.0 International Licence
https://creativecommons.org/licenses/by-sa/4.0/
The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.d-nb.de.

Edition Open Sources
Sources 14

Preface

Al-Hawārī was not an important mathematician. His only known work, the Essential Commentary on a short book of Ibn al-Bannāʾ, contains no groundbreaking developments nor any original insights. His textbook is simply an introduction to practical arithmetic which enjoyed a modest circulation in the Western part of the Islamic world in late medieval times. And yet this unremarkable book has attracted our interest, and it should attract your interest, too.

The Essential Commentary is worth a modern edition, translation and commentary first of all for what it tells us about the state of Arabic arithmetic in the early fourteenth century. From an historical angle, it is remarkable how techniques and ideas from Greece, the Middle East, and India came to be selected and combined to form such an eclectic whole. Ibn al-Bannāʾ and al-Hawārī were not the first to present shortcuts for mental calculation in a book ostensibly devoted to rules for operating with Indian (i.e., “Arabic”) numerals, and by their time the integration of portions of Greek number theory, from definitions to rules of calculation, was already an established part of arithmetical instruction. The very real tensions created by combining these disparate techniques and ideas could be disregarded by those interested in their utility, and could serve as a starting-points for philosophical discussion by people more interested in theory.

From the conceptual angle, we get to see how the numbers of Arabic arithmeticians differ radically from both Greek and modern numbers. Mathematicians today define real numbers as the elements of a set which, together with binary operations, satisfy the ordered field axioms. Historians of Greek mathematics, for their part, can point to Euclid’s definition of number as a multitude of indivisible units. Arabic numbers were neither of these. For medieval arithmeticians numbers derive from the acts of counting, measuring, and weighing, and as a consequence they worked with any positive quantity that arises in calculation, including fractions and irrational roots. These numbers were validated through practice, so the people who worked with them had no need to ground them in definitions or axioms. Nevertheless, several Arabic authors, Ibn al-Bannāʾ included, attempted to give practical arithmetic some kind of foundation modeled on or based in Greek mathematics, usually in Euclid’s Elements.

It is not just the concept of number that differentiates medieval from modern arithmetic. If we pay attention to how Ibn al-Bannāʾ, al-Hawārī, and other authors phrased their explanations and examples, we discover that they expressed them differently, too. Like the numbers in Greek logistic, numbers in Arabic arithmetic were regarded as being numbers of something. Rather than belonging all to the same abstract set, they come in different kinds: a three might be three units (3), three fifths ( 3\over 5), three roots of five (a trio of \sqrt{5}’s), three cubes ( 3x^3, again a trio), or, in problem-solving, three men or three dirhams. Further, their way of expressing binomials and apotomes, whether of fractions, roots, or algebraic terms, is without parallel in modern arithmetic and algebra. These and other seemingly innocent curiosities of language had a significant impact on how calculations were performed.

Becoming familiar with the ways numbers were conceived and expressed is not only interesting in itself, but it provides the necessary background for the study of the works of more brilliant Arabic writers on algebra and arithmetic such as al-Karajī, al-Khayyām, and al-Fārisī. These and other scholars took their inspiration and developed their concepts directly from the practical tradition. And because medieval Europeans learned much of their arithmetic (including algebra) from Arabic sources, it is only natural that the concepts and modes of expression exhibited in al-Hawārī’s book should be found in their works, too. In fact, they are abundantly evident in Fibonacci, Italian abacus authors, Jean de Murs, Luca Pacioli, and others, and they persisted in sixteenth-century Europe even as significant conceptual developments were taking place. To pick just two examples: Michael Stifel (1550) conforms to the practice of Arabic algebraists by insisting on rational coefficients in algebra, like multiplying \sqrt{6} by x to get \sqrt{6x^2} instead of \sqrt{6}x; and François Viète (1590s), like his Arabic predecessors, always writes single roots, such as \sqrt{1200} and \sqrt{45\over 16}, instead of 20\sqrt{3} and 3\sqrt{5}\over 4.[1] These premodern ways of thinking about and expressing numbers and polynomials only truly began to give way with Descartes, whose 1637 La Geometrie gave a numerical reinterpretation to Viète’s new geometric algebra.[2]

Finally, it is worthwhile to actually work through the numerical operations and to solve problems by the methods described in this book. For you adventurous types willing to perform a division of fractions or to work through a problem by double false position, you will leave paper and pencil behind and take up the dust board (a chalkboard will suffice). And by reading aloud the instructions provided by our authors, you can gain some insight into the prominent role played by recitation and memorization in student-teacher interactions. Supplementing notational board calculations with rhetorical presentation not only helps one remember the rules, but it also causes one to think about the meanings of the signs, which, on reflection, ultimately accounts for their curious (to us) ways of expressing arithmetical and algebraic quantities.

We have written this book with different readers in mind, mainly historians of mathematics and related fields, mathematicians, and mathematics educators. Please keep in mind that, for each group, we explain different aspects of al-Hawārī’s book that people in other groups might find too elementary.

This book is divided into five parts. The introduction covers the basic historical, textual, social, and conceptual aspects of Arabic arithmetic that form the backdrop to al-Hawārī’s book. This is followed by our translation of the Essential Commentary. Because books written in another language many centuries ago cannot speak entirely for themselves, we have written a passage-by-passage commentary to al-Hawārī’s commentary which turned out to be longer than the translation itself. Appendices follow, which include a conspectus of problems, sample worked-out problems solved by various methods from other Arabic books, a brief chronological list of ancient and medieval mathematicians and other scholars, and a glossary of Arabic terms. The English portion finishes with the bibliography and an index of names. Last (or first, if you are an Arabic reader) is the corrected critical edition of the Arabic text together with its own introduction and conspectus, which we originally published in Tunisia in 2013.

Each passage of the translation and our commentary is linked to the Arabic edition by a reference to page and line number. For example, “201.14” in the margin of the translation, or in a box in our commentary, refers to page 201, line 14 of the Arabic edition. Cross references are given throughout. For readers who will not consult the original Arabic, these references still serve to link our commentary with the translation. Hyperlinks connect the different parts of the introduction, translation, commentary, and appendices of the online version of this book. We were unable to make hyperlinks with the Arabic part because the English part was typeset in XeLaTeX while the Arabic part was typeset in MSWord.

Most of what we write in the commentary is directed specifically at the passage in question. But many times we offer remarks to clarify the mathematics from a more general point of view, whether mathematical, conceptual, linguistic, or historical. In our comments at 68.18, for example, we pause to give a breakdown of the vocabulary used in the writing of numbers; at 74.14 we devote a page to how zero was dealt with in calculations; and at 219.1 and 219.4, we devote more than five pages to an explanation of how Arabic ways of expressing operations and their results differ from ours, and so on.

Two bio-bibliographic works are indispensable for sorting out details of the works of medieval Arabic mathematicians. B. A. Rosenfeld’s and E. Ihsanoğlu’s 2003 Mathematicians, Astronomers, and Other Scholars of Islamic Civilisation and Their Works (7th-19th c.) lists the names and works of 1,711 scholars known to have written books in the mathematical sciences, and Driss Lamrabet’s 2014 Introduction à l’Histoire des Mathématiques Maghrébines does much the same for 983 scholars who worked in the western part of the Islamic world. References after names are to these two works. For example, for al-Ḥaṣṣār we write (#532, M55), which means that he is scholar #532 in Rosenfeld and Ihsanoğlu and scholar M55 in Lamrabet. The “M” stands for “Maghreb”. For others, like Ibn al-Samḥ, the “A” stands for “al-Andalus”. Numbers are given for each scholar listed in Appendix C. For authors not listed in Appendix C, numbers are given on mentioning their names.

We follow the International Journal of Middle East Studies system for transliteration of Arabic words.[3]

It was Mahdi’s idea to write this book. With some input from me, he collected the manuscripts and produced the critical edition. We each then translated the book independently, and I compared the two translations to produce the version you see here. I then wrote the commentary, the introduction, and the appendices, with valuable input from Mahdi.

Finally, we thank Len Berggren, Margaret Gaida, and Robert Morrison for their comments and suggestions on an earlier version of this book.

Jeff Oaks,

Indianapolis, May 4, 2019.

Footnotes

[1] (Stifel 1544, fol. 283a); (Viète 1646, 140, 309). In Stifel’s notation, he multiplies \sqrt{}Too late6 by 1Too late to get \sqrt{}Too late6Too late. See (Oaks 2017).

[2] (Oaks 2018b).

[3] https://www.cambridge.org/core/services/aop-file-manager/file/57d83390f6ea5a022234b400/TransChart.pdf

1. Introduction

1.1 Ibn al-Bannāʾ and al-Hawārī

1.1.1 Ibn al-Bannāʾ’s textbook and al-Hawārī’s commentary

Shortly before the year 1301 CE, in the western part of North Africa, the celebrated mathematician and astronomer Ibn al-Bannāʾ (1256-1321) wrote what was to become a very popular arithmetic textbook. His Condensed [Book] on the Operations of Arithmetic (Talkhīṣ aʿmāl al-ḥisāb) describes rules for operating with Arabic numerals and methods of arithmetical problem-solving. One factor that made the book appealing was that it is brief. Condensed expositions were popular because students then were often expected to memorize textbooks, and this one covered just the right amount of material.

But brevity is also a drawback. Ibn al-Bannāʾ’s little book contained no numerical examples to illustrate how the calculations were to be performed, and his concise exposition left no room for proofs or supplementary remarks. So, over the course of the next three centuries, the Condensed Book inspired a number of people to write commentaries elaborating on its contents in various ways. Ibn al-Bannāʾ himself became the first to write such a commentary in 1301 CE (701H). In Lifting the Veil from the Face of the Operations of Arithmetic (Raf ʿ al-ḥijāb ʿan wujūh aʿmāl al-ḥisāb) he expounds on his short work by providing philosophical remarks, proofs, numerical examples, and expanded explanations.

Al-Hawārī was among Ibn al-Bannāʾ’s students in Marrakesh early in 1305 (704H). Ibn al-Bannāʾ was teaching his Condensed Book, and already during the course of the lectures al-Hawārī was writing his own commentary. As he tells us in his introduction, al-Hawārī’s specific purpose was to provide numerical examples for Ibn al-Bannāʾ’s rules. The present book contains our edition, translation, and commentary of his Essential Commentary on the Condensed [Book] on the Operations of Arithmetic (al-Lubāb fī sharḥ Talkhīṣ aʿmāl al-ḥisāb), which he completed June 12, 1305. In it, al-Hawārī reproduces passage by passage the entire text of the Condensed Book. After each of Ibn al-Bannāʾ’s rules he gives his own numerical examples, and, where he finds it necessary, he supplements Ibn al-Bannāʾ’s explanations with further remarks often extracted from his teacher’s Lifting the Veil.

1.1.2 The authors and their works

Ibn al-Bannāʾ’s full name is Abū al-ʿAbbās Aḥmad ibn Muḥammad ibn ʿUthmān al-Azdī al-Marrākushī. For those not familiar with the forms of Arabic names, we will dissect this one. His given name is Aḥmad, and he gained the sobriquet Abū al-ʿAbbās when his wife gave birth to a son they named al-ʿAbbās (Abū means “father”). The word ibn means “son”, so he was the son of Muḥammad, who in turn was the son of ʿUthmān al-Azdī. The designation al-Marrākushī indicates that he was from Marrakesh, a city now in modern Morocco. He is best known today by the nickname Ibn al-Bannāʾ, which means “son of the builder”.

Ibn al-Bannāʾ was a renowned scholar known to have written nearly 120 works on topics ranging from the rational sciences (mathematics, astronomy, logic) to religion (including law), language (rhetoric and prosody), the occult sciences (including astrology), agronomy, philosophy, and medicine.[1] He was born in Marrakesh and remained in Morocco his whole life.

Four of his books are relevant to us, and the Arabic text of each has been published:

Condensed [Book] on the Operations of Arithmetic (Talkhīṣ aʿmāl al-ḥisāb, [M1]).[2]

This is Ibn al-Bannāʾ’s brief introduction to calculation with Indian (i.e. Arabic) numerals, including methods for finding unknown numbers. Because it does not include examples, the book does not show the numerals. It was completed before 1301. Souissi’s 1964 edition also contains a French translation.

Lifting the Veil from the Face of the Operations of Arithmetic (Raf ʿ al-ḥijāb ʿan wujūh aʿmāl al-ḥisāb, [M8]).[3]

Ibn al-Bannāʾ’s own commentary to his Condensed Book was completed in 1301. Although it contains many numerical examples that would be of help to students, the theoretical and philosophical nature of many of his remarks suggest it was written for a more sophisticated audience. Al-Hawārī copied many passages from this commentary into his own book, mainly those giving further explanations of the rules.

Essays on Arithmetic (Maqālāt fī l-ḥisāb, [M2]).[4]

This is another arithmetic book that covers much of the same material as the Condensed Book. In it, Ibn al-Bannāʾ sparingly shows the Indian numerals. He does not cover algebra or double false position, but he does include a collection of problems solved by single false position and proportion. Problems [1] and [2] in Appendix B are translated from this book.

Book on the Fundamentals and Preliminaries in Algebra (Kitāb al-uṣūl wa l-muqaddimāt fī l-jabr wa l-muqābala, [M6]).[5]

Ibn al-Bannāʾ’s book on algebra, dating from the late thirteenth century, builds on Abū Kāmil’s late ninth century book on the same topic. This book explains problem-solving by algebra at a level appropriate for students. Problem [6] in Appendix B is translated from this book.

Al-Hawārī’s full name is ʿAbd al-ʿAzīz ibn ʿAlī ibn Dāwud al-Hawārī al-Miṣrātī. The designation “al-Miṣrātī” indicates that he was descended from the Libyan tribe of Miṣrāta. The “al-Hawārī” tells us that he hailed from the Berber tribe named Hawārī, which we know emigrated from Libya to Morocco in the ninth century.[6]

The only secure dates we have on al-Hawārī are found in his book. The earliest extant manuscript, that of Medina (described below), reports that he completed it on Saturday, 18 Dhū al-Qaʿda, 704H, which is the Julian date 12 June 1305. Also, at line 204.3 he relates that Ibn al-Bannāʾ was dictating to him on Wednesday, the 28th of the month of Rajab. This can only be in 704H, which corresponds to 24 February 1305. Because we know that Ibn al-Bannāʾ was in Marrakesh at that time, it places al-Hawārī there as well. We know nothing else about his dates, his locations, or his career. His commentary on the Condensed Book is his only known work: Essential Commentary on the Condensed [Book] on the Operations of Arithmetic (al-Lubāb fī sharḥ Talkhīṣ aʿmāl al-ḥisāb, [M1]).[7]

Al-Hawārī delivered just what he said he would in his introduction: a book supplementing each of Ibn al-Bannāʾ’s rules with numerical examples. There is nothing innovative or theoretical about it. Like the Condensed Book, al-Hawārī’s commentary makes no advances in either ideas or techniques. He gives no proofs, and he generally avoids philosophical discussions. And yet, the very lack of innovation in this book makes it a good source for learning about the nature of arithmetic in the epoch of its author, especially when it is compared with other Arabic arithmetic books. By reading how these authors worked with and expressed numbers and equations, we gain insights into the concepts underlying their arithmetic. Furthermore, the apparent lack of interest of both Ibn al-Bannāʾ and al-Hawārī in prior mathematical traditions, when coupled again with an examination of other books, allows us to identify influences from Greece, the Middle East, and India on the practical arithmetical tradition in Arabic.

1.1.3 Manuscripts of al-Hawārī’s Essential Commentary

The fourteen known surviving manuscripts of al-Hawārī’s book were copied between the fourteenth and nineteenth centuries CE. We published our Arabic edition based on the following five manuscripts:

Medina, MS Ḥikmat 21 ḥisāb. 63 ff, 16 lines per page, 16 × 21 cm. The copyist completed it on 18 Rabīʿ I, 746H (Julian 18 July 1345). This manuscript does not distinguish between Ibn al-Bannāʾ’s and al-Hawārī’s words.

Oxford, MS Marsh 378/3, fols. 109a-162a. Copied in 1444 according to Woepcke.[8] In this manuscript a word or two is written in red ink to denote the beginning of a new idea, or what we might consider as a new paragraph. This often corresponds to a shift in author (Ibn al-Bannāʾ to al-Hawārī, or vice-versa), but many times it does not.

Istanbul, Süleymaniye Library, MS Şehit Ali Paşa 1977/2 (Türkiye Yazma Eserler Kurumu Başkanlığı Süleymaniye Yazma Eser Kütüphanesi, Şehit Ali Paşa Collection 01977), fols. 54a-103b. Copied 20 Ramaḍān 880H/Julian 16 January 1476 in Constantinople. The copyist distinguished Ibn al-Bannāʾ’s text from al-Hawārī’s comments by placing in front of Ibn al-Bannāʾ’s extracts the letter “Too late” (ṣād), which stands for muṣannif (“Author”), while passages from al-Hawārī are preceded by the letter “Too late” (shīn), which is the first letter of the word sharḥ (“commentary”).

Tehran, Library of Parliament MS 2672/2, fols. 10a-56b, copied before 972H/1564. Red ink is employed here like in the Oxford manuscript.

Tunis, National Library of Tunis MS 9940. 32 ff., 22 × 26 cm, 29 lines per page. Copied 4 Jumādā II, 1082H/Julian 28 September 1671 in Damascus. In this manuscript Ibn al-Bannāʾ’s passages are preceded by the letter “Too late” (mīm), which either stands for matn or muʾallif, both of which mean “[original] text”, and al-Hawārī’s comments start with a “Too late” (shīn) for sharḥ (“commentary”).

For those not familiar with the notation, the Tehran manuscript is number 2672 in the collection, and the “/2” indicates that al-Hawārī’s treatise is the second treatise contained in that manuscript. In this manuscript, al-Hawārī’s treatise begins on the front (a) of folio (sheet) 10 and ends on the back (b) of folio 56. The Tunis manuscript 9940 contains only al-Hawārī’s treatise, and it is 32 folios long.

As is common with manuscripts, there are variations from one copy to another. In the case of al-Hawārī’s commentary, these differences are all minor. Some are clearly errors, while for others it is not easy to tell which variation the author originally wrote. When in doubt we have sided with the oldest manuscript, that of Medina, copied only four decades after the book was finished. There are, however, greater differences between the figures drawn in the manuscripts. We discuss them below.

On the following pages are samples from each of the five manuscripts. We have chosen the page that contains the double false position diagram just before 208.8. (Double false position is described below in chapter 1.5.)

Fig. 1.1: Medina Manuscript fol. 55b, covering the text from 207.16 to 209.2.

Fig. 1.2: Oxford manuscript fol. 154b, covering the text from 208.6 to 211.15. The scan is in black and white. Lighter shaded portions, which can be seen in the text in the middle of the page, in the lines of the diagram, and as bars over some words, are in red ink.

Fig. 1.3: Istanbul manuscript fol. 97b, covering the text from 208.2 to 211.15. This manuscript does not show the diagram.

Fig. 1.4: Tehran manuscript fol. 50b, covering the text from 207.1 to 208.9.

Fig. 1.5: Tunis manuscript fol 28a, covering the text from 208.2 to 213.5. This scan is in black and white, so it is difficult to tell if there is anything written in red ink.

1.2 Historical setting

Even a quick look over the translation reveals that al-Hawārī’s arithmetic textbook differs from its modern counterparts in format, contents, and methods. These differences run deeper than what might be seen as accidental variations. Medieval Islam was an intellectual melting-pot of ideas from the various cultures it was in contact with, and al-Hawārī’s book exhibits these different influences, each with qualities of its own, and which were not always amenable to merging. Further, medieval people conceived of numbers differently than we do today, and their system of education and their very attitude toward books affected the way they presented their calculations. These historical, mathematical, and conceptual issues warrant a few words to put the Essential Commentary in context, and we begin in this section with history.

When the prophet Muḥammad died in 632 CE, his empire already covered most of the Arabian peninsula. The military push by his successors resulted in rapid conquests both east and west, so that by the time the Umayyad dynasty fell to the ʿAbbāsids in 750 CE the empire stretched from the Iberian peninsula in Western Europe all the way to the Indus River. The Arab conquerors found themselves ruling over territory once controlled by Greece to the west, Persia in the Middle East, and bordering India to the east.

One remarkable cultural phenomenon that arose from this expansion is known as the Arabic translation movement. From roughly the middle of the eighth century until the end of the tenth, the ruling class and other wealthy patrons subsidized the large-scale collection, translation, and study of scientific and other knowledge. Books were sought out from any available source, whether in Greek, Sanskrit, Persian, or Syriac. Topics included the mathematical sciences (arithmetic, geometry, optics, mathematical astronomy, etc.) as well as philosophy, astrology, geography, mechanics, medicine, agriculture, alchemy, and assorted other topics. Concurrent with this translation activity, several authors wrote down in books the arithmetic (including algebra), mensuration, and folk astronomy that had previously been transmitted orally among people working in the trades in the Middle East.

The causes of the Arabic translation movement are complex and are still being debated. We will only note that it appears to have been spurred by the demands of managing such a large empire combined with the demands of political expediency involving Persian influence.[9] The causes of its demise around the latter tenth century are necessarily equally complex, but one factor seems to have been that they had just about run out of relevant books to translate, and most of those that were left had been surpassed by original works in Arabic.

In conjunction with the translation movement, scholars composed original works in various branches of mathematics. The circulation in this singular environment of books originating from different cultures naturally gave rise to the mixing of ideas and techniques. To pick just a few examples: in astronomy, Indian trigonometry was adopted in books based on Greek planetary theory; in geometry, definitions and constructions from Greek texts were introduced to books on practical mensuration; and we find applications of algebra to Greek geometry and of Greek geometry to algebra. This combining of elements from different traditions was a lasting characteristic of Arabic mathematics and science, despite their occasional incompatibility.

The prevalence of the word “Greek” in this last paragraph is not accidental. The majority of works translated and studied by Arabic scholars were originally written in that language. And, just as important, Arabic authors consciously structured their mathematical works, at least the more theoretical ones, on Greek models. Arabic scholars were drawn in particular to the logical-deductive structures that guided Euclid’s Elements in geometry and Ptolemy’s Almagest in astronomy. The philosophical underpinnings of Greek-inspired mathematics in Arabic were also Greek in origin, and were drawn mainly from the works of Aristotle and some late antique neoplatonists. While mathematicians in other cultures had also engaged in stating definitions, outlining principles, and writing proofs, it was the unified construction of a science based on these elements, together with underlying philosophical principles, that for Arabic scholars set Greek mathematics apart.[10]

After the translation movement had run its course, the production of new books continued unabated. Innovative work continued in algebra, geometry, optics, mechanics, and various aspects of astronomy that included planetary models, trigonometry, and numerical interpolation. In some branches of learning this momentum persisted longer than in others. After the thirteenth century, for example, we see little advancement in algebra from a theoretical perspective (including the algebra in al-Hawārī’s book), while new and innovative books on planetary theory continued to appear at least to the middle of the sixteenth century.[11]

Even if Arabic scholars distinguished between practical (ʿamalī) and theoretical (naẓarī) mathematics,[12] it was common for the two to appear together in the same book. A good example in which the two approaches mingle is the Completion of Arithmetic of al-Baghdādī (died 1037).[13] This book gives rules for practical calculation while at the same time incorporating and building on various aspects of theoretical Greek number theory. Practical topics could also serve theoretical needs: in the latter eleventh century, al-Khayyām adapted the algebra of merchants and surveyors to aid in theoretical geometric problem-solving by providing solutions to irreducible cubic equations via conic sections.[14] Often, practical techniques attracted the attention of mathematicians who crafted Greek-style proofs for them. One example from the late ninth century is Qusṭā ibn Lūqā’s short work proving the validity of the method of double false position in the manner of Euclid’s Data.[15]

Ibn al-Bannāʾ states his practical intent when he writes that his Condensed Book “is useful for inheritance and business transactions and other [purposes]”.[16] The commentaries it spawned, however, frequently take it in philosophical and theoretical directions. For example, Ibn al-Bannāʾ ensured that the nature of the unit would become a source for philosophical debate by the elaboration he gave the issue in his Lifting the Veil.[17] And many commentators, Ibn al-Bannāʾ included, provided proofs to many of the rules in the Condensed Book. Al-Hawārī’s commentary, on the other hand, retains the practical orientation by focusing almost exclusively on numerical examples.

Like many Arabic mathematics books, Ibn al-Bannāʾ’s Condensed Book is a kind of hybrid, exhibiting techniques originating from Indian, Greek, and Middle Eastern sources. Part I of the book is devoted to calculation with Indian numerals (we call them Arabic numerals), which originated in India. Certain chapters contain material extracted ultimately from Greek sources, like Euclid’s treatment of quadratic irrationals recast in numerical terms (beginning at 173.4), and portions of theoretical number theory borrowed from Nicomachus of Gerasa (starting at 65.10 and 127.10). Last, the methods of mental multiplication (beginning at 108.9) and the techniques of problem solving by proportion, double false position, and algebra comprising Part II derive ultimately from local oral tradition. These different elements will now be described.

1.3 Arabic arithmetic

The two common notations for writing numbers in medieval Arabic were jummal and Indian. Jummal notation is sometimes called abjad, after the first four letters in the traditional ordering of the Arabic alphabet: alif, bā, jīm, and dāl. In this system, the 28 letters of the alphabet are assigned values 1, 2, 3,…, 9, 10, 20, 30,…, 90, 100, 200, 300,…, 900, and 1000.[18] It is an additive system, so to express the number 542, for example, one writes the letters whose values are 500 (Too late), 40 (Too late), and 2 (Too late): Too late. Jummal notation makes its appearance in al-Hawārī’s book only in the rule for casting out sevens, in the figures just before 89.10 and 90.1 (see our commentary at 88.10). Jummal numerals were associated with two forms of calculation, finger reckoning and sexagesimal arithmetic, which will be described below.

We acquired our “Arabic” numerals from Arabic sources, but Arab-speaking people called them “Indian” (al-hindī) numerals because they learned them from Indian sources. Indian numerals consist of the signs for 1, 2, 3, …, 9, together with a 0 for the empty place. This is the system taught in Ibn al-Bannāʾ’s textbook and illustrated by al-Hawārī’s examples. The shapes of the numerals have varied greatly over time and place, even among the manuscripts of al-Hawārī’s book. See our commentary at 69.2 for some of the forms.

There was another way of writing numbers that was practiced mainly in the far western part of the Islamic world. The earliest extant mention of rūmī (“Roman”) signs is a brief description in the Chapters on Indian Arithmetic by al-Uqlīdisī, completed in Damascus in 952-53 CE (341H).[19] Other texts mentioning these numerals date from the latter twelfth century or later, and were written in al-Andalus and Morocco.[20] Ibn al-Bannāʾ wrote a book on calculation with these numerals, which were popular with public administrators. They were also called “Fez signs”, after the Moroccan city. Like abjad numerals, this is a decimal and additive system, with 27 individual signs for 1, 2, 3,…, 9, 10, 20, 30,…, 90, 100, 200, 300,…, 900. The peculiar shapes of the numerals cannot be securely said to derive from any other known system. No examples are known in al-Andalus after the thirteenth century, but the system remained popular with accountants in Morocco until about a century ago. Ibn al-Bannāʾ and al-Hawārī mention rūmī signs in passing at 85.16 and 104.1.

The three main methods of calculation in medieval Islam were finger reckoning, sexagesimal arithmetic, and Indian arithmetic. Sexagesimal arithmetic and finger reckoning had already been in use in the Middle East prior to the rise of Islam. And although Indian numerals are attested in Syria by the mid-seventh century CE, it seems that rules for calculating with them became known to Muslims about a century later through direct contact with Indian scholars, at the start of the translation movement. The three systems remained popular throughout the medieval period, and were often described together in the same book.

Finger reckoning was a method popular among merchants, government secretaries, and surveyors. Calculations were preformed mentally in base ten, with intermediate results “stored” by positioning the fingers in particular ways. It was called ḥisāb al-yadd (hand arithmetic), al-ḥisāb al-hawāʾī (aerial calculation), al-ḥisāb al-maftūḥ (open calculation), or sometimes ḥisāb al-rūm wa l-ʿarab (calculation of the Romans (i.e., Byzantines) and Arabs). In one of many examples, Abū l-Wafāʾ (tenth century) explains how to multiply 46 by 28:

We begin by multiplying forty by twenty, to get eight hundred, and by eight to get three hundred twenty. Hold them. Then we multiply the six by twenty to get one hundred twenty, and by eight to get forty-eight . Then we add them all to get one thousand, two hundred eighty-eight.[21]

The command “hold them” means “store the intermediate sum” 1,120 by positioning the fingers. Then, after the next sum 168 is calculated, the two can be added to get the answer. When the result of such a calculation was recorded on paper, jummal notation was used. Finger reckoning and jummal numerals played roles similar to the abacus (counters manipulated on tabletops) and Roman numerals in the West: the first was for calculating and the second for recording the numbers.[22]

Most Arabic arithmetic books covering mental calculation do not address the actual positioning of the fingers. One that does is ʿAlī ibn al-Maghribī’s fifteenth-century Poem on Reckoning with Finger-Joints.[23] Units were stored by positioning the last three fingers on the right hand, tens with the right-hand thumb and index finger, hundreds with the left hand thumb and index finger, and thousands with the last three fingers on the left hand. The earliest illustration we know for the positioning of the fingers is found in an Italian textbook, the 1494 Summa de Arithmetica of Luca Pacioli, shown below.[24] It differs from ʿAlī ibn al-Maghribī’s system mainly by switching the hundreds and thousands, and by switching the left and right hands.

Fig. 1.6: Finger positions in Luca Pacioli’s 1494 Summa de Arithmetica.

We know from literary references that finger reckoning had been practiced by the ancient Greeks and Romans, and it was probably known in other parts of the ancient world as well.[25] The passage from Abū l-Wafāʾ translated above is from his Book of What is Necessary for Scribes, Businessmen, and Others in the Science of Arithmetic.[26] This book, composed in the period 961-976 CE, is the earliest extant Arabic book explaining the techniques of mental calculation. This and later books on the topic cover mainly multiplication, division, and ratios, including some computational shortcuts. They also typically have chapters on mercantile and other practical problems in which the rules are applied. The more common methods for solving these problems are the rule of three, single false position, double false position, and algebra. Abū l-Wafāʾ’s book, like many others on finger reckoning, also gives rules for base sixty calculations. Ibn al-Bannāʾ covers several rules for mental multiplication, most copied from Ibn al-Yāsamīn’s late twelfth century Grafting of Opinions, beginning at 108.9; and his rules on division from 119.18 through 124.19 are also from finger reckoning.

Sexagesimal arithmetic is calculation in base sixty. This place-value system is like our base ten system, with sixty taking the place of ten. Where, for example, our 9062 is 2 + 6\cdot 10+0\cdot 10^2+9\cdot 10^3, the sexagesimal number 12,0,45,1 is 1+45\cdot 60+0\cdot 60^2+12\cdot 60^3 or, in base ten, 2,594,701. Sexagesimal arithmetic dates back to late third millennium BC Mesopotamia, where the numbers were recorded in cuneiform on clay tablets. It became the standard system for Babylonian astronomers, and later it was adopted by astronomers writing in Greek, Arabic, Sanskrit, and other languages. It is because of sexagesimal arithmetic that today we still have sixty seconds in a minute and sixty minutes in an hour. In Arabic the numbers 1 through 59 were written in jummal form so, reversing our 12,0,45,1, the number would have looked like Too lateToo lateōToo late. The “ō” is the sexagesimal zero, or empty place. Sexagesimal arithmetic is not covered in al-Hawārī’s book.

Indian arithmetic is calculation in our base-ten place-value system using the nine figures 1, 2, …, 9 and the zero (0). This system originated in India probably around the first century CE, though the earliest Indian text showing the numerals dates from the late sixth century. The numerals had reached Syria by 662 CE when the bishop Severus Sebokht mentioned them, and they were known to the Muslims by 760 CE.

The earliest known Arabic treatise on Indian numerals is the Book on Indian Calculation,[27] written by Muḥammad ibn Mūsā al-Khwārazmī in Baghdad in the first half of the ninth century. Both the Arabic original of this book and a Latin translation made in the twelfth century are lost, but we do have a reworking of the Latin translation. It appears that al-Khwārazmī’s original treatise described how numbers are formed by the nine figures, and then covered in order the operations of addition, subtraction, doubling, halving, multiplication, division, all for whole numbers; then operations on fractions were discussed, and finally square root extraction. This book is already a hybrid, since the fractions are taken from finger reckoning and sexagesimal arithmetic.

Al-Khwārazmī’s rules are intended to be worked out on a dust board or some other ephemeral surface that allows for the easy erasing and shifting of digits. A dust board is a flat board covered in dust or fine sand on which one wrote with a finger or a stylus. It was a medium for working through calculations much like today’s chalkboard or whiteboard. It is because of this association with the dust board that “Indian arithmetic” (al-ḥisāb al-hindī) was also called “board arithmetic” (ḥisāb al-takht) or “dust arithmetic” (ḥisāb al-ghubār). Another erasable writing surface was the wax tablet, with wax instead of sand covering the board.[28]

The earliest extant Arabic book on Indian reckoning is al-Uqlīdisī’s mid-tenth century Chapters on Indian Arithmetic, mentioned above.[29] This book covers much the same material as al-Khwārazmī’s, but with some innovations. It is the earliest we know that explains decimal fractions, like 4.75 instead of 4{3\over 4}. Al-Uqlīdisī also gave rules for working out the computations with pen and paper, without any erasing. He famously explains that one advantage of switching to ink and paper is because the dust board is associated with “the misbehaved who earn their living by astrology on the streets”.[30]

Al-Uqlīdisī’s innovations did not catch on quickly. The dust board remained popular for centuries, only disappearing in some parts of the Muslim world at the beginning of the 1900s. Most of Ibn al-Bannāʾ’s rules, in fact, are intended for the dust board or wax tablet, since they call for the erasing of digits. He does give three rules for multiplication that require no erasing, at 101.16, 103.14, and 106.11. These rules will work for pen and paper, but they may have been intended for the lawḥa (board; tablet). This was a board covered in soft clay on which one wrote with a cane stick dipped in ink. Because the lawḥa is reusable it would have been more economical than paper. Decimal fractions did not catch on quickly, either. Ibn al-Bannāʾ does not describe them at all, but instead works with the common forms of fractions from finger-reckoning.

1.4 Greek mathematics and Arabic arithmetic

1.4.1 Greek and Arabic concepts of number

Some Arabic arithmeticians gave their subject the semblance of a rigorous foundation by adding definitions from Greek number theory. This might have been a good way to merge different approaches to arithmetic were it not for the incongruity of their number concepts. The numbers of the Arabic arithmeticians (ḥussāb) include any positive quantity that arises in calculation, including fractions and irrational roots. But numbers in Euclid’s Elements Books VII-IX and Nicomachus’s Arithmetical Introduction, the two main Greek sources, are restricted to positive integers.[31]

Aristotle’s works provided the main philosophical backdrop to this particular Greek concept of number, at least as it circulated in medieval Islam. In his Categories, Aristotle divided the genus of “quantity” into discrete and continuous: “Of quantities some are discrete, others continuous…Discrete are number and language; continuous are lines, surfaces, bodies, and also, besides these, time and place”.[32] Geometric magnitudes are continuous because they are infinitely divisible. Any line, surface, or body can be divided into as many parts as one wishes. By contrast, numbers are discrete because the unit (1) is necessarily indivisible. How can it be a unified whole if it can be split into parts? Euclid was repeating the accepted definition when he wrote “[a] number is a multitude composed of units”.[33] There are two important consequences of these characterizations of the unit and number. First, “one” cannot be regarded as a number because it is not a multitude. One is instead the origin or cause of number, and the sequence of numbers begins with two. Second, there can be no fractions due to the indivisibility of the unit, so the only true numbers are integers.[34]

Although the relevant Greek works had been translated into Arabic by the end of the first quarter of the ninth century, the Greek notions of unit and number took some time to make their way into practical Arabic arithmetic books. The earliest books we consulted give no definition of number, and instead begin with instructions on how to perform operations.[35] Philosophical treatments of arithmetic could not ignore the issue, however. Two early books whose chapters on arithmetic depend heavily on Nicomachus are the Treatises of the Brethren of Purity (10th c.) and the Book of Instruction of the Elements of the Art of Astrology (11th c.) by al-Bīrūnī. These two books get around the problem of the incompatibility of Greek and Arabic numbers by asserting that although the unit is properly speaking indivisible, in practice we work with a divisible quantity that we call a “unit”.[36] Some later authors cited both views without trying to reconcile them, like the Persian al-Fārisī (died 1319) in his Foundation of Rules on Elements of Benefits. Al-Fārisī first reviews Aristotle’s classification of continuous and discrete quantity before he presents the Greek definitions of the unit and of number. He then cites the definition of the practical arithmeticians (ḥussāb) of his time: “Number is a quantity you obtain from one by repetition or partition or both, and it is clear by this meaning that the type is divided into whole numbers and their fractions”.[37]

Evidence of Greek influence on practical arithmetic in Western Islam appears in the earliest extant texts from the late 1100s. Al-Ḥaṣṣār, who probably worked in Salé in Morocco, added to his descriptions of the digits that one “is the origin of number” and that two “is the first number”, as did his contemporary Ibn al-Yāsamīn.[38] About a century later, Ibn al-Bannāʾ became the first arithmetician we know of to give Euclid’s definition of number in a book on practical calculation. He begins his Essays on Arithmetic with

Number is a multitude composed of units…Its origin emerges from the arithmetical one. And the arithmetical one is by its essence not a number, since it is the cause and number is the effect.[39]

Ibn al-Bannāʾ took a different approach in the Condensed Book by condensing the definitions of Euclid and the arithmeticians in one short statement: “A number is a collection of units, and it is divided according to how it is produced into two kinds: whole and fraction”. (See our commentary at 65.2.) The friction between the Greek and Arabic number concepts also comes into play in Ibn al-Bannāʾ’s definitions of multiplication (95.2), division (117.2), and fractions (133.1).

The incompatibility problem was solved by the Persian mathematician and poet ʿUmar al-Khayyām (Omar Khayyam), though his work does not seem to have been noticed by arithmeticians. Where geometric magnitudes in Euclid and other early Greek geometers do not possess any numerical measure, al-Khayyām worked with arithmetized lines, planar regions, and bodies, and he identified the numbers of the arithmeticians with the dimensionless measures of such continuous magnitudes.[40] This way a number like \sqrt{10} can be the length of a line, the area of a planar figure, or the volume of a body. Consequently there are two different units, the indivisible arithmetical unit and the divisible geometric unit.

1.4.2 Greek number theory and geometry in Arabic arithmetic

Books VII to IX of Euclid’s Elements and Nicomachus’s Arithmetical Introduction describe theoretical arithmetic for positive integers, or what we would call elementary number theory. Topics include the classification of numbers into even, odd, and their sub-species; prime numbers and divisibility; amicable and perfect numbers; ratio and proportion; figurate numbers; and series summation.

Ibn al-Bannāʾ’s Condensed Book covers the classification of even numbers (65.10) and the sieve of Eratosthenes (127.10) for finding prime numbers. Both are taken directly or indirectly from the Arabic translation of Nicomachus, as is the classification of different kinds of proportion that al-Hawārī copied from Lifting the Veil (195.2). Several definitions from Book VII of the Elements found their way into Ibn al-Bannāʾ’s and al-Hawārī’s books, certainly through some intermediary Arabic source. Besides definition VII.2 for “number” at 65.2, these include definition VII.16 for “side” and “surface” at 66.17 and 67.3, VII.3-4 for “part” and “parts” at 133.1, and also possibly VII.10 for “oddly odd” at 66.7 and VII.15 for “multiplication” at 95.2. Like the words we translate as “square” and “cube”, “side” (ḍilʿ) and “surface” (saṭḥ) are arithmetical terms borrowed from geometry.

Rules for summing finite series were known in both India and Greece before the advent of Islam, and probably in other parts of the old world as well.[41] The rules we read in Arabic books ultimately derive from Greek sources, with some innovations introduced by the Arabic authors.[42] Ibn al-Bannāʾ likely took his rules (starting at 79.13) from another Arabic source.

It is largely because numbers could take only integer values that geometric magnitudes in Euclid, Apollonius, and Archimedes are without numerical measure. For the Pythagorean Theorem, for example, Euclid drew literal squares on the three sides of a right triangle and showed that the square on the hypotenuse is the same size as the other two together.[43] While working in the Greek tradition, Arabic mathematicians properly kept numbers out of geometry. One example is Thābit ibn Qurra’s treatise proving that the volume of a section of a paraboloid is half the volume of the enclosing cylinder.[44]

By contrast, Arabic practitioners held a concept of number deriving as much from measurement as from counting, allowing them to work freely with fractions and irrationals. By routinely assigning numerical values to geometric magnitudes in surveying work and in architecture, they were able to foster the intimate connection between arithmetical and geometrical calculation. So when Euclid’s Elements became available in Arabic translation in the late eighth century, mathematicians writing in a practical vein did not hesitate to reinterpret propositions from the geometric books in terms of arithmetic.[45] Thābit ibn Qurra, for example, gave a numerical reading to Elements propositions II.5 and II.6 in his proofs for the rules to solve three-term algebraic equations.[46]

Euclid’s treatment of quadratic irrational lines in Book X was particularly amenable to numerical interpretation. One treatise from the ninth century, probably written by al-Māhānī, gives numerical calculations of the square roots of binomials and apotomes.[47] For example, a line in Euclid that is divided into two incommensurable parts satisfying a certain condition could now be identified with the number 5+\sqrt{45}. Ibn al-Bannāʾ briefly presents the arithmetical version of the theory of quadratic irrationals in his Condensed Book, and al-Hawārī gives it a thorough review with the help of some passages from Lifting the Veil (starting at 173.4). Al-Hawārī also copied from Lifting the Veil the manipulations of geometric proportion that Ibn al-Bannāʾ took, once again directly or indirectly, from definitions 12-15 in Book V of an Arabic translation of Euclid’s Elements (196.16).

1.5 Arithmetical problem solving

Part II of the Condensed Book covers arithmetical problem-solving techniques that were originally associated with finger reckoning. Here, instead of solving problems that ask for the result of a calculation on given numbers, Ibn al-Bannāʾ explains methods for finding unknown numbers. For instance, instead of working out a calculation like “divide thirty-five by fifteen” (at 120.16), al-Hawārī poses the problem “a quantity: taking away its third and its fourth leaves ten. How much is the quantity?” (at 199.1).

Today, a problem asking for an unknown number is typically viewed as an algebra problem. We would solve the problem just mentioned by naming the quantity x, and then setting up and solving the equation x-{1\over 3}x-{1\over 4}x=10. But problems like this were regarded by medieval mathematicians as belonging to arithmetic, since algebra was just one of several methods available to solve them. In fact, al-Hawārī works out this particular problem not by algebra, but by double false position (described below). The independence of problems from the methods of solution is most evident in other books that show the same problem solved by two or three different methods. One example is a problem of al-Ḥaṣṣār translated in Appendix B, problem [7].

Here is a brief list of the problem solving methods covered by Ibn al-Bannāʾ. These are the most common among the methods described in Arabic arithmetic books:

The method Arabic authors called “the four proportional numbers” is known to us also as “the rule of three”. Given a proportion a:b::c:d, this method shows how to find one of the values given the other three. It is explained starting at 195.9. Al-Hawārī does not solve any problems by this method, so we translate one from Ibn al-Bannāʾ’s Essays on Arithmetic in Appendix B, problem [1].

Single false position is a method that directly applies the rule of three. One posits a convenient, but probably incorrect, value for the solution to a problem, then the answer is calculated from the value and the error via proportion. This method is not taught in Ibn al-Bannāʾ’s Condensed Book, but it was very common in Arabic arithmetic. Problems [2] and [7] translated in Appendix B are solved by single false position. The first of these is from Ibn al-Bannāʾ’s Essays on Arithmetic.

Double false position. Two (false) values are posited for the solution to a problem, and the correct answer is found from the values and their respective errors. This method is also based on proportion, and is explained beginning at 198.2 with examples at 199.1, 200.1, 201.1, 202.3, 203.1, 205.5, 207.6, and Appendix B, problem [7].

Algebra. In Arabic algebra the powers of the unknown are assigned particular names corresponding to our x, x^2, x^3, etc. To solve a problem an unknown is assigned one of these names, or sometimes a combination of them. The conditions of the enunciation are worked out to set up an equation, which is then simplified and solved. The basic rules of algebra are explained starting at 209.1, though no sample problems are provided. Examples worked out by algebra from other books are given in problems [3] through [7] in Appendix B.

The rule of three was practiced across the Old World in antiquity, from Europe to China. As mentioned, it is the basis for the method of single false position, which dates back at least to a problem solved in the Rhind Papyrus in Egypt, ca. 1650 BC. A version of single false position is also attested to in cuneiform tablets from the end of the Old Babylonian period (that is, around 1700-1600 BC), and it appears in Sanskrit sources before the advent of Islam.[48] This method was probably as widespread in antiquity as the rule of three. It was practiced and taught among merchants and others over the course of many centuries and over a vast territory before we encounter it in medieval Arabic books.

Double false position also circulated among people who solve problems on the job. The only evidence we have of its use before the ninth century comes from a few problems in a single ancient Chinese text, the Nine Chapters on Mathematical Procedures, written some two thousand years ago.[49] Contrary to some reports, double false position is absent from Hero of Alexandria’s approximation of \root 3 \of {100} (ca. 100 CE), nor has it been found in Babylonian or Sanskrit sources.[50] This does not mean that it was not practiced in those places. We just have no positive evidence that it was.

Because of confusion in many modern accounts, algebra will require a bit more explanation. Algebra in medieval Arabic was called al-jabr wa l-muqābala, literally “restoration and confrontation”.[51] The phrase was often shortened to just al-jabr, and transliterations into Latin and Italian eventually led to our word “algebra”. But algebra then was quite different from algebra now. Many of the various meanings we impart to the word “algebra” today are due to modern developments and to the disappearance of other problem-solving methods. Today, people often apply the words “algebra” and “algebraic” to any kind of formal, abstract reasoning, or to any technique of finding unknown numbers. Two current definitions of “algebra” in the Oxford English dictionary support the former view:

2a. As a mass noun: (originally) the branch of mathematics in which letters are used to represent numbers in formulae and equations; (in later use more widely) that in which symbols are used to represent quantities, relations, operations, and other concepts, and operations may be applied only a finite number of times.

3. In extended use and fig. Something, esp. a system or process, that resembles algebra in substituting one thing for another, or in using symbols, signs, etc., to represent ideas and concepts.[52]

Neither of these characterizations apply to Arabic algebra. Letters and other symbols are only incidentally employed in Maghrebi texts (see below), and in any case they represent kinds of numbers and not numbers themselves.[53] Further, there is no abstract reasoning that would take the art beyond arithmetic. These modern views of algebra have muddled the attempts of some historians to properly identify just what al-jabr wa l-muqābala is, and to distinguish it from other techniques and modes of reasoning in Arabic and other premodern mathematics.

By “Arabic algebra” we mean the art of al-jabr wa l-muqābala as it was understood by those who practiced it. This was a specific technique of numerical problem-solving, with it own vocabulary and rules. As Ibn al-Bannāʾ and al-Hawārī explain, the powers of the unknown are given names. The first power of the unknown is called a “thing” or a “root”, akin to our x. The second power is called a “māl ”, an Arabic word which ordinarily means “sum of money” or “wealth”, and which corresponds to our x^2.[54] The cube of the thing, our x^3, is called a “cube”, and higher powers are expressed by combinations of māl and cube, such as māl māl for the fourth power and māl cube for the fifth.

In most problems solved by al-jabr wa l-muqābala an unknown number is named “a thing”, though sometimes it is named as one of the other powers or some combination of them. The conditions of the problem are then applied to set up a polynomial equation expressed in terms of the names of the powers, and the equation is then simplified and solved. One important distinction between al-jabr and other methods like single and double false position is that, in the latter, operations are performed only on known numbers, while in algebra calculations can be performed on the names of the powers. In other words, in algebra one operates on the unknown.

The earliest known Arabic books on algebra are the Book of Algebra by al-Khwārazmī and the book of the same title by Ibn Turk.[55] Both were written in Baghdad during the reign of the caliph al-Maʾmūn, 813-833 CE. Some historians have thought that because no Arabic book on this topic is known before al-Khwārazmī’s that he must have invented algebra. This view does not take into account the oral traditions of calculation before the start of the translation movement. In fact, evidence of oral transmission is manifest even in al-Khwārazmī’s book.[56] Analogously, the earliest known work on double false position is the Book on Proof of the Method of Double False Position by Qusṭā ibn Lūqā (ca. 820-ca. 912/3).[57] No one would claim that Qusṭā invented double false position. He was merely among the first we know to write down a book describing the method.

With a clear idea that al-jabr must predate the time of al-Khwārazmī, historians have looked for the technique in the mathematics of previous civilizations. The most prominent candidates are (a) the Brāhma-spuṭa-siddhānta of Brahmagupta and other works from India (seventh century CE and later), (b) the Arithmetica of Diophantus of Alexandria (ca. early fourth century CE), and (c) ancient Babylonian cuneiform tablets (mainly ca. 2000-1600 BC). The method practiced in India is certainly a kind of algebra – names are given to unknowns and equations are formed and solved. But because the vocabulary is of an entirely different nature and the technique is more advanced in some respects than what we find in Arabic algebra, we can exclude direct influence of this sophisticated Indian algebra on the latter.[58] What is called Babylonian algebra is likewise distant from Arabic algebra. There the known and unknown quantities are represented by lines and rectangles and the manipulations take place in the context of a diagram. Although operations can be performed on the unknown, there is no simplification to a canonical set of equations, so each problem requires its own trick to solve.[59] Babylonian algebra, then, seems to be a precursor to Arabic algebra.

Diophantus’s way of solving problems, on the other hand, matches Arabic algebra in its vocabulary, overall structure, and concepts. More specifically, they agree in their systems of naming the powers of the unknown, in their basic procedures of setting up, simplifying, and solving equations, and in their conceptions of monomials, polynomials and equations.[60] Diophantus wrote in Alexandria in late antiquity. His book was translated into Arabic in the latter ninth century, after the time of al-Khwārazmī.[61] Arabic authors called the Arithmetica a book on al-jabr wa l-muqābala or al-jabr,[62] and many of Diophantus’s problems found their way into Arabic algebra books.

We also know from two passages in Ibn al-Nadīm’s Fihrist that the Greek astronomer Hipparchus of Bithynia wrote a book on algebra in the second century BC that was also translated into Arabic.[63] This book is unfortunately lost in both languages. Neither Hipparchus nor Diophantus can be said to have invented algebra, however. Like al-Khwārazmī some centuries later, both most likely composed books inspired by the algebra of practitioners that was already circulating orally in their times. The identity and even the nationality of the person or people who first practiced this premodern algebra are lost to history. There is certainly some historical link between Greek and Arabic algebra, either through texts, through oral practice among those who performed the calculations in their work, or both.

1.6 Education, books, and notation

One peculiar feature of medieval mathematics books, at least to modern readers, is their general lack of notation. Ibn al-Bannāʾ wrote all numbers in his Condensed Book in words, even if he was teaching the use of Indian numerals! And al-Hawārī, whose purpose was to give numerical examples, still writes the numbers and operations in words before showing the notation. This is perhaps the most glaring of the many differences between medieval and modern arithmetic books. The causes of these differences have to do largely with how the material was taught, and more generally with how people in al-Hawārī’s time related to books.

1.6.1 Education in medieval Islam

Where today education is structured around institutions, curricula, and degrees, in medieval Islamicate societies it was structured around individual teachers, books, and licenses (ijāzāt).[64] Schools did exist and provided support for both teachers and students, and from what we can gather they often had curricula; but throughout the medieval period education remained centered on the student-teacher relationship. To learn a topic, a student chose a teacher (shaykh) with a good reputation to study a particular book. Once the student had learned the book to the satisfaction of the shaykh, he (or rarely, she) received an ijāza, or license, granting him authority over the text. The ijāza not only certified mastery of the contents of the book, it gave its possessor authorization to teach it himself. The ijāza recorded the names in the chain of transmission of that particular text, listing the student, his teacher, his teacher’s teacher, ideally all the way back to the author of the book.

Instruction was centered around dictation and memorization. The shaykh read the book to the student, who memorized it and dictated it back to the shaykh. This emphasis on speaking and listening reflects the notion that books properly reside in the minds of those who had memorized them, and not in manuscript copies. Students recited aloud even when studying alone after the lecture, “for what the ear hears becomes firmly established in the heart”.[65] Manuscripts played an essential role in this oral environment. Copying the words of a teacher on paper aided memorization, and written books gave the student access to the text when studying alone. But true learning was thought to take place in the presence of the teacher, who possessed an understanding of the book that manuscripts lacked.

Instruction was not restricted to dictation of the book under study. The shaykh would often supplement it with examples, illustrative remarks, and material from other sources. Al-Hawārī cites additional procedures for double false position that Ibn al-Bannāʾ dictated to him at 204.3, and he signals other added material at 163.11 and 232.9. Many other passages are attributed in the manuscripts to Ibn al-Bannāʾ that are not in the Condensed Book, and most likely they were dictated to al-Hawārī as well.[66] Whatever a textbook lacked in examples and explanations would have been covered somehow by the shaykh teaching it.

Many authors wrote their books concisely to make memorization easier, which is certainly the case for Ibn al-Bannāʾ’s Condensed Book. Some authors even put down the mathematical ideas in verse, since poetry is easier to memorize than prose. Ibn al-Yāsamīn’s famous Poem on Algebra covers the basic rules of algebraic problem-solving in just 54 lines,[67] and Ibn Ghāzī reduced the contents of Ibn al-Bannāʾ’s Condensed Book to 461 verses in his Desire of Reckoners.[68] In the other direction, many authors filled their books with detailed explanations and many worked-out exercises, making them too long to have been memorized in their entirety. For example, Ibn al-Hāʾim’s 1387 Commentary on the Poem of al-Yāsamīn amounts to an encyclopedia of algebraic knowledge that takes up 257 pages in the modern printed edition,[69] and Ibn Ghāzī himself expounded on his arithmetical poem to make a commentary that occupies 323 pages of the modern edition.[70] Al-Hawārī’s commentary falls into this category, too.

1.6.2 The role of notation

Because of the oral nature of learning in medieval Islam, books read like transcriptions of lectures.[71] Notation serves no purpose to the student listening to a lecture, so there is no advantage in including it in the running text of a book. This is why Ibn al-Bannāʾ wrote his book entirely in words, and why the notation for the calculations in al-Hawārī’s book only appear as figures or illustrations. These figures play a role much like geometric diagrams or other illustrations, and were intended to show the student what is to be put down on the dust-board or other surface. For example, he begins this calculation in the chapter on addition, at 75.9:

Suppose we want to add nine hundred seventy-eight to four hundred fifty-six. We put them down on two lines, as mentioned, as in this figure: \begin{matrix}978\\[-.5mm]456\end{matrix}.

The visual notation (here \begin{matrix}978\\[-.5mm]456\end{matrix}) is always set apart from the audible text in al-Hawārī’s book with some phrase like “and its figure is” or “so we write it down like this”.

Some Arabic authors included the Indian notation in the spoken parts of their arithmetic books, but there the numbers were still meant to be recited aloud like the words that surround them. Naṣīr al-Dīn al-Ṭūsī, for example, begins this problem in his Gathering of Arithmetic by Means of Board and Dust (thirteenth century):[72]

For example, we want to multiply 123 by 456. We write them in two lines like this: \begin{matrix} 4&1 \\ 5&2 \\ 6&3 \end{matrix}  .

Here the 123 and 456 are still intended to be spoken, while the same numbers shown in columns form a figure that was to be apprehended visually, like any illustration.

There were thus two ways of writing arithmetic. To solve a problem, one wrote in notation on a dust-board or other temporary surface. Then, if one wished to communicate the result to others, a rhetorical version was composed. For this reason, the numerals tend to appear in arithmetic books only to show students what is to be put down on the board.

Because board calculations were for personal, immediate use, the notation can tolerate ambiguities. For example, the notation for excluded fractions of both connected and disconnected types were often represented the same way in notation. But this does not matter, because the person working out the calculation knows which is which while doing so (see our commentary at 140.14). Ambiguity would only be a problem if one wanted to consult the book later or have others read it. Dust does not permit the former, and for the latter a rhetorical version in unambiguous prose was composed.

1.6.3 Developments in Indian notation: fractions, roots, and algebra

Arabic practitioners devised notations to extend Indian numerals to show fractions and roots in arithmetic and polynomials and equations in algebra.[73] There are differences in these notations that can usually be ascribed to geography, mainly between east and west. The Persian mathematician al-Fārisī, a contemporary of Ibn al-Bannāʾ, shows fractions with the denominator over the numerator, and without the bar. His “three sevenths” is shown as 7\atop 3, for example.[74] By contrast, al-Kāshī, working in Samarkand in eastern Persia in 1427, put the numerator over the denominator. His “eight and four sevenths” is shown in notation as \overset{\scriptstyle {8}}{\underset{\scriptstyle {7}}{\scriptstyle {4}}}.[75] The earliest known book from the Maghreb showing fractions in notation is al-Ḥaṣṣār’s late twelfth century Book of Demonstration and Recollection in the Art of Dust Board Reckoning. There, fractions are shown just the way we write them today, with the numerator over the denominator separated by a horizontal line. Here, for example, is his 9\over 11 from a manuscript copied in 1194 CE: Too late.[76] Other books from the west, including al-Hawārī’s commentary, also show fractions this way.

Ibn al-Yāsamīn’s late twelfth century Grafting of Opinions is the earliest known book that indicates the square root by placing an elongated letter jīm (Too late) above the affected numbers, much like the way we use our elastic sign “ \sqrt{~~~~~}”. This notation became common in the Islamic West, though al-Hawārī does not show it. We have not (yet) seen a notation for roots in Arabic books written east of Egypt.

Initially, there seems to have been no notation specific to Arabic algebra. The little evidence we have indicates that polynomials were written on a board as a list of coefficients. In the early eleventh century, al-Karajī, working in Baghdad, wrote out in words the polynomial that we would express as x^{12}+4x^{11}+10x^{10}+20x^9+35x^8+56x^7+84x^6+104x^5+115x^4+116x^3+106x^2+84x+49, and then wrote “you put it down in this figure: 1 4 10 20 35 56 84 104 115 116 106 84 49”.[77] A century and a half later al-Samawʾal, working in Persia, shows polynomials in his Dazzling [Book] on the Science of Calculation the same way, but with the names of the powers written in words above the Indian numerals for each coefficient to keep straight which number goes with which power.[78] The large majority of problems solved by algebra do not require anything higher than the second power, so when working out a problem on a board there was little need to indicate the name of the power when working through the calculations. We have found one Western manuscript that shows an equation in the margin in Eastern notation: Too late. We can transcribe this as “ 1~2=99”, and in modern notation we write it as x^2+2x=99.[79]

Later, in the western part of the Islamic world, a notation came into common use in which the first letter of the name of the power was placed above the coefficient. Our earliest glimpses are two brief figures in Ibn al-Yāsamīn’s late twelfth century Grafting of Opinions.[80] Most authors working in the Maghreb or al-Andalus in the fourteenth and fifteenth centuries also show it, while some in the region, Ibn al-Bannāʾ and al-Hawārī included, do not. Like the Indian notation, it is only presented to instruct students on how to write it to perform calculations on a board. In almost every instance, we are treated to a single expression or equation in notation here and there during the course of a rhetorical explanation or solution to a problem. We are fortunate, then, that in one place Ibn Ghāzī shows an entire problem worked out using this algebraic notation in nearly a dozen lines in his 1483 Aim of the Students.[81] Incidentally, it is on that page in the Library of Congress manuscript that the Eastern equation shown above appears. The Western version is written like this: Too late. Algebraic notation is neither mentioned nor shown in more advanced books on algebra, like Ibn al-Hāʾim’s Commentary mentioned above. Because it is better suited to represent medieval algebraic calculations than modern notation, we describe the Arabic notation in our commentary beginning at 211.13 and further at 219.1, and we use it to explain al-Hawārī’s calculations.

1.6.4 Nesselmann’s stages revisited

Addressing Nesselmann will help some readers apprehend better the algebra in al-Hawārī’s book, and indeed Arabic algebra in general. To make sense of the historical development of algebra, the German orientalist G.H.F. Nesselmann devised a three-stage scheme that he described in his 1842 book Versuch Kritischen Geschichte der Algebra (Critical Essay on the History of Algebra).[82] He dubs the most primitive stage “rhetorical algebra”, in which all calculations are written verbally. Next is “syncopated algebra”, which was still essentially rhetorical but with some recurring abbreviations. The third stage is “symbolic algebra”, where calculations are represented in a language independent of oral presentation. Nesselmann identified Arabic algebra, quoting al-Khwārazmī in particular, as belonging to “rhetorical algebra”, and Diophantus’s Arithmetica as belonging to “syncopated algebra”.

Although historians of mathematics in the past few decades have moved beyond Nesselmann’s classification,[83] it is still frequently cited in popular accounts of the history of mathematics and in studies in mathematics education. Unfortunately, it is misleading as a scheme for describing historical developments in algebra or for identifying conceptual shifts. Nesselmann was unaware of board calculations when he classified Arabic algebra as “rhetorical”. Although the algebraic calculations we find in the books of al-Khwārazmī, al-Khayyām, and others fall into his “rhetorical” category, the same work on a dust-board or wax tablet in the Arabic notation would be safely classified as “symbolic” according to his definition.[84] The mode of presentation is thus not an indication of a stage in algebraic development. Diophantus’s instructions for abbreviating algebraic vocabulary read much like the instructions for writing algebra in notation in the later Arabic books. His abbreviations were probably used when working through the calculations on a wax tablet, so in this respect it is no different from the later Arabic practice. In manuscripts of the Arithmetica the abbreviations were evidently intended to be expanded and pronounced, like the numbers in Naṣīr al-Dīn al-Ṭūsī’s arithmetic book quoted above. And finally, Nesselmann did not notice the difference between premodern and modern algebraic notations, where the letters or signs in the former designate types, and in the latter designate values.[85]

1.6.5 The figures in al-Hawārī’s book

One consequence of the distinction between visual figure and audible text is that in European languages we read numbers expressed with Indian numerals backwards. Arabic is written right-to-left, so when reading a number like “214” in that language one starts from the 4 in the units place. When Europeans translated Arabic texts on arithmetic into Latin in the Middle Ages they preserved the orientation of figures. This is why “214” did not become “412”. Because European languages are written left-to-right, we read the number 214 starting from 2 in the hundreds place.

Like the Latin translators, we have preserved the orientation of the Arabic figures in our translation. We could not reverse figures showing numbers horizontally, like \begin{matrix}4043\\[0mm]2685\end{matrix} (at 74.17), or else the “four thousand forty-three” would look like 3404. It would not make sense to reverse some figures and not others, so we kept the orientation of all figures as they appear in the manuscripts. Keep in mind when reading the translation that figures would have been viewed right to left.

Although copyists of al-Hawārī’s book transcribed his words more or less faithfully, the same cannot be said for the figures. For example, the figure we typeset for the passage after 103.3 is shown below, followed by the figures from the Tehran, Oxford, and Tunis manuscripts.

Fig. 1.7: Comparing figures in the manuscripts

We deem our translation to be correct because it agrees with the verbal description of the operations in the text. Of the five manuscripts we consulted, only the Tehran manuscript shows this figure correctly. The Oxford and Medina manuscripts leave off the last round of calculations, and although not incorrect, the Oxford manuscript is the only one that crosses off numbers. Most of the numbers in the Tunis manuscript are wrong, and in the Istanbul manuscript (not shown here) the numbers between the lines are shifted up one row. All five manuscripts, including the Tehran manuscript, have errors in many figures, and it is rare to find a mistake that is common to all manuscripts. In our translation we show figures in the style found in the manuscripts which agree with the calculations described in the text.

1.7 The legacies of the Condensed Book and al-Hawārī’s commentary

The popularity of Ibn al-Bannāʾ’s Condensed Book is evident in the number of commentaries, poems, and abridgments it inspired. After his own Lifting the Veil and al-Hawārī’s Essential Commentary, we know of ten people who wrote commentaries on the Condensed Book between the fourteenth and sixteenth centuries.[86] In the other direction, Ibn al-Hāʾim (ca. 1400) wrote an even more condensed version of Ibn al-Bannāʾ’s little book, and four authors, including Ibn Ghāzī (15th c.), set the contents of the Condensed Book to verse.[87]

Although by no means as well known as his professor’s Condensed Book, the number of surviving manuscripts of al-Hawārī’s Essential Commentary is testimony to its popularity across several centuries. We have found four later works that reference or copy from al-Hawārī. ʿIzz al-Dīn al-Ḥanbalī (d. 1409) copied many numerical examples from al-Hawārī’s book into his own commentary on Ibn al-Bannāʾ’s Condensed Book, but without citing his source. Later, Ibn Ghāzī likewise copied many numerical examples into his Aim of the Students (1483). He mentions al-Hawārī by name twice, though not in connection with his copying. Sibṭ al-Māridīnī’s (1423-1506) Student’s Guide to the Way of Arithmetic paraphrases al-Hawārī’s remarks on the ontology of ratios (at 133.3), and cites the source.[88] And while still a student in the 1740s in Istanbul, future mathematics teacher Şeker Zāde put together a notebook titled Examples from Ibn al-Bannāʾ’s Condensed [Book] and Ibn al-Hāʾim’s Contents [of Calculation].[89] Şeker Zāde copied over a hundred numerical examples from al-Hawārī’s commentary, always citing his source. This notebook is remarkable because the problems are all worked out in the Arabic notation, showing what would have been put down on the dust-board. And finally, a two-page extract from al-Hawārī’s book, consisting of the subsection “on finding deaf parts” (127.9-128.16), appears in the middle of MS Berlin Landberg 199, on folios 31b-32a. This manuscript contains several works on arithmetic and algebra by other authors, including Ibn Fallūs (d. 1239) and al-Khwārazmī. Neither al-Hawārī’s name nor the title of his book are given on those two pages.

We last hear of al-Hawārī in 1875, in connection with the proposed mathematics curriculum for the Zaytūna mosque school in Tunis. Until then, mathematics instruction at the mosque had been limited to five stylistic commentaries of a poem on inheritance and arithmetic. In an attempt to modernize instruction, Prime Minister Hayreddin Pasha oversaw the creation of an entirely new curriculum made public on December 25, 1875. Although a great improvement over the dismal curriculum in place at the time, the ten recommended texts on mathematics were themselves already hopelessly out of date. They included al-Hawārī’s Essential Commentary in the upper division, even though it was already over 500 years old! The proposed curriculum received harsh resistance from the ulama, the religious authorities, and it was likely never implemented.

Footnotes

[1] (Lamrabet 2014, 164ff); (Samsó 2007).

[2] Published in (Ibn al-Bannāʾ 1969).

[3] Published in (Ibn al-Bannāʾ 1994).

[4] Published in (Ibn al-Bannāʾ 1984).

[5] Published in (Saidan 1986, vol. 2).

[6] (Abdeljaouad and Oaks 2013, 11).

[7] Published in (al-Hawārī 2013) and in the present volume.

[8] (Ibn al-Bannāʾ 1969, 8).

[9] (Gutas 1998); (Saliba 2007).

[10] The classic work on Greek influence is (Rosenthal 1975). For mathematics and philosophy, see (Endress 2003).

[11] See (Saliba 2007); (Ragep 2007); (Brentjes 2007).

[12] As, for example, al-Fārābī does in his Enumeration of the Sciences (first half of 10th c.) (al-Fārābī 1953, 54.7).

[13] Al-Takmila fī l-ḥisāb. Saidan published the Arabic text in (al-Baghdādī 1985). See also (Saidan 1987).

[14] (Oaks 2011a). Al-Khayyām himself used algebra this way to solve a particular problem of cutting a quadrant of a circle to produce a particular ratio.

[15] A German translation is published in (Suter 1908–1909).

[16] (Ibn al-Bannāʾ 1994, 202.5).

[17] See our commentary at 65.2.

[18] There were minor differences between east and west in the Islamic world in the schemes used.

[19] (al-Uqlīdisī 1978, 310); (al-Uqlīdisī 1984, 386). The signs are likely related in some way with special signs appearing on Roman coins minted in Siscia in the period 348-350 CE (Kent 1981, 343, 359, 364-67).

[20] Al-Ḥaṣṣār (d. before 1194), writing in the Maghreb, mentions them in his Complete [Book] on the Art of Number (Guergour 2000, 68). The signs, which varied between east and west, are also mentioned in the writings of Mozarabs (Christians writing in Arabic) in Toledo around the same time (Colin 1933, 204).

[21] (Saidan 1971, 142.17).

[22] For the Roman and later European abacus, see (Pullan 1968).

[23] Saidan describes the poem and translates the relevant parts in (Saidan 1968). Pellat gives an edition and French translation of a similar poem from the previous century, by Abū ʿAbd Allāh Muḥammad ibn Aḥmad al-Mawṣilī al-Ḥanbalī (#768), in (Pellat 1977, 52-59).

[24] The illustration is in (Pacioli 1494, fol. 36b). We reproduce a scan from the 1523 printing from the Columbia University Library copy. This illustration is identical to that in the 1494 printing.

[25] (Smith 1958, vol. II, 196ff); (Pellat 1977).

[26] Edited in (Saidan 1971). See (Saidan 1974) for a description of the contents.

[27] Kitāb al-ḥisāb al-hindī. Following (Saidan 1987, 44), we attribute this and some other books listed in Ibn al-Nadīm’s Fihrist under Sanad ibn ʿAlī (#48) to al-Khwārazmī. This is our source for the Arabic title of the book.

[28] For more on the dust-board, and on the transmission of Arabic numerals generally, see (Kunitzsch 2003).

[29] The Arabic text is published in (al-Uqlīdisī 1984), and Saidan’s English translation is published in (al-Uqlīdisī 1978).

[30] Saidan’s translation, from (al-Uqlīdisī 1978, 247).

[31] (Djebbar 2004). There was no single Greek concept of number. Diophantus’s Arithmetica (ca. fourth century CE) shows many fractions, and the works of Hero of Alexandria (first century CE) give many rational approximations to irrational numbers. Also, because number concepts, whether in Greek or Arabic, were based in counting or measuring, it would not have occurred to ancient or medieval mathematicians that negative or complex numbers could exist.

[32] Categories 4b20-24, translated in (Aristotle 1963, 12). Language is discrete because it consists of indivisible syllables.

[33] (Euclid 1956, vol. 2, 277).

[34] For these ideas in Aristotle, see (Cleary 1995, 345ff).

[35] We checked the arithmetic books of al-Khwārazmī (Latin redaction), al-Uqlīdisī, Abū l-Wafāʾ, Kūshyār ibn Labbān, al-Karajī, al-Baghdādī, Ibn al-Samḥ, and Ibn al-Haytham.

[36] The Brethren of Purity write “The ‘one’ is said in two ways: in its proper usage, and by way of metaphor. In its proper usage it is a thing that cannot be partitioned or divided…As for ‘one’ in metaphor, it is every aggregate that is considered a unity. So, for example, ten is called a ‘unit’, and a hundred is called a ‘unit’, and a thousand is called a ‘unit’” (translated in (El-Bizri 2012, 67). See also (Goldstein 1964, 136)). Al-Bīrūnī writes “Although ‘one’ is in reality indivisible, nevertheless the unit, one as a technical expression, employed in dealing with sense-objects, whether by weighing, measuring by bulk, or length or number, or merely in thought, is obviously capable of sub-division” (translated in (al-Bīrūnī 1934, 24)). Al-Baghdādī tacks onto his Completion of Arithmetic a chapter on number theory based in Nicomachus (Chapter 6), but he does not address the nature of the unit.

[37] (al-Fārisī 1994, 71.8). Sibṭ al-Māridīnī (died 1506), working in Egypt, similarly contrasted Euclid’s definition with that of the “Arithmeticians” (Sibṭ al-Māridīnī 2004, 73.5). See also (Djebbar 2004, 315-318).

[38] (al-Ḥaṣṣār manuscript, fol. 3b); (Zemouli n.d., 5).

[39] (Ibn al-Bannāʾ 1984, 121.4).

[40] (Oaks 2011a, 59ff).

[41] Two Indian books containing such rules are the Āryabhaṭīya of Āryabhaṭa (ca. 500 CE) and the Pāṭī-gaṇita of Śrīdhara (8th or 9th c.) (Plofker 2009, 131-132); (Sridhara Acarya 1959).

[42] (Saidan 1996, 341); (Djebbar 2004, 310).

[43] Proposition I.47.

[44] (Rashed 1996, 319ff).

[45] Even Greek geometers in late antiquity had begun to assign numbers to magnitudes.

[46] (al-Khwārizmī 2009, 34-42). We deliberately cited the same author in this paragraph and the last. For Thābit ibn Qurra, a proposition in the tradition of Greek geometry should be free of any arithmetizations, but for algebra this rigor is relaxed. It is the setting, not the mathematician, that determined which approach is appropriate.

[47] (Ben Miled 2005).

[48] For Egypt: (Imhausen 2003, 37, 51); (Gillings 1972, Chapter 14). For Babylonia: (Høyrup 2002, 59-60, 102, 311-313). For Sanskrit: (Hayashi 1995, 396-399).

[49] (Chemla 1997).

[50] (Høyrup 2002, 103); (Plofker 2002, 182-183); (Plofker 2009, 259).

[51] The uses of these two terms in algebra are described briefly at 211.1, and through examples in the section beginning at 223.1.

[52] http://www.oed.com/, accessed August 26, 2018.

[53] See our commentary at 229.8; for a more thorough explanation, see (Oaks 2017).

[54] There is no good English translation of māl, so we leave it transliterated.

[55] Rashed’s edition of al-Khwārazmī’s book, with an English translation, is published in (al-Khwārizmī 2009). Only a portion of Ibn Turk’s book survives. It is edited with an English translation in (Sayili 1962). In his tenth-century Kitāb al-Fihrist, Ibn al-Nadīm reports a book on algebra by Sanad ibn ʿAlī (#48), a contemporary of al-Khwārazmī and Ibn Turk. This attribution is probably an error, and the book belongs instead to al-Khwārazmī. See (Saidan 1987, 439-440).

[56] For evidence of oral transmission, see (Oaks 2012b); (Saidan 1974, 369); (King 1988).

[57] Kitāb al-burhān ʿalā ʿamal ḥisāb al-khaṭaʿān. Suter published a German translation in (Suter 1908–1909). We have also consulted the British Library, Oxford, and Cairo manuscripts. Abū Kāmil (late ninth century) is also reported to have written a book on double false position about the same time, but it is not extant.

[58] Léon Rodet made a detailed analysis of these differences in (Rodet 1878).

[59] (Høyrup 2002).

[60] (Christianidis and Oaks 2013).

[61] The four surviving books of the Arabic translation have been edited and translated twice, in (Sesiano 1982) and (Diophantus 1984).

[62] (Sesiano 1982, 8-13).

[63] (Ibn al-Nadīm 1871–1872, 269, 283); (Ibn al-Nadīm 1970, 642, 668). Both references mention Abū l-Wafāʾ’s commentary on Hipparchus’s “book on the art of algebra (al-jabr)”. In addition, Abū l-Wafāʾ himself mentions this commentary in his Book of What is Necessary for Scribes, Businessmen, and Others in the Science of Arithmetic (Saidan 1971, 126.7).

[64] For Islamic education, see (Makdisi 1981), (Berkey 1992), and (Chamberlain 1994).

[65] Abū l-Hilāl al-Ḥasan al-ʿAskarī, quoted in (Berkey 1992, 27).

[66] These passages are shown in small caps in the translation.

[67] Published in (Abdeljaouad 2005a). Ibn al-Yāsamīn also wrote a 55-line poem on root extraction and an 8-line poem on double false position.

[68] Note also the short poems quoted by al-Hawārī describing the shapes of the Indian numerals at 69.2, the abjad numerals for casting out sevens at 88.14, and ʿAlī ibn al-Maghribī’s Poem on Reckoning with Finger-Joints mentioned above.

[69] (Ibn al-Hāʾim 2003).

[70] The commentary is titled Aim of the Students in Commentary on Desire of Reckoners (Ibn Ghāzī 1983).

[71] (Chamberlain 1994); (Berkey 1992).

[72] (Saidan 1967, 125). Some other books that show notation in the spoken parts of the text include al-Uqlīdisī’s Chapters on Indian Arithmetic (952/3 CE), Ibn Munʿim’s Understanding Calculation (12th-13th c.), and al-Fārisī’s Foundation of Rules on Elements of Benefits (early 14th c.).

[73] For algebraic notation, see (Abdeljaouad 2005a) and (Oaks 2012a).

[74] (al-Fārisī 1994, 199).

[75] (al-Kāshī 1969, 89). Al-Baghdādī writes fractions the same way (al-Baghdādī 1985, 102ff).

[76] (al-Ḥaṣṣār manuscript, fol. 43b).

[77] (al-Karajī 1964, 52). The same polynomial is written similarly by al-Samawʾal (al-Samawʾal 1972, 67).

[78] (al-Samawʾal 1972, 45ff). Here al-Samawʾal is performing operations on the coefficients in tables.

[79] (Ibn Ghāzī manuscript, fol. 108b). It appears that the Eastern marginal annotator felt the need to clarify the final calculations shown in the manuscript, which were written sloppily in Western notation.

[80] (Zemouli 1993, 137, 231).

[81] (Ibn Ghāzī 1983, 302); (Ibn Ghāzī manuscript, fol. 108b); (Oaks 2012a, 62).

[82] (Nesselmann 1842, 301ff).

[83] Albrecht Heeffer, in particular, has recently criticized Nesselmann (Heeffer 2009).

[84] Ibn Ghāzī’s page showing an entire problem worked out in notation is the one example we have of what must have been commonly written on a board. See (Oaks 2012a, 62).

[85] See (Oaks 2017) for an explanation of this difference. We also discuss it briefly in our commentary at 229.8.

[86] Al-Ghurbī (M158, 2nd half 14th c.), al-Mawāḥidī (M176, ca. 1382), Ibn Zakariyyā (A336, #793, died 1403-4), Ibn Qunfudh (died 1407-8), al-ʿUqbānī (died 1408), Ibn Haydūr (M196, died 1413), Ibn Majdī (#815, died 1447), al-Ḥabbāk (M219, #831, died 1463), al-Qalaṣādī (died 1486), and Muḥammad al-Ghazzī (#998, 16th c.).

[87] The others are Ibn Marzūq (M205, died 1438), al-Wansharīsī (M257, died 1548-9), and Ibn al-Qāḍī (M315, died 1616).

[88] (Sibṭ al-Māridīnī 2004, 145).

[89] (Abdeljaouad 2011); (Abdeljaouad and Oaks 2013). Ibn al-Hāʾim’s Contents of Calculation is his abridgment of the Condensed Book mentioned above. Incidentally, Şeker Zāde’s figure for the passage after 103.3 is shown correctly.

Source

Contents of al-Hawārī’s book

Pages for the translation and our commentary are given below, followed by the page and line number of the Arabic edition. The Arabic portion of this book receives a separate pagination. It is placed after the English part, or at the beginning if you are an Arabic reader. The chapter and section titles in this index are sometimes paraphrased from the Arabic.

| Trans. | Comm. | Edition
[Introduction] | 59.1 | 59.1 | 59.1
Part I. Known numbers | 61.9 | 61.9 | 61.9
  I.1 Whole numbers | 63.1 | 63.1 | 63.1
    I.1.1 The divisions of numbers and their ranks | 65.1 | 65.1 | 65.1
      § Knowing the index of the repeated number | 71.12 | 71.12 | 71.12
    I.1.2 Addition | 73.1 | 73.1 | 73.1
    I.1.3 Subtraction | 83.1 | 83.1 | 83.1
      § Checking calculations by casting out | 91.1 | 91.1 | 91.1
    I.1.4 Multiplication | 95.1 | 95.1 | 95.1
    I.1.5 Division | 117.1 | 117.1 | 117.1
      § Finding deaf parts | 127.9 | 127.9 | 127.9
    I.1.6 Restoration and reduction | 129.1 | 129.1 | 129.1
  I.2 Fractions | 131.1 | 131.1 | 131.1
    I.2.1 Names of fractions and numerating them | 134.1 | 134.1 | 134.1
    I.2.2 Adding and subtracting fractions | 147.1 | 147.1 | 147.1
    I.2.3 Multiplying fractions | 149.1 | 149.1 | 149.1
    I.2.4 Division and denomination | 151.1 | 151.1 | 151.1
    I.2.5 Restoration and reduction | 154.1 | 154.1 | 154.1
    I.2.6 Converting | 157.1 | 157.1 | 157.1
  I.3 Roots | 161.1 | 161.1 | 161.1
    I.3.1 Taking roots of whole numbers
           and fractions | 163.1 | 163.1 | 163.1
      [Binomials and apotomes] | 173.4 | 173.4 | 173.4
    I.3.2 Adding and subtracting roots | 179.1 | 179.1 | 179.1
    I.3.3 Multiplying roots | 183.1 | 183.1 | 183.1
    I.3.4 Dividing and denominating roots | 187.1 | 187.1 | 187.1
Part II. Finding unknown numbers | 191.1 | 191.1 | 191.1
  II.1. Solving problems with proportion | 193.1 | 193.1 | 193.1
    [The four proportional numbers] | 195.7 | 195.7 | 195.7
    [The method of scales (double false position)] | 198.2 | 198.2 | 198.2
  II.2 Algebra | 209.1 | 209.1 | 209.1
    II.2.1 The meaning of algebra | 211.1 | 211.1 | 211.1
    II.2.2 Solving the six types [of equation] | 213.1 | 213.1 | 213.1
    II.2.3 Addition and subtraction | 219.1 | 219.1 | 219.1
          § Examples with the two sides
            of an equation | 223.1 | 223.1 | 223.1
    II.2.4 Multiplication and knowing
           the power and the term | 225.1 | 225.1 | 225.1
    II.2.5 Division | 229.1 | 229.1 | 229.1
  § Secret numbers | 231.1 | 231.1 | 231.1

Translation

Note: The Istanbul and Tunis manuscripts differentiate between passages attributed to Ibn al-Bannāʾ and those attributed to al-Hawārī, while the other three manuscripts we consulted do not make this distinction. In our translation, passages attributed to Ibn al-Bannāʾ that are taken from his Condensed Book are in bold font, while passages attributed to al-Hawārī are not in bold font. Passages borrowed from Ibn al-Bannāʾ’s Lifting the Veil, often with minor changes in wording, are in italics: bold italics for those attributed to Ibn al-Bannāʾ, and regular italics for those attributed to al-Hawārī. Passages attributed to Ibn al-Bannāʾ that are neither in the Condensed Book nor in Lifting the Veil are in small caps. Words and phrases added by us to b make the meaning clearer are placed in [square brackets]. Arabic words behind our translation are placed in parentheses, like “cube (mukaʿʿab)”.

link to parallel text 59.1 In the name of God, the Merciful and Compassionate, the humble and submissive servant to his Lord, hoping for His reward and His pardon for his sin, ʿAbd al-ʿAzīz ibn ʿAlī ibn Dawūd al-Hawārī al-Miṣrātī – may God forgive him – said: Praise be to God, the provider of graces and creator of life, who brings things into existence from nothingness, with praises the counting of which is uninterrupted. And full prayers on Muḥammad, His prophet and servant, and for His acceptance of the ancestors, who followed the revealed sunnas while not overstepping their boundary. And a prayer of supplication to our lord, the Commander of the Muslims, son of the Commander of the Muslims, Abū Yaʿqūb, deliverer of cherished victories from God. Then, may God prolong the term [in office] of our sovereign minister, the exalted, the radiant, the noble, the blessed, the honored, the jurisprudent, the learned, the all-knowing, the venerated of lofty virtue and superior intention, the superior, the proud, the issuer of laws guided by the light of the auspicious Marinid state, Abū Muḥammad ʿAbdallāh, son of our magistrate, the Shaykh Jurist, the virtuous, the pious, the ascetic, of excellent conduct and behavior, the late, sanctified, purest Ibn Madyan, who was made one of God’s intercessors at the Gathering Place. May God perpetuate their memories among the pious, and protect the era, their return, and their place of honor from disgrace.

link to parallel text 59.15 I begged almighty God to help me in my task of commenting on the book named Condensed Operations in Arithmetic, written by the incomparable scholar, our most eminent and open-minded master, Sheikh Abū l-ʿAbbās Aḥmad ibn Muḥammad ibn ʿUthmān al-Azdī, may God continue to bestow his divine blessing, care, and grace on him. I enhance my work with your name and embellish it with the splendor of your merits, so that I may deserve your blessing and your protection. To this end I asked our above-mentioned master, the scholar Abū l-ʿAbbās, for permission to undertake this work, to which request he kindly assented. Since the book I undertake to comment on has already been enriched by its own author, God bless him, with Lifting the Veil, containing all that is needed but providing only a few examples, I shall, almighty God willing, illustrate them with examples in the right place and when necessary. I call it the Essential Commentary on the Condensed [Book] on the Operations of Arithmetic. With God’s help and assistance in all circumstances, it is time for me to begin the work, imploring the Almighty to guide and bless my endeavor.

link to parallel text 61.1 Our Master, the jurisprudent, the teacher, the leader, the learned, the radiant, the all-knowing, the guardian [of religion], Abū Aḥmad ibn Muḥammad ibn ʿUthmān al-Azdī, may God forgive him, said:

61.3 The goal of this book is to condense the operations of arithmetic, and to bring within reach its sections and its concepts, and to master its rules and its structures. It is comprised of two parts: the first is on operating with known numbers, and the second is on the basic rules through which the access to knowledge of the required unknown from the posited known is established, whenever a connection between them is provided. And I ask of God, praise be to Him, for assistance and success, and guidance toward the correct path.

link to parallel text 61.9 Part One, on known numbers, which is divided into three chapters. The first is on working with whole numbers, the second on working with fractions, and the third on working with roots.

link to parallel text 63.1 Chapter One, on whole numbers. For our purpose, this chapter is divided into six sections.

link to parallel text 65.1 Section One, on the divisions of numbers and their ranks.

link to parallel text 65.2 A number is a collection of units, and it is divided according to how it is produced into two kinds: whole and fractional. Examples of whole numbers are fifteen, eighteen, and the like. Examples of fractions are a half, three eighths and half an eighth, a ninth and a fourth, six sevenths of seven eighths, and five sixths less a ninth.

link to parallel text 65.6 Whole numbers come in two varieties: even and odd. An even number begins with a two or a four or a six or an eight, or does not begin with units, such as ten or fifty or the like. An odd number begins with a one or a three or a five or a seven or a nine.

link to parallel text 65.10 Even numbers come in three species: evenly-even, evenly-odd, and evenly-evenly-odd. As for the evenly-even number, it is any number that can be halved, and each of its halves can be halved, until the halving reaches one. For example: thirty-two can be halved, and its half is sixteen. And each of the sixteens can be halved, and its half is eight. Half of the eight is four, half of the four is two, and half of the two is one, and the same for similar examples.

65.17 As for the evenly-odd number, it is any number that can first be halved and gives an odd number other than one. For example, fourteen can be halved, and its half is seven, which is an odd number other than one. The same goes for similar numbers. For this reason the [number] two is of the first species. So know it.

66.1 An evenly-evenly-odd number is any number that can be halved in such a way that each of its halves can be halved repeatedly until one reaches an odd number other than one. For example, twenty-eight can be halved, and its half is fourteen. Each of these fourteens can be halved, and its half is seven, which is an odd number, as we saw before. Since it can first be divided into evens, it may look like an evenly-even number, and since it reaches an odd number other than one, it may look like an evenly-odd number. So think it over.

link to parallel text 66.7 Odd numbers come in two species: prime and oddly-odd. As for the odd prime number, it is any number that cannot be counted except by one, like eleven and twenty-nine and the like. They are also called deaf parts or simple, which will become clear in the work on the sieve.[1] As for the oddly-odd number, it is any number that can be counted by odd numbers, like fifteen, which is composed from the product of three by five. The same for similar examples.

link to parallel text 66.13 I say we need to provide here an introduction in which we mention the names of the composite numbers according to their differences, and we will give examples of them. We say that the number, in relation to its composition, is either even or odd. Even numbers come in two varieties, either non-composite simple prime, which is uniquely two, or composite, which are of three kinds.

link to parallel text 66.17 The composition of two equal numbers is called a square, or a number that has a root, and each of the two numbers is called a side or a root. For example, thirty-six is composed from six by six. The whole thirty-six is called a square or a number that has a root. Each of the sixes is called a side or a root.

67.3 The composition of two or more different numbers is called a surface, and each of these numbers is called a side. An example of the composition of two different numbers is eighteen, which is composed from three by six, or two by nine. The whole eighteen is called a surface. Each of the two and the nine, or the three and the six, is called a side. An example of the composition of three numbers is twenty-four, which is composed from three by four by two. So the whole twenty-four is called a surface, and each of the three and the four and the two is called a side. Likewise for more [numbers].

link to parallel text 67.12 And the composition of three equal numbers is called a cube, and each of these numbers is a side and a cube root. For example, sixty-four is composed from the product of four by four by four. The whole sixty-four is called a cube, and each of these fours is called a side or a cube root. Likewise for similar situations. Some call the cube (mukaʿʿab) a cube root (kaʿb), which is a name for its side.

67.17 Odd numbers also come in two varieties: either simple, as we saw before, or composite, which are of three kinds, like the even numbers:

67.19 The composition of two equal numbers is called a square, or a number that has a root, and each of these numbers is a side or a root. For example, twenty-five is composed from five by five.

68.1 The composition of two or more different numbers is called a surface, and each of these numbers is a side. An example of the composition of two different numbers is thirty-five, composed from five by seven. An example of the composition of three different numbers is one hundred five, composed from three by five by seven. Likewise for more [numbers].

68.8 The composition of three equal numbers is called a cube, and each of these numbers is called a side, or cube root. For example, twenty-seven is composed from three by three by three.

link to parallel text 68.11 Subsection. The side and the cube root have come to have the same meaning, just as the root and the side also have the same meaning. They differ only as general and particular [substances] differ formally, but they are equal in usage. Taking a side of the cube[2] is a lengthy work of little benefit, which is why he[3] did not mention it, may God be satisfied with him. However, it can be found by means of decomposition, which is easy. You decompose the cube into the numbers of which it is composed. With them, you piece together three equal numbers by composition such that one of them is the required side. Study this with attention. You shall attain your aim, by God’s will.

link to parallel text 68.18 As numbers increase indefinitely, they are placed in three ranks. They are called ranks because one of them follows another, and the units in each rank are greater than the units that came before them and smaller than the units that follow.[4] They are also called places, because the number resides in them. The places of each number repeat periodically.

link to parallel text 69.2 In each rank are nine numbers. The first rank consists of one to nine, and is called the units rank. Their figures are 1, 2, 3, 4, 5, 6, 7, 8, 9. Someone has written a poem about them:

Alif and ḥā then ḥajja followed by ʿuw

              and after the ʿuw by an ʿayn. Draw

Hā followed by a distinct figure

              looking like an anchor, and also you position

Two zeros for eight with an alif [between them]

              and wāw is the ninth [digit]; so understand it.

69.9 The poet is right. He is also being clever when he states that eight is in the form of two zeros, since this helps us know the [figure for] zero.

69.10 The second [rank] consists of ten to ninety, and is called the tens rank. Their figures are 10, 20, 30, 40, 50, 60, 70, 80, 90. And the third consists of one hundred to nine hundred, and is called the hundreds rank. Their figures are 100, 200, 300, 400, 500, 600, 700, 800, 900.

69.15 The names of numbers are formed from twelve simple names.[5] The first nine are those of the units, the tenth is for the tens, the eleventh is for the hundreds, and the twelfth is for the thousands, which is the units place, in that it is the first of the three ranks, just as the units were first before its tens and its hundreds. And from there the cycle begins again.

link to parallel text 70.2 And of the fourth rank, which is the thousands, we say: units, tens, hundreds, which all are thousands, and together they differ with the first three ranks only by the word “thousands”. Likewise for the three third ranks, which are thousands of thousands: they are also units and tens and hundreds, and differ with the preceding only by the word “thousand” twice. And likewise for the three fourth ranks, which [differ] with the preceding according to the previous description. And likewise for succeeding numbers. So know it.

link to parallel text 70.8 For example,[6] five and twenty and two hundred and four and eighty thousands and a hundred thousands and seven and sixty thousands thousands and three hundred thousands thousands and nine thousands thousands thousands, and its figure is 9367184225. The four and eighty thousands and the hundred thousands are the units, tens, and hundreds like the first three ranks, differing only by the particular word “thousands”. Likewise, the seven and sixty thousands thousands and three hundred thousands thousands are units, tens, and hundreds, and they vary from the preceding in being thousands of thousands. Likewise for the nine thousands thousands thousands. They are units, tens, and hundreds,[7] which differ from the preceding only by the repetition of “thousands” three times. So know it.

70.15 It should be known that each number can be indicated by its index and its name. The index is a term for the rank of the number. The index of the units is one, since they are in the first rank. The index of the tens is two, since they are in the second rank. The index of the hundreds is three, since they are in the third rank, and so on.

70.23 Take, for example, five and twenty and seven hundred and four and eighty thousands. Its figure is 84725. You find the five, which are units as mentioned before, in the first rank, and that is its index and the index of the other units. Likewise, the twenty belongs to the tens, which is in the second rank, and that is its index and the index of the other tens. So if it were, for example, thirty or sixty or eighty, it could have replaced the twenty, since tens have index two and are in the second rank, as mentioned. Likewise, the seven hundred is in the third rank, and that is its index, and the index of other [hundreds]. Likewise the four thousands belongs to the fourth rank, which is its index, and the index of other [thousands]. Likewise the eighty thousands is in the fifth rank, which is its index, and the index of other [tens of thousands]. So understand this. The same holds for greater numbers and for comparable numbers.

71.6 The name is the term for the number that occupies some rank. The name of the one is units, of two is tens, and of three is hundreds. Take for example three and forty and a hundred. Its figure is 143. You find that it has three places. The index of the first is one, and the name of this one is units, since it is the rank of the units rank. The index of the second is two, and the name of this two is tens, since it is the rank of the tens rank. The index of the third is three, and the name of this three is hundreds, since it is the rank of the hundreds rank. So understand it.

link to parallel text 71.12 Subsection on knowing the index of the repeated number.[8]

71.13 You multiply the number of repetitions by three, and you add to the result the index of the species of that number to get the required number. And the repetition is the number of times you say “thousand”.

71.16 For example, suppose someone said to us, “What is the index of ten thousands thousands?” You find the number of repetitions to be two. You multiply it by three, giving six. Likewise if the repetitions are more or less than two: you must always multiply it by three. Then you add to this six the index of the species of ten thousands thousands, which is two, since it was stated before that the units of the thousands thousands are in the units place and the tens are in the tens place, and the hundreds are in the hundreds place. So the sum is eight, which is the index of the given number. So know it. Its representation is made of seven zeros and a one, as in this figure: 10000000.

link to parallel text 72.1 Conversely, if you have a great many places and you want its name, then divide it by three. The division leaves you with three or less. The quotient is the number of repetitions of the number obtained in the remainder.

72.4 For example, suppose someone said to you, “What is the name of a number falling in the tenth place?” You divide the ten, which is the number of places, by three. Our quotient is three, and the remainder is one. Likewise if the number of places is greater or smaller than ten: you must always divide it by three. The resulting three is the number of repetitions for the name of the remainder one. And the name of the remainder one is units. The tens rankis the index of units of thousands thousands of thousands, and its figure is 1000000000.

72.10 If the remainder from the division were two, it would be tens of thousands of thousands of a thousand, and if it were three, it would be hundreds of thousands of thousands of a thousand.[9] So know it.

link to parallel text 73.1 Section two, on addition.

link to parallel text 73.2 Addition is the joining of numbers one to the other in order to express them with one expression. This is divided into five types. One of them is addition with no known relation, and the second is addition with a known disparity[10], which are divided into two kinds.

link to parallel text 73.7 [The first is] a disparity in quality, which occurs when the numbers are in geometric progression. The disparity of the numbers consists of different numbers which are equal in quality in the sense that the ratio of one number to the next is equal to the ratio of a half or a third or something else.[11] This kind comes in two types, since the ratio can either be the ratio of a half, which is what he means when he says “addition with a disparity like that of the squares of the chessboard and similar [problems]”,[12] where the one in the first square is half of the two in the second square, and the two is half of the four in the third square, and so on to the end; or the ratio can be a third or a fourth or a seventh or any other ratio, which is what he means when he says “if the numbers have another disparity”.[13] Examples of these will be presented later, almighty God willing.

73.17 The second kind is a disparity in quantity, in which the numbers are in arithmetic progression, such as consecutive numbers with a disparity of one, or consecutive odd numbers with a disparity of two, and the like. So the numbers differ by equal numbers, and it is a difference in quality when considered as the ratio of one number to the next.[14] This is what he means when he says “If the disparity of the numbers is a known number other than doubling”.[15] He, may God be satisfied with him, said “known disparity” and not “known relation” because whenever the word “relation” is uttered, it suggests the well-known geometric [progression].[16]

74.1 The third [type] is the addition of consecutive numbers, their squares, and their cubes, the fourth [type] is the addition of consecutive odd numbers, their squares, and their cubes, and the fifth [type] is the addition of consecutive even numbers, their squares, and their cubes.[17]

74.4 Note: Adding consecutive numbers, consecutive odd numbers, and consecutive even numbers are truly [conducted in] arithmetic progression. The specific aim of these last three types is to find sums of [consecutive] squares and cubes. He included in these [types] the sums of [consecutive] numbers, even though they were of the preceding type, because they are a foundation for adding the squares and the cubes, and because working with a particular case makes the approach to the general one easier.[18]

74.9 For addition with no known relation, the aim is for you to add a number made up of several digits to a similar number. You should put one of the addends on a line, and below it you put the other addend, with each place below its counterpart. Then you add each digit of one of the addends to its counterpart in the other. If there is no counterpart, then the answer is the addend, as if it had a counterpart. So the sum is the answer.[19] One can start adding from the first digits or the last, but choosing the first is most orderly.[20]

link to parallel text 74.17 Here is an example of adding from the first digit. Suppose we want to add four thousand forty-three to two thousand six hundred eighty-five. We write the addend on a line and the augend below it on another line parallel to it, as mentioned, as in this figure: \begin{matrix}4043\\[-.5mm]2685\end{matrix}.

74.20 We add the five, which is in the first [place] of the lower line, to its counterpart in the upper line, which is the three, giving eight. We put it above them, since they are units of their species. Then we add the eight that is in the tens rank of the lower line to its counterpart in the upper line, which is the four, giving twelve. We put the two above them, since they are also the same species, and we add the ten, in the form of a one, to the six in the hundreds rank of the lower line, giving seven, since for every rank, the units which are after it are the tens before it.[21] Nothing corresponds to the seven in the upper line, so it is considered to be the sum of that rank and that of its counterpart as if it had something. We put the seven above the zero. Then we add the two in the thousands rank of the lower line to its counterpart in the upper line, which is the four, giving six. We put it above them, and this completes the work. The sum is six thousand seven hundred twenty-eight, and this is the figure for that: 6728.

link to parallel text 75.9 Here is another example, adding from the last rank. Suppose we want to add nine hundred seventy-eight to four hundred fifty-six. We put them down on two lines, as mentioned, as in this figure: \begin{matrix}978\\[-.5mm]456\end{matrix}.

75.11 We add the four in the hundreds rank of the lower line to its counterpart in the upper line, which is the nine, giving thirteen. We put the three above them, and the ten, in the form of a one, after the three. Then we add the five in the tens rank of the lower line to its counterpart in the upper line, and that is seven, giving twelve. We put the two above them and we add the ten, in the form of a one, to the three that is above the addends, giving four. We replace it with it. Then we add the six in the units rank of the lower line to its counterpart in the upper line, which is the eight, giving fourteen. We put the four above them, and we also add the ten, in the form of a one, to the two that is above the addends, giving three. We replace it with it, and this completes the work. The sum is one thousand four hundred thirty-four, and the figure for that is 1434.

75.20 The most one can gain by addition is one place.[22] For example, suppose we want to add nine to nine. These are the maximum numbers that can occur in those two places on the two lines. So we say nine [added] to nine gives eighteen, and its figure is 18. The sum gains one place.

76.4 To check the addition, you subtract one of the two lines from the answer. This leaves the other line. For example, if we want to check this problem, we subtract the nine, which is one of the addends, from eighteen, which is the answer. This leaves the other nine. So understand.

link to parallel text 76.7 For addition with a disparity like that of the chessboard squares and similar [problems], a one [is placed] in the first square, then one proceeds by doubling from the first [square] to another assigned [square]. You add one to the one that is in the first square to get what is in the second square. Then you multiply that by itself, so the outcome is what is in the second square and what is before it, with an added one. Then you also multiply that by itself, so the outcome is what is in the fourth square and what is before it, with an added one.[23] Continuing, you also multiply the result by itself, and you double the squares,[24] until you reach the assigned [square], and you drop the one from the sum. The remainder is the required number.

76.15 For example, suppose we want to add what is in sixteen squares arranged as described. We add one to the one in the first square, giving two. We multiply it by itself, giving four. This is what is in the second square and what is before it, with an added one. It is also what is in the third square alone. Because the places of the squares are like the places of the number, no doubt that if we multiply a place by another place, the index of the result is always the index of the two multiplicands[25] less one, according to what we will show in [the section on] multiplication.[26] So when we multiply what is in the second square by itself, it most certainly results in what is in the third. So know it.

76.22 Then we likewise multiply the four by itself, giving sixteen, which is what is in the fourth square and what is before it, with an added one. It is also what is in the fifth square alone. Then we multiply the sixteen by itself, giving two hundred fifty-six, whose figure is 256. This is what is in the eighth square and what is before it, with an added one, and it is also what is in the ninth square alone. Then we likewise multiply the two hundred fifty-six by itself, giving sixty-five thousand five hundred thirty-six, whose figure is 65536. This is what is in the sixteenth square and all of what is before it, with an added one, and it is also all of what is in the seventeenth square alone. So it is clear that each number in a square exceeds all of what is before it by one. So you drop the added one, leaving sixty-five thousand five hundred thirty-five, which is the required number. Do it the same way if the required [number of squares] is greater or smaller. So know it. This is the figure for the table:

link to parallel text 77.9 If the situation is different, then multiply the remainder by the first square to get the required number.[27] A different situation is when the first square is something other than one.

77.11 For example, suppose we want to add what is in eight squares, and a four is in the first square. We suppose that a one is in the first square, and we find the sum as before, and we drop the one. This is what he meant when he said “the remainder”. It yields two hundred fifty-five, and its figure is 255. We multiply it by the four that is in the first square, which gives one thousand twenty. Its figure is 1020, and it is the required number.

link to parallel text 78.1 If the numbers have another disparity, then multiply the smallest by how much the greatest exceeds it, and divide [the result] by the difference between the smallest and the number that follows it. Then add the result to the greatest. This gives the required number.[28]

78.4 For example, suppose we want to add five numbers with a ratio of, say, two thirds, like sixteen, twenty-four, thirty-six, fifty-four, and eighty-one. We put them on a line and we put dots between them, as in this figure: 81 \therefore 54 \therefore 36 \therefore 24 \therefore 16. We multiply the smallest, which is the sixteen, by how much the eighty-one exceeds it, since it is the greatest of the numbers, and that is sixty-five. The result is one thousand forty, whose figure is 1040. Then we divide it by the difference between the sixteen and the number after it, which is the twenty-four, and that is eight. The result of the division is one hundred thirty, whose figure is 130. We add it to the greatest to get two hundred eleven, whose figure is 211, and it is the required number.

78.13 This procedure is common to all numbers that are in a geometric progression, by generalization with the ratio of a half or anything else. The preceding [solution] is particular. And the particular is a basis for the technique as much as the general is a basis, since working it out in general instead of the particular becomes unbeneficial due to its prolixity. And from this general procedure it is clear to you that any number in the squares of the chessboard exceeds the sum of what is before it by one.[29] So know it.

link to parallel text 79.1 If the disparity of the numbers is through a known number other than doubling, then multiply the disparity by the number of numbers less one. Adding the first number to the result gives the last of the numbers. Add it to the first, and multiply it by half of the number of numbers. It yields the required answer.[30]

79.4 For example, suppose we want to add six numbers, the first being ten, which is the smaller extreme, with a disparity of three down to the last. There are two unknowns in this problem: the greater extreme, which is the last, and the sum. We multiply the three, which is the disparity, by the five, the number of numbers less one, giving fifteen. We add to it the ten, which is the first number, giving twenty-five, which is the last of the numbers. We also add it to the first, yielding thirty-five. We then multiply it by three, which is half of the number of numbers, giving one hundred five. Its figure is 105, and it is the total of these numbers, I mean their sum. In this way the sum is clear to you, and it is necessarily the way to do it. So know it.

link to parallel text 79.13 For the addition of consecutive numbers, you multiply half of the upper extreme by the upper extreme and one.[31] For example, suppose we want to add from one to ten consecutively. We add one to the ten, the upper extreme, and that is eleven. We multiply it by half of the ten, giving fifty-five, which is the sum, and its figure is 55.

79.18 And squaring them is given by multiplying two-thirds of the upper extreme increased by a third of one, by the sum. For example, suppose we want to add from a square of one to a square of ten consecutively. We take two thirds of the ten, the upper extreme, giving six and two-thirds. We always add a third of one to it, giving seven. We multiply it by the sum, and that is fifty-five. The result is three hundred eighty-five, which is the required number, and its figure is 385.

80.1 And cubing them is given by squaring the sum. For example, suppose we want to add from a cube of one to a cube of ten consecutively. We multiply the sum, which is fifty-five, by itself. The result is three thousand twenty-five, which is the required number, and its figure is 3025.

link to parallel text 80.5 For the addition of consecutive odd numbers you square half of the established upper extreme with the one. For example, suppose we want to add the odd numbers from one to nine consecutively. We add one to the nine, the upper extreme, giving ten. We then multiply half of the ten by itself, giving twenty-five, which is the required number, and its figure is 25.

80.10 And squaring them is given by multiplying a sixth of the upper extreme by the surface of the two numbers that come after it. For example, suppose we want to add the odd numbers from a square of one to a square of nine consecutively. We multiply a sixth of the nine, the upper extreme, which is one and a half, by the surface of the ten by the eleven, and that is one hundred ten, since they are the two numbers which are after the nine. The result is one hundred sixty-five, which is the required number, and its figure is 165.

80.15 And cubing them is given by multiplying the sum by its double less one. For example, suppose we want to add the odd numbers from a cube of one to a cube of nine consecutively. We multiply the sum, which is twenty-five, by its double less one, which is forty-nine. The result is one thousand two hundred twenty-five, which is the required number, and its figure is 1225.

link to parallel text 80.20 For the addition of consecutive even numbers, you always add two to the upper extreme, and you multiply half of the sum by half of the upper extreme. For example, suppose we want to add the even numbers from two to ten consecutively. We add two to the ten, the upper extreme, giving twelve. We multiply its half by half of the ten, giving thirty, which is the sum, and its figure is 30.

81.4 And squaring them is given by multiplying two-thirds of the upper extreme and two thirds of one by the sum. For example, suppose we want to add the even numbers from a square of two to a square of ten consecutively. We take two-thirds of the ten, the upper extreme, giving six and two-thirds, and we always add to it two thirds of one, giving seven and a third. We multiply it by the sum, which is thirty, so the result is two hundred twenty, which is the required number, and its figure is 220.

81.9 And if you wish, multiply a sixth of the upper extreme by the surface of the two numbers that come after it. For example, suppose we want to add from a square of two to a square of twelve. We take a sixth of the twelve, the upper extreme, giving two. We multiply it by the surface of thirteen by fourteen, which is one hundred eighty-two, since they are the next two numbers after the twelve. The result is three hundred sixty-four, which is the required number, and its figure is 364.

81.15 And cubing them is given by multiplying the sum by its double. For example, suppose we want to add the even numbers from a cube of two to a cube of ten consecutively. We multiply the sum, which is thirty, by its double, which is sixty. The result is one thousand eight hundred, which is the required number, and its figure is 1800.

82.1 For the preceding three types, whenever it begins with something other than one, you add from one to the upper extreme, then from one to the number before the beginning number. Then you drop the smaller from the greater. For the even numbers the two takes the place of the one.[32] By understanding all of this and managing it you shall succeed, almighty God willing.

link to parallel text 83.1 Section Three, on subtraction.

link to parallel text 83.2 Subtraction is the search for the remainder after the dropping of one of two numbers from the other. It comes in two types. [One] type is subtracting the smaller from the greater one time. And [another] type is subtracting the smaller from the greater more than one time, until the greater vanishes or it leaves a remainder less than the smaller. This type is called testing by casting out.

83.6 For the first type you should write the minuend on a line, and below it the subtrahend, arranged as in addition. You subtract each digit from its corresponding digit if you find it has a counterpart. If you do not find a counterpart, or if it is smaller than the subtrahend, then subtract the minuend from the subtrahend, and subtract the remainder from the next digit, and you then put the remainder in the place which agrees with its given rank.

83.11 If you wish, you can always add ten to the counterpart and subtract [the digit in the subtrahend] from the sum, and add one to the next digit of the subtrahend. Then continue in the same way until you have finished all of the subtrahend and minuend. You can start a subtraction from the first of the ranks or from the last. It is preferred to start from the last, contrary to what is preferred in addition.

83.16 Here is an example of subtraction beginning from the last rank. Suppose we want to subtract four thousand nine hundred sixty-eight from five thousand thirty-five. We put them down on two parallel lines, like we mentioned for addition, as in this figure: \begin{matrix}5035\\[-.5mm]4968\end{matrix}.

83.19 We subtract the four which is in the thousands rank from its counterpart in the minuend, which is five. The remainder is one, so we put it above it. Then we likewise subtract the nine that is in the hundreds rank from its counterpart. There is nothing but a zero there, and the zero is nothing. So we subtract this nothing of the minuend from the nine of the subtrahend, leaving nine. We subtract it from the one that is above the five, since it is ten with respect to the zero, as mentioned. This leaves one. We put it above the zero, and nothing is above the five. Then we likewise subtract the six that is in the tens rank from its counterpart, which is three. [This is] smaller than the six of the subtrahend, so we subtract the three of the minuend from the six of the subtrahend, leaving three. We subtract it from the ten that is above the zero, leaving seven. We put it above the three, and nothing above the zero. Then we likewise subtract the eight that is in the units rank from its counterpart, which is five. [This is] smaller than the eight, so we subtract the five of the minuend from the eight of the subtrahend, leaving three. We subtract it from the seventy that is above the three. The remainder is sixty-seven. We put the seven above the five, and the sixty, in the form of a six, above the three. It is the remainder, and the figure for that is: \begin{matrix}{0067}\\[-.5mm]5035\\[-.5mm]4968\end{matrix}.

link to parallel text 84.13 Here is another example of subtraction from the first [rank]. Suppose we want to subtract three thousand four hundred sixty-nine from six thousand five hundred forty-three. We put them down in two parallel lines, as mentioned, as in this figure: \begin{matrix}6543\\[-.5mm]3469\end{matrix}. We subtract the nine that is in the units rank from its counterpart in the minuend, which is three. [This is] smaller than the nine, so we add ten to the three of the minuend, giving thirteen. We subtract the nine from it. The remainder is four. We put it above the three, then we add one to the six that is in the tens rank of the subtrahend, giving seven. We likewise subtract it from its counterpart, which is four. [This is] smaller than the seven, so we add ten to the four, giving fourteen. We subtract the seven from it, leaving seven. We put it above the four. Then we add one to the four that is in the hundreds rank, giving five. We likewise subtract it from its counterpart, which is five, and it vanishes. So we put a zero above the five. Then we likewise subtract the three that is in the thousands rank – with nothing added to it, since nothing remains from the previous position – from its counterpart, which is the six. Three remains, and we put it above the six. This completes the work. The remainder is three thousand seventy-four, and the figure for that is 3074.[33]

85.8 The most one can reduce by subtraction is one place. For example, suppose someone said to you, “Subtract one from ten”. The remainder is nine. The minuend is reduced by a place. So understand it.

85.11 To check the subtraction, you add the remainder to the subtrahend. The result is the minuend. Or you subtract the remainder from the minuend, leaving the subtrahend. For example, in the above-mentioned problem the remainder is nine. We add it to the one that is the subtrahend, giving ten. This is the minuend. And if we subtract the remainder, which is the nine, from the minuend, which is the ten, there remains one, which is equal to the subtrahend.

link to parallel text 85.16 In the first way of checking[34] a subtraction by addition you should look for a number which, if you add it to the subtrahend, then [the sum] is equal to the minuend. One begins this from the first of the ranks, just like those who work with rūmī signs do.

link to parallel text 86.1 Whenever you are faced with subtracting a number from a number, and then the remainder from another number, and so on, then you work it out in the steps as indicated in this section.

86.3 If we wish, we can piece the problem together with addition and subtraction. We collect the even terms from among the subtrahends, namely the second, the fourth, the sixth, and so on, and we add them together with the minuend. We likewise add the odd terms from among the subtrahends, namely the first, the third, the fifth, and so on, and we drop their sum from the first sum. The numbering with the subtrahends truly begins after the minuend [and continues] to the last.[35]

86.9 For example, subtract two from five, and the remainder from seven, and the remainder from eight, and the remainder from ten. Sometimes this is expressed with exclusions, so one says: ten less eight less seven less five less two. We put it down in a line like this: 2~\scalebox {-1}[1]{\ell }~5~\scalebox {-1}[1]{\ell }~7~\scalebox {-1}[1]{\ell }~8~\scalebox {-1}[1]{\ell }~10. We subtract the two from the five, and the remainder from the seven, and the remainder from the eight, and the remainder from the ten, leaving six, which is the required number.

link to parallel text 86.15 If we wish, we can add the even subtrahends, which are the seven and the two, with the minuend ten, to get nineteen. Then we add the odd subtrahends, which are the eight and the five, giving thirteen. We subtract it from the nineteen, leaving six.

link to parallel text 86.18 And if we wish, we can consider three successive terms. We subtract the middle from the sum of the extremes, leaving the remainder as one number. We insert it in [their] place in the subtrahend. Then we likewise consider it and the two remaining numbers. Take for example the ten and the eight and the seven. We drop the eight from seventeen, the sum of the extremes, leaving nine which, with the two and the five, are three successive numbers. We drop the five from the sum of the extremes, leaving six.

87.3 Or we first consider the seven and the five and the two. We drop the five from the nine, the sum of the extremes, leaving four which, with the eight and the ten, are three successive numbers. We drop the middle number, eight, from the sum of the extremes, leaving six.

87.6 Or we consider the eight and the seven and the five first. We drop the middle seven from the sum of the extremes, leaving six. This is now the middle term between the ten and the two. We drop it from their sum, leaving six.

link to parallel text 87.9 And if we wish, we can subtract the eight from the ten, and add the remainder to the seven, and subtract the five from it, and add the remainder to the two, yielding six.

link to parallel text 87.11 The cause of this is that when subtracting the deleted from the appended, it is deleted, but when subtracting the deleted from another deleted, it is appended. Thus the deleted second and fourth and sixth of the even terms are always appended, since each of them is deleted from a deleted. And the odd terms are always deleted, since they are deleted from an appended.[36] So know it.

link to parallel text 87.15 The second type [of subtraction] consists of three [kinds of] subtractions. These are frequently used in checking work. One is casting out nines, the second is casting out eights, and the third is casting out sevens.

87.17 Casting out nines leaves one from each power of ten. You collect the number from its digits as if they were units, and then you cast out nines from them.

87.19 For example, suppose we want to cast out [nines from] six thousand four hundred thirty-five. We put the number on a line like this: 6435. We add the five to the three, giving eight. We add it to the four, giving twelve. We subtract nine from it, leaving three. We add it with the six, giving nine. This is cast out entirely, and it is the answer.

88.1 Casting out eights leaves two from each ten, four from each hundred, and a pair of hundreds and what is above it are cast out entirely. So four is the remainder from odd hundreds, and you multiply the tens by two, since the remainder from each ten is two. You add that with the four and with the units, and you cast out eights from it.

88.5 For example, suppose we want to cast out [eights from] five thousand three hundred ninety-three. We put down the number on a line like this: 5393. Five thousand is cast out entirely, and the remainder from the three hundred is four. Keep it in mind. Then we multiply the ninety[37] by two, giving eighteen. We add it to the remembered four with the three units, to get twenty-five. Casting out eights leaves one, which is the answer.

link to parallel text 88.10 In casting out sevens, three remains from each ten, two from each hundred, six from each thousand, four from each ten thousand, five from each hundred thousand, one from each million, and from there the cycle begins again. You can come to know it with these letters: A J B W D H, repeated below the digits.

88.14 The A is one, the J is three, the B is two, the W is six, the D is four, and the H is five. Someone arranged these elements into verse. He said, “Three and two and six and four * and five and one; that is casting out sevens”.

link to parallel text 88.17 If you wish, you can write down the letters as mentioned, or if you wish you can make them dust numerals. You multiply each digit by what is below it in number – sound out the letters – and you cast out sevens. You leave its residue above it. Then you add all the digits of the remainders as if they were units and you cast out sevens.

link to parallel text 89.4 For example, suppose we want to cast out [sevens from] twenty-three million seven hundred eighty-six thousand four hundred thirty-five. We write it down on a line and we draw a line above it. We write the letters below it, each letter below a number consecutively. If we run out of letters and we have not run out of digits, we repeat the letters for the remaining [digits]. This is what he meant when he said “and from there the cycle begins again”. That is, after the H, and for the rest of the number, the letters are repeated along with the quantity with it. The figure always looks like this:

89.10 We multiply the five by the number of the A below it, giving five. We put it over the line above the five. Then we likewise multiply the three by the number of the J below it, giving nine. We cast out sevens, leaving two, and we put it over the line above the three. Then we multiply the four by the number of the B, giving eight, leaving one. We put it above the four. Then we multiply the six by the number of the W, giving thirty-six, leaving one. We put it above the six. Then we multiply the eight by the number of the D, giving thirty-two, leaving four. We put it above the eight. Then we multiply the seven by the H, giving thirty-five, which vanishes. So we put a zero above the seven. Then we likewise multiply the three by the repeated A, giving three. We do not cast out, so we put it above the three. Then we multiply the two by the J, giving six. We do not cast out here, either. We put it above the two. Its figure is:

90.1 Now we work with the remainders five, two, one, one, four, three, and six. We add them as if they were units, and we cast out sevens. The remainder is the answer, and that is one.

link to parallel text 90.3 If you wish, multiply what is in the last place by three. You cast out sevens, and you add the remainder to what is before it. If there is no number in the place before it, then you multiply the accumulated residue by three and you cast out sevens. Keep doing this until you reach the units [place].

90.7 For example, suppose we want to cast out [sevens from] fifty-eight thousand sixty-four. We put it down on a line like this: 58064. Then we multiply what is in the last place, which is five, by three, giving fifteen. We cast out sevens, leaving one. We add it to the eight that is before it, giving nine. We multiply it by three, giving twenty-seven, whose remainder is six. There is nothing in the preceding place, so this six is considered to be the sum of the residue and the rank as if it had something. We multiply it by three, giving eighteen, and its remainder is four. We likewise add it to the six that is before it, giving ten. We multiply it by three, giving thirty, and its remainder is two. We add it to the four in the units place, giving six. This completes the work, and the six is the answer. So know it.

link to parallel text 90.15 If you wish, make the last digit tens and add what is before it as if it were units, and then cast out sevens. Then you make the remainder tens and you add it to what is before it as if it were units, and you cast out again.

90.18 For example, in our preceding example we make the last five tens, and we add to it the eight that is before it as if it were units, giving fifty-eight. We cast out sevens, and the remainder is two. Then we make it tens and we add to it the zero before it. Since there is no number there, it gives twenty, whose remainder is six. We make it tens, and we add to it the six before it, giving sixty-six. Its remainder is three. We make it tens and we add to it the four that is before it, giving thirty-four. Its remainder is six, which is the answer.

link to parallel text 91.1 Subsection on the way to test [calculations] by casting-out.

91.2 For addition, you cast out each line, add their remainders, and cast it out. The remainder is the answer. Then you cast out the sum in the problem, which will agree with the answer.

link to parallel text 91.4 For example, we added forty-three, whose figure is 43, to sixty-four, whose figure is 64. The sum is one hundred seven, whose figure is 107. We can check this problem and other problems by casting out nines or eights or sevens. However, many people cast out sevens. We want to check this problem likewise, as well as the other examples. We add the one, the remainder of the addend, to the one, the remainder of the augend, giving two, which is the answer. If the sum of the two remainders can be cast out, we cast it out also, and its remainder is the answer. Then we cast out the sum, leaving two, which agrees with the answer. So know it.

91.12 For subtraction, you cast out the minuend and you keep the remainder in mind. Then you cast out the subtrahend and you drop its residue from the remembered number. If it is smaller, add the modulus to it and drop it from the sum. The remainder is the answer. Then cast out the remainder in the problem. This should agree with the answer. Or, you add the residue of the subtrahend to the residue of the remainder. This should agree with the residue of the minuend.

91.16 For example, we subtracted seventy-four, whose figure is 74, from ninety-six, whose figure is 96. The remainder is twenty-two, whose figure is 22. If we want to check it, we keep in mind the residual five of the minuend. Then we cast out the subtrahend, whose remainder is four. We drop it from the remembered five, leaving one, which is the answer.

link to parallel text 92.3 And if this residual four of the subtrahend were greater than the residual five of the minuend, then we add [the modulus] to the five, the number of the minuend: nine if it is nine, eight if it is eight, or seven if it is seven. Then we cast out the remainder [in the problem], leaving one, which agrees with the answer. And if we add the four, the residue of the subtrahend, to the one, the residue of the remainder, it is the same as the five, the residue of the minuend. So know it.

92.8 For multiplication, you cast out the two multiplicands and you multiply the remainder of one of them by the remainder of the other, and you cast it out. The remainder is the answer. Then you cast out the result of the multiplication, which should agree with the answer.

92.10 For example, we multiply twelve, whose figure is 12, by sixteen, whose figure is 16. The result of the multiplication is one hundred ninety-two, whose figure is 192. We multiply the residual five of the multiplicand by the residual two of the multiplier, giving ten. Its remainder is three, which is the answer. Then we cast out the result of the multiplication, leaving three, which is equal to the answer.

92.15 This generalizes to whole numbers and fractions after numerating them, that is, when each of them in the problem becomes of a type [that is] one fraction.

link to parallel text 92.17 For example, suppose someone said, “Multiply a third by fourteen and a fourth”. The result of the multiplication is four and three fourths, according to what will come in the work on fractions,[38] almighty God willing. If we want to check it [by casting out sevens], we multiply the third, which is the remainder of one of the two multiplicands, by the fourth, the remainder of the other multiplicand, resulting in a third of a fourth. Its remainder after its numeration is one, which is a third of a fourth, which is the answer. Then we numerate the result of the multiplication, which gives nineteen fourths. Its remainder is five fourths. We numerate it by a third by multiplying it by three, giving fifteen. Its remainder is one, which is a third of a fourth, equal to the answer in quantity and quality.[39]

link to parallel text 93.6 For division and denomination, you cast out the result and the divisor or denominating number. You multiply the remainder of one of them by the remainder of the other, and keep it in mind. The remainder is the answer. Then you cast out the dividend or denominated number, which should agree with the answer. This method also generalizes to whole numbers and to fractions, after numerating them.

link to parallel text 93.10 Here is an example of division. We divide one thousand four hundred eighty-eight, whose figure is 1488, by twelve. The result of the division is one hundred twenty-four, whose figure is 124. If we want to check it, we multiply the residual five of the result by the residual five of the divisor, giving twenty-five. Its remainder is four, which is the answer. Then we cast out the dividend, whose remainder is four, which agrees with the answer.

link to parallel text 93.15 Here is an example with fractions. Suppose someone said to you, “Divide five sixths and three fourths by a half”. The result of the division is three and a sixth, and its remainder, after numerating it, is five sixths. We multiply it by one, the numerator of the half, which is the divisor, giving five halves of a sixth. We then multiply it by two so it becomes fourths of sixths, agreeing in the numeration with the numerated dividend. It gives ten. Its remainder is three fourths of a sixth, which is the answer. Then we cast out the numerated dividend, which is thirty-eight fourths of a sixth. Its remainder is three fourths of a sixth, which is equal to the answer.[40]

link to parallel text 94.1 Here is an example of denomination. Suppose someone said, “Denominate eleven with fifteen”. The result of the denomination is three fifths and two thirds of a fifth. The remainder after numerating it is four. We multiply it by the one, the remainder of the denominating number, giving four, which is the answer. Then we cast out the denominated number. Its remainder is four, equal to the answer.

link to parallel text 94.5 Here is an example with fractions. Suppose someone said, “Denominate two sixths and two thirds of a sixth with five eighths and a third of an eighth”. It results in two-thirds, and the remainder of its numerator is two. We multiply it by the two, the remainder from numerating the denominating number, giving four thirds of a third of an eighth. Then [we multiply it] by the six, giving twenty-four thirds of a third of a sixth of an eighth. Its remainder is three, which is the answer. Then we cast out the numeration of the numerator. Its remainder is one. We multiply it by the three, then by the eight. It gives twenty-four thirds of a third of a sixth of an eighth. Its remainder is three, which equals the answer.

94.11 It is necessary that everything in the problem be reduced to a finer fraction, which is the part named with all of the denominators. The meaning of numeration shall be clarified later, with the help of almighty God, may he be exalted.[41]

link to parallel text 95.1 Section Four, on multiplication and understanding its subtleties.

link to parallel text 95.2 Multiplication consists of the duplication of one of two numbers by however many units are in the other.

link to parallel text 95.3 This section covers two types. [In one] type, in putting down the multiplier, each one of them is equal to the one of the multiplicand. Here the duplication clearly occurs in both the term and the meaning.

95.6 In the second type, all of what is in the multiplier in units is equal to the one of the multiplicand. So the [number of] units in the multiplier is the number of what is in one of the multiplicand in parts. This type is called conversion, and the duplication occurs in the term but not in the meaning.

95.10 Suppose someone said, “Three men: each of them has five dirhams”. You multiply five by three, which gives fifteen dirhams. This is duplication in the term and in the meaning.

95.12 Suppose someone said, “Five dirhams: how many thirds does it contain?” You multiply five by three, which gives fifteen thirds. Here the duplication is only in the term, and as for the meaning, fifteen thirds are exactly five.[42]

link to parallel text 95.15 [Multiplication] is divided into three kinds; that is, with regard to procedure. The first kind is by shifting, the second by half-shifting, and the third without any shifting.

link to parallel text 95.17 The first kind, which is multiplication by shifting, calls for erasing, and is called sleeper [multiplication]. You write the multiplicand and the multiplier on two lines so that the first digit of the multiplier is below the last digit of the multiplicand. Then you multiply it by all the digits of the multiplier. You begin by writing the result in its place, traversing the line, continuing along the line of the multiplicand. Then you shift the number of the multiplier so that it is below the digit that follows those before it. Then you multiply it by all the digits of the lower number, like the first time. Whenever you multiply by a number, you add the result with what is over the head of that number from the previous result, and you put it where it belongs. This procedure is universal for all problems of multiplication.

link to parallel text 96.3 For example, if we want to multiply forty-three by fifty-four, we write down the forty-three, the multiplicand, on a line, and the fifty-four, the multiplier, on another line, so that the first digit of the multiplier is below the last digit of the multiplicand, as mentioned. Here is the figure:

We multiply the last digit[43] of the multiplicand, which is the four, by the five, the last digit of the multiplier, giving twenty. We put a zero above the five, and the twenty, in the form of a two, after the zero. We also multiply it by the four below it, giving sixteen. We put the six in its place, and the ten, in the form of a one, in place of the zero. These are included with the line of the multiplicand. Then we shift the multiplier back one place, so the four is below the three, and the five is below the six. This is the figure:

96.13 Then we likewise multiply the shifted three below it[44] by the entire multiplier, and we add that to what is above it. We first multiply it by the five, giving fifteen. We add it to the sixteen above it, giving thirty-one. We put the one in place of the six, and the thirty, in the form of a three, in place of the ten. Then we likewise multiply it by the four, giving twelve. We put the two in its place, and we add the ten, in the form of a one, to the one that is in the second rank, giving two. We replace it with it, and this completes the work. So the result is two thousand three hundred twenty-two, and its figure is 2322.

link to parallel text 97.1 The other type is known as vertical [multiplication]. You set up the two multiplicands in two vertical lines so that the first digit of the multiplier is opposite the last digit of the multiplicand. You proceed in multiplying them just as you did in the sleeper method, by shifting and erasing.

97.4 For example, suppose we want to multiply forty-two by thirty-seven. We put down the multiplicand in a column, as mentioned, and the multiplier as well, so that its first digit is next to the last digit of the multiplicand, and this is the figure:

97.7 Then we multiply the last digit of the multiplicand, which is four, by all the digits of the multiplier, as before. We first multiply it by the three, giving twelve. We put the two adjacent to the three in the column of the multiplicand, as before, and the ten, in the form of a one, below the two. Then we likewise multiply it by the seven next to it, giving twenty-eight. We put the eight in its place, and we add the twenty, in the form of a two, to the two that is below that rank, giving four. We replace it with it. Then we shift the multiplier back one place, so the seven is next to the two, and the three is next to the eight, as in this figure:

97.14 Next we likewise multiply the two by the whole multiplier, as before. We first multiply it by the three, giving six. We add it to the eight next to it, giving fourteen. We put the four in place of the eight, and we add the ten, in the form of a one, to the four in the rank below it, giving five. We replace it with it. Then we likewise multiply the two by the seven, giving fourteen. We put the four in its place, and we add the ten, in the form of a one, to the four that is in the rank below it, giving five. We replace it with it. This completes the work. The result is one thousand five hundred fifty-four, whose figure is 1554.

link to parallel text 98.4 The second kind is multiplication by half-shifting, which only works for two equal numbers. In this scheme you write down one of the equal numbers on a line, and you put marks in the form of dots between its digits. Then you multiply the last digit by itself and you put the result above it. Then you double it and you shift it by writing it down in place of the dot preceding it. Then you multiply the preceding digit by the shifted number and by itself, and you write the results of each multiplication above them. Then you double the digit that you just multiplied, as you did before. Then you shift it in place of the dot that precedes it. Then you shift the first doubled number, [adding] to its calculation. Then you multiply the digit preceding the dot that was replaced by all the doubled number, then by itself, as you did before. You continue in the same way, doubling, shifting, and multiplying until you reach the end of the line.[45]

98.15 For example, if we want to multiply four hundred sixty-three by itself, we put it down on a line and we separate the numbers with dots as mentioned, like in this figure: 4~\bullet ~6~\bullet ~3. We multiply the last four by itself, giving sixteen. We put the six above the four, and the ten, in the form of a one, after the six. Then we double it, giving eight, and we put it in place of the dot preceding it. So it yields this figure:

98.20 Then we multiply the preceding six facing it by the shifted eight, giving forty-eight. We put the eight above the eight of the multiplier, and we add the forty, in the form of a four, to the six in the next rank, giving ten. We put a zero in its place and we add the ten, in the form of a one, to the one in the next place, giving two. We replace it with it. Then we likewise multiply the six by itself, giving thirty-six. We put this six above it, and we add the thirty, in the form of a three, to the eight that is in the next rank, giving eleven. We put the one in place of the eight, and the ten, in the form of a one, in place of the zero. We then move the six back, also doubling, giving twelve. We put the two in place of the dot before it, and we add the eight to the ten, which resulted from doubling [the six], in the form of a one, giving nine. We put it down in place of the six, as in this figure:

99.7 Then we likewise multiply the preceding three facing it by the whole shifted number and by itself, as before. We multiply it first by the nine, giving twenty-seven. We add to it the sixteen above it, giving forty-three. We put the three in place of the six which is above it, giving three, and the forty, in the form of a four, in place of the ten. Then we also multiply it by the two, giving six. We put it above it. Then we also multiply it by itself, giving nine. We put it above it. And this completes the work. The result is two hundred fourteen thousand three hundred sixty-nine, and its figure is 214369. So understand it and pursue similar problems the same way. Proceed by the power of almighty God.

link to parallel text 99.14 The third kind is multiplication without shifting, of which there are several types. One is table multiplication, in which you draw a quadrilateral surface, extending length and width according to the ranks in the two numbers to be multiplied. You draw diagonals through its squares from the lower right to the upper left, and you write the multiplicand above the quadrilateral, matching each digit with a column. Then you write the multiplier down the left or right side of the quadrilateral, so that each digit also corresponds to a row. Then you multiply digit after digit of the multiplicand by all the digits of the multiplier, and you put the digits of each rank in the intersecting square. And the meaning of “intersection” is where they meet. You put the units above the diagonal and the tens below it. Then you begin adding from the upper right corner. You add what is between the diagonals, without erasing, and you write each number in its rank, and you add the tens of each sum to the next diagonal. You then put it together, and the sum that you obtain is the result.

100.5 For example, suppose we want to multiply four hundred thirty-five by two hundred eighty-seven. We draw the quadrilateral as he mentioned, with the multiplier on the left or right of the quadrilateral, and the multiplicand above the quadrilateral. Each number is above a column of the quadrilateral, and similarly we write the multiplier on the right of the quadrilateral, as in this figure:

100.9 We multiply the five by the seven, giving thirty-five. We put down the five above the diagonal in the corresponding square, and the thirty, in the form of a three, below it. Then we likewise multiply the five by the eight, giving forty. We put the zero above the diagonal of the corresponding square, and the forty, in the form of a four, below it. Then we likewise multiply the five by the two, giving ten. We put the zero above the corresponding square, and the ten, in the form of a one, below it. This is the figure:

101.1 Then we do the same for the three. We multiply it by all the digits of the multiplier and we put the results of each of them in the corresponding square as before, and we do likewise for the remaining four of the multiplicand. Then we finish the multiplication. The numbers are situated in the quadrilateral in its entirety as in this figure:

101.5 Then we begin adding. We raise the five that is above the first diagonal in the upper right corner, as mentioned. Then we likewise add what is between the first and second diagonals, which are three and one, which add to four. We raise it after the five that we raised first. Then we likewise add what is between the second and third diagonals, which are four and four and two and eight. They add to eighteen. So we raise the eight after the raised four, and we add the ten, in the form of a one, with what is between the third and fourth diagonals, which are one and six and two and two and two. They add to fourteen. So we raise the four after the raised eight, and we add the ten, in the form of a one, with what is between the fourth and fifth diagonals, which are eight and three, giving twelve. We raise the two after the raised four, and the ten, in the form of a one, after it. This completes the work. All of this together is the required number, which is one hundred twenty-four thousand eight hundred forty-five, and its figure is 124845.

link to parallel text 101.16 Another [type] is vertical multiplication, in which you draw two vertical lines with some empty space between them, and you draw the two multiplicands beside them. Then you multiply one of them, digit by digit, by all the digits of the other, and you put the results in the space between the lines, taking into account the ranks of the indexes.

102.1 For example, suppose we want to multiply one hundred eighty-three by three hundred forty-seven. We put down the multiplicand vertically in a line, and the multiplier parallel to the multiplicand, and a line after the multiplicand with another line before the multiplier as mentioned, as in this figure:

102.4 We multiply the three, which is the first [digit] of the multiplicand, by all the digits of the multiplier. We multiply it first by the seven, giving twenty-one. Then we put the one down next to the three beside the line, and the twenty, in the form of a two, below it. Then we multiply it likewise by the four, giving twelve. We add to it the two that is in the second rank of the result, giving fourteen. We put the four down next to the two, and the ten, in the form of a one, below it. Then we multiply it likewise by the three, giving nine, and we add it to the one that is in the third rank of the result, giving ten. We put the zero down next to the one, and the ten, in the form of a one, below it. This is the figure:

102.11 Then we similarly multiply the eight, which is in the second rank of the multiplicand, by all the digits of the multiplier. We multiply it first by the seven, giving fifty-six. We add it to the four that is in the second rank of the result, giving sixty. We put the zero down next to the four, and the sixty, in the form of a six, next to the zero that is after the second [rank]. Then we multiply it also by the four, giving thirty-two. We add it to the six that is in the third rank of the result, giving thirty-eight. We put the eight down next to it, and we add the thirty, in the form of a three, to the one that is after that rank, giving four. We put it down next to it. Then we multiply it also by the three, giving twenty-four. We add it to the four that is in the fourth rank of the result, giving twenty-eight. We put the eight down next to it, and the twenty, in the form of a two, after that digit.

103.3 Then we multiply it also by the one, which is in the third rank of the multiplicand, again by all of the multiplier. We multiply it first by the seven, giving seven. We add it to the eight that is in the third rank of the result, giving fifteen. We put the five next to it, and we add the ten, in the form of a one, to the eight that is after that rank, giving nine. We put it down next to it. Then we multiply it also by the four, giving four. We add it to the nine that is in the fourth rank of the result, giving thirteen. We put the three down next to it, and we add the ten, in the form of a one, to the two that is after that rank, giving three. We put it down next to it. Then we multiply it also by the three, giving three. We add it to the three that is in the fifth rank of the result, giving six. We put it down next to it. This completes the work, and this is the figure:

103.12 The result is sixty-three thousand five hundred one. This is the vertical line adjacent to the multiplier. So know it.

link to parallel text 103.14 [Another type] is sleeper multiplication. You put the two multiplicands in two parallel lines. Then you multiply each digit of one of them by each digit of the other, and you put down the result taking into account the ranks of the indexes. You may start the multiplication from the first digit or the last. This type is also called multiplication by indexes.

link to parallel text 104.1 I say that we need here an introductory remark for this type that may also be useful for rūmī multiplication. The result of multiplying the units is units, since it comes from multiplying a number by a number. The index of the result is the sum of the indexes of the two multiplied numbers less one, as shown by the author, God be satisfied by him, in the section on addition in Lifting the Veil.[46] Certainly the index of the multiplicand is one and the index of the multiplier is one, so their sum is two. Dropping one leaves one, and the name of the one, according to what [was said] before, is units.[47] So the result of multiplying the units by the units is units. Likewise, their product by the tens is also tens, and by the hundreds is hundreds. And multiplying the tens by the tens is hundreds, and by the hundreds is thousands. And the hundreds by the hundreds is tens of thousands. This is the end of the introductory remark.

104.10 Suppose we want to multiply two hundred fifty-three by nine hundred eighty-seven. We put them down in two parallel lines, as mentioned. This is the figure:

104.12 We multiply the three, which is the units digit of the multiplicand, by all the ranks of the multiplier. We multiply it first by the seven, giving twenty-one. We put down the one in the units rank, as before, and the twenty, in the form of a two, after it. Then we multiply it also by the eight, giving twenty-four. We add it to the two that is in the tens rank of the result, since they are of the same species, as [mentioned] before, giving twenty-six. We put the six above the two, and the twenty, in the form of a two, after it. Then we multiply it also by the nine, giving twenty-seven. We add to it the two that is in the hundreds rank of the result, because they are also of the same species, giving twenty-nine. We put the nine above the two, and the twenty, in the form of a two, after it. This is the figure:

105.1 We likewise multiply the five that is in the tens rank of the multiplicand by all the ranks of the multiplier. We multiply it first by the seven, giving thirty-five. We add to it the six that is in the tens rank of the result, since it is also of the same species, giving forty-one. We put the one above the six, and we add the forty, in the form of a four, to the nine that is after that rank, giving thirteen. We put the three above the nine, and we add the ten, in the form of a one, to the two that is after that rank, giving three. We put it down above the two. Then we multiply it also by the eight, giving forty. We add to it the three that is in the hundreds rank of the result, giving forty-three. We put the three in its place[48] and we add the forty, in the form of a four, to the three that is after that rank, giving seven. We put it above the three. Then we multiply it also by the nine, giving forty-five. We add to it the seven that is in the thousands rank of the result, giving fifty-two. We put the two above the seven, and the fifty, in the form of a five, after it. This is the figure:

105.14 Similarly we multiply the two that is in the hundreds rank of the multiplicand by all the ranks of the multiplier. We multiply it first by the seven, giving fourteen. We add it to the three that is in the hundreds rank of the result, since they are also of the same species, giving seventeen. We put the seven above the three, and we add the ten, in the form of a one, to the two that is after that rank, giving three. We put it above the two. Then we multiply it also by the eight, giving sixteen. We add it to the three that is in the thousands rank of the result, giving nineteen. We put the nine above the three, and we add the ten, in the form of a one, to the five that is after that rank, giving six. We put it above the five. Then we multiply it also by the nine, giving eighteen. We add it to the six that is in the ten thousands rank of the result, giving twenty-four. We put the four above the six, and the twenty, in the form of a two, after it. This completes the work, and its figure is:

106.8 The result of the multiplication is two hundred forty-nine thousand seven hundred eleven, which is the required number. And if we wish, we can follow the method by starting from the last places. So know it.

link to parallel text 106.11 Another type [of multiplication] requires that the [numbers of] digits of the multiplicands be equal, and that the digits in each rank of the ranks in each line also be equal. The way to write it down is similar to the way [for the method] by erasing. Then you write a one below the first rank of the ranks of the upper line, and a two below the second [rank], likewise increasing by one until you reach the last rank of the multiplicand. What is below it is shared with the first rank of the multiplier. From the second rank of the multiplier you begin decreasing one by one until you reach the last rank of the multiplier. These written numbers are all on a third line. The indexes of the ranks of the multiplicand are in the correct order, but the indexes of the ranks of the multiplier are reversed. Then you multiply the number of the rank of the multiplicand by the number of the rank of the multiplier. The result is multiplied by what is in the written line, and the result is the required number. This type of multiplication is called “by repetition”.

107.6 For example, suppose someone said, “Multiply four hundred forty-four by three hundred thirty-three”. We put them down on two lines, as mentioned. This gives the figure:

Then we write a one below the first four, a two below the second, and a three below the third. This is the first rank of the multiplier. Then we write a two below the second three, since from this point we begin decreasing one by one, as mentioned, and a one below the third three, as in this figure:

As mentioned, the indexes of the ranks of the multiplicand are in the correct order, and the indexes of the ranks of the multiplier are reversed, since the index of the three in the first [rank] of the multiplier is one, and we wrote it below the third three, which is its last rank.

107.15 Then we multiply the number of a rank of the multiplicand, which in this example is four, by the number of a rank of the multiplier, which in this example is three, giving twelve. We multiply it by the new line. We multiply it first by the one, which is the last in the line, giving twelve. We put the two above it, and the ten, in the form of a one, after it. Then we multiply it also by the two in the fourth [position] of the line, giving twenty-four. We put the four above the two, and we add the twenty, in the form of a two, to the two after that rank, giving four. We put it above it. Then we multiply it also by the three in the third [position] of the line, giving thirty-six. We put the six above the three, and we add the thirty, in the form of a three, to the four that is after that rank, giving seven. We put it above it. Then we multiply it also by the two in the second [position] of the line, giving twenty-four. We put the four above the two, and we add the twenty, in the form of a two, to the six that is after that rank, giving eight. We put it above it. Then we multiply it also by the one in the first [position] of the line, giving twelve. We put the two above it, and we add the ten, in the form of a one, to the four that is after that rank, giving five. We put it above it. That completes the work, and this is the figure:

108.7 The result of the multiplication is one hundred forty-seven thousand eight hundred fifty-two, which is the required number.

link to parallel text 108.9 Another [type] is multiplication by excess, in which you denominate the excess over ten of one of the two multiplicands with the ten. Then you take that ratio of the other [multiplicand]. You add it to it and you make it tens. If the ratio has a fractional part, you take it with respect to the ten[49] and you put it in the units place.

108.13 For example, suppose we want to multiply twelve by fifteen. We denominate the two, the excess over the ten in the multiplicand, with the ten, yielding a fifth. We take a fifth of the fifteen, the multiplier, giving three. We add it to it, giving eighteen, and we make it tens. So it is one hundred eighty. This is the result, which is the required product, and its figure is 180. And if we denominate the five, the excess over the ten in the multiplier, with the ten, it yields a half. We take half of the twelve, giving six. We add it to it, again getting eighteen. We make it tens to get one hundred eighty, which is the result, as before.

109.1 Another example: suppose we want to multiply thirteen by seventeen. We denominate the three, the excess over the ten in the multiplicand, with the ten, yielding three tenths. We take three tenths of seventeen, the multiplier, giving five and a tenth. We add it to it, giving twenty-two and a tenth. We make it tens, and we take the fraction of the ten, which is a tenth, giving one. This we put it in the units place, as mentioned. That yields two hundred twenty-one, which is the required number, and its figure is 221.

link to parallel text 109.7 Another type is known as denomination. Here you add the two multiplicands, then you denominate one of them with the sum. Then you take that ratio of the other [multiplicand] and you multiply it by the sum, so the result is the required number.[50]

109.10 For example, suppose we want to multiply six by twelve. We add them, giving eighteen. Then we denominate one of them with it. We denominate the six with it, giving a third. We take a third of the multiplier, which is the twelve, giving four. We multiply it by the sum. The result is the required number, which is seventy-two, and its figure is 72. And if we denominate the twelve with the sum, we get two-thirds. So we take two thirds of the six, giving four. We multiply it by the sum, and the result is the required number, as before.

link to parallel text 109.16 And recall what the jurist Abū Muḥammad ibn Ḥajjāj, known as Ibn al-Yāsamīn, said about this method: “if we denominate one of them with the sum, then we drop that ratio of the number from itself and we multiply the remainder by the sum, it results in the required number”.

link to parallel text 109.19 Another type is also known as denomination. You denominate the most convenient of the two multiplicands with whatever simple power of ten you wish, or you divide it by it. Then you multiply the result of the denomination or division by the other [multiplicand]. Then you raise each [digit] of the result by the power of ten of the divisor. The outcome is the required number.

110.1 If the result of dividing or denominating gives a whole number only by adding something to it or subtracting it from it, then do it. Then you multiply the added quantity by the number you did not add it to, and you subtract the outcome from the result. If you worked it by subtraction, then add the outcome to the result.[51]

link to parallel text 110.5 And by “simple power of ten” we mean that the first [non-zero] rank is equal to the ten or the hundred or the like.

110.6 Example problem: we want to multiply twenty-four by eight. So we denominate the eight, the multiplier, with whatever simple power of ten we wish. Supposing it is ten, it gives four-fifths. We multiply it by the twenty-four, resulting in nineteen and a fifth. We raise each [digit] by ten, the denominated power of ten, giving one hundred ninety, and a fifth of one [raised by] ten gives two. We add it to it, so the outcome is the required number, and that is one hundred ninety-two, and its figure is 192.

link to parallel text 110.12 And another example: suppose we want to multiply twelve by fifteen. We divide the fifteen by the ten, resulting in one and a half. We multiply it by the twelve. The result of the multiplication is eighteen. We raise each [digit] by the ten, the divided power of ten, yielding one hundred eighty, which is the required number.

110.16 And if we wish, we can instead drop the five from the fifteen and divide the remaining ten by the ten, resulting in one. We multiply it by the twelve, giving twelve. We make it tens and we add to it the product of the subtracted five by the twelve, since it was not subtracted from it. The sum is one hundred eighty, which is the required number, as before.

link to parallel text 111.1 Another example: we want to multiply three by fifteen. So we add two to the three, giving five. We denominate it with the ten, giving a half. We multiply it by the fifteen. The result of the multiplication is seven and a half. We raise each [digit] by ten, the power of ten, and half of the ten, to get seventy-five. We drop from it the product of the added two by the fifteen, since it was not added to it. The remainder is the required number, and that is forty-five, and its figure is 45.

link to parallel text 111.7 [Another type] is multiplication of “nines”, with the requirements that the [numbers of] ranks in the two lines be equal, one of them consist of all nines, and the digits of the other be equal. A description of the procedure is that you write down the two parallel lines, one of them below the other, and you put dots above them, as many as there are places in them. You multiply the digit of the place of one of them by the digit of the place of the other. You put the units of the result in the first of the dots, and its tens in the middle of the remaining dots. You note the difference between the nine and the digit of the multiplier. Then with [this difference] you fill in what is between the two digits of the results, I mean the units and the tens, and you fill the remaining dots with the digit that is different from the nine. What this yields is the answer.[52]

111.15 For example, we want to multiply four hundred forty-four by nine hundred ninety-nine. We put them down in two parallel lines, as mentioned, and we put dots above them, as many dots as there are places in both of them, and that is six. This is the figure:

111.18 We multiply the number of a rank of the multiplicand by the number of a rank of the multiplier, which is nine by four, giving thirty-six. We put down the six on the first of the dots, and five [dots] remain. We put down the thirty, in the form of a three, on the third of the five [remaining dots], since it is in the middle. Then we take the difference between the nine and the four, which is five. We fill in with it both of the dots between the units, which is six, and the tens, which is three, and we fill in the remaining dots with the four. This is what he meant when he said “with the number that is different from the nine”. This is the figure:

112.3The result is the number that is above the two multiplicands, and that is four hundred forty-three thousand five hundred fifty-six. So know it, and solve similar examples the same way.

link to parallel text 112.5 Another type of multiplication of nines has no condition. Instead, the digits in one of the lines are nines, and the digits in the other line can be whatever they are, and the number of places can also be whatever they are. To work it out, you add as many zeros to the ranks of the other line as the number of ranks of the nines. Then you subtract from the total the number different from the nines, leaving the answer.[53]

112.10 For example, we want to multiply nine hundred ninety-nine by nine thousand three hundred fifty-four. We add to the multiplier as many zeros as the number of ranks of the nines, which is three. So it comes to nine million three hundred fifty-four thousand, whose figure is 9354000. Then we drop the number that is not nines from it, which is the multiplier. The remainder is the required number, which is nine million three hundred forty-four thousand six hundred forty-six. Its figure is 9344646.

link to parallel text 112.16 Another type is known as squaring. You take half of the sum of the multiplicands and you square it. You subtract from the result a square of half of the difference between them. The remainder is the result of the multiplication.[54]

113.1 For example, we want to multiply seventeen by nineteen. We take half of their sum, which is eighteen. We multiply it by itself, which is the meaning of “squaring”, as [mentioned] before, giving three hundred twenty-four. We drop from it one, which is a square of half of the difference between the multiplicands, leaving three hundred twenty-three, which is the required number, and its figure is 323.

link to parallel text 113.6 Another type is also known as squaring, which is that you multiply a square of one of the two multiplicands by what results from the ratio of the other to the number you squared, or you divide a square of one of them by the result of dividing the squared one by the other one.

113.9 For example, we want to multiply twenty-five by fifteen. So we multiply a square of twenty-five, which is six hundred twenty-five, by three fifths, which is the ratio of the fifteen to the twenty-five. The result is the required number, and that is three hundred seventy-five, and its figure is 375. Or we divide a square of the fifteen, which is two hundred twenty-five, by the three fifths, which is the result of dividing the fifteen that we squared by the twenty-five. The result is the required number, like before. So know it.

link to parallel text 113.16 Another type is that you multiply the difference between the two multiplicands by the greater of them, and you drop the result from a square of the greater. Or you multiply the difference by the smaller of them and you add the result to a square of the smaller. The result is the required number.[55]

113.19 For example, suppose someone said, “Multiply thirty-six by fourteen”. We multiply the twenty-two, which is the difference between the multiplicands, by thirty-six, the greater of the multiplicands, resulting in seven hundred ninety-two. We drop that from a square of the greater, which is one thousand two hundred ninety-six. The remainder is the required number, which is five hundred four, and its figure is 504. Or we multiply the twenty-two, the difference, by the fourteen, the smaller multiplicand, resulting in three hundred eight. We add it to a square of the smaller, which is one hundred ninety-six. The sum is the required number, like before. So know it.

link to parallel text 114.4 And if you multiply a number with zeros by a number with zeros, multiply the parts of one by the parts of the other, stripped of the zeros. Then you dress the result with all the zeros. The outcome is the required number,[56] since multiplying the number by the zero or the zero by the number is identical. It comes from voiding the number or duplicating zero. Neither of these gives a number, so its sign is always a zero.

114.8 For example, suppose someone said, “Multiply thirty by one hundred forty”. We remove the zero from the thirty and we set it aside, leaving three. Likewise we remove it from the one hundred forty and we set it aside, leaving fourteen. We multiply it by the three and we add the two zeros that were set aside to the result. The outcome is the required number, which is four thousand two hundred, and its figure is 4200.

114.12 The upper limit on the [number of] ranks of the result is the sum of the ranks of the two multiplicands, since the greatest [number] one can have in a single rank is a nine, and multiplying nine by nine is eighty-one. Thus multiplication in the units rank can give tens.

114.15 To check it, you divide the result by one of the multiplicands. The result is the other [multiplicand]. For example, in the previous example we divide the eighty-one by the nine, which is one of the multiplicands. The result is nine, which is the other multiplicand.

link to parallel text 115.1 The student should memorize the following breakdown and master it, which is:

115.2 Multiplying a number by one or multiplying one by it leaves the number unduplicated, like two by one gives two, or one by five gives five.

115.5 And two by two gives four, and by each number that follows gives an additional two. For successive numbers, multiplying two by three gives six, and by four gives eight, and by five gives ten, and so on up to the ten, since this is the traditional limit of the breakdown.

115.9 And three by three gives nine, and by each number that follows gives an additional three. He means “for successive numbers” here, too. Thus multiplying it by four gives twelve, and by five gives fifteen, and so on up to ten.

115.12 And four by four gives sixteen, and by each number that follows gives an additional four. And five by five gives twenty-five, and by each number that follows gives an additional five. And six by six gives thirty-six, and by each number that follows gives an additional six. And seven by seven gives forty-nine, and by each number that follows gives an additional seven. And eight by eight gives sixty-four, and by each number that follows gives an additional eight. And for each of them for successive [numbers], just as the ones before. And nine by nine gives eighty-one, and by ten gives ninety. And ten by ten gives a hundred. So know it. Proceed by the power of almighty God.

link to parallel text 117.1 Section Five, on division.

link to parallel text 117.2 Division is the decomposition of the dividend into equal parts in such a way that its number is equal to what is in the divisor in units.[57] This applies to discrete quantities.[58]

117.5 Division is also viewed as the ratio of one of two numbers with respect to the other.[59] And this concerns continuous quantities.[60]

117.7 Most people view division in all circumstances as knowing how many whole units of the divisor are in the dividend.

link to parallel text 117.9 Division has two meanings. One of them is described first, which concerns the division of a type by another type, like dirhams by men. The other is described second, and concerns the division of a type by the same type. So the word “division” has two meanings. And it should not be given a single description without regard to circumstances, as most people do. The meaning many people give it is specifically the first meaning, and they ignore the second meaning. Their description without regard to circumstances misleads the general public to lump the two meanings together, or to think that division truly has one meaning, and this is not so.[61]

link to parallel text 117.16 Here is an example of division with the first meaning. Divide fifteen dirhams among three men. We decompose the fifteen into three equal parts, which is how many units are in the divisor. So each part consists of five dirhams, which is how many whole units there are of that three, the divisor.

118.1 Here is an example of the second meaning. Divide a piece of wood of fifteen spans by a piece of wood of three spans. The intention here is, how many copies of the divisor are in the dividend? So we cut up the dividend into copies of the divisor. There are five parts in the dividend of copies of the divisor, each part equal to the divisor.

118.6 So the result of working out the division in each of the two meanings is five. But the units of the result of five in the first meaning is different from the units of the result of five in the second meaning, since in the first meaning it is the number that is in a part of the parts of the dividend in units, and in the second meaning it is the number of parts in the dividend. So, in the first meaning the dividend is divided into a given number of parts, and what is in each of these parts comes to be known through the division. And, in the second meaning, what is in each of the parts of the dividend is given in units, and the number of parts into which it is divided is what comes to be known through the division. Thus the second meaning is the opposite of the first meaning.[62] So know it.

link to parallel text 118.14 Division comes in two types: dividing a greater number by a smaller number and dividing a smaller number by a greater number. Dividing the smaller by the greater specifically is called “denomination”, and the word “division” is reserved for dividing the greater by the smaller.

link to parallel text 118.17 The usual way of dividing the greater by the smaller is that you write the dividend on a line, and you write the divisor below it, making sure that the greater is not below the smaller. Then look for a number to put below the first digit of the digits of the divisor so that when you multiply it by all of its digits, it either exhausts the entire dividend or it leaves a remainder smaller than the divisor, in which case you denominate it with it.

119.1 For example, suppose we want to divide two hundred forty-five by twelve. We put the dividend down on a line, and the divisor twelve on a line below it, with the two below the four, and the ten, in the form of a one, below the two, as in this figure:

So, it happens that the divisor is below the twenty-four. If it were smaller, say for example eleven or ten or something similar, we would still put down the last digit of the divisor below the last digit of the dividend.

119.7 In our example, we look for a number to put below the two of the divisor, since it is the first digit. We multiply it by all of it, so that it either exhausts the twenty-four above it, or it leaves a remainder less than the twelve. We find that it is two. We multiply this two first by the one that is in the last rank of the divisor, giving two. This exhausts the two above it. Then we likewise multiply it by the two below it, giving four, and it exhausts the four above it. Then we shift the twelve back one place below the five, and we look for a number to put below the two of the divisor so that when we multiply them, it exhausts what is above it. We find nothing, so we put a zero in its place, and five remains from the twelve. We denominate it with it, to get two sixths and half a sixth. We add it to the whole number, which gives the required number, and that is twenty and two sixths and half a sixth, and its figure is {1~\,2\over 2~\,6}\,20. Similar problems are worked out the same way.

link to parallel text 119.18 If you want, you can divide the dividend into parts and add the quotients to get the result. Or you can decompose the divisor into the numbers of which it is composed and make them denominators, and divide the dividend by them. Or you can reconcile the dividend and the divisor and divide the reconciled divisor by the reconciled dividend.

120.5 An example of the first: suppose we want to divide forty-four equally among eleven men. We split the forty-four into twenty-two and twenty-two. If we wish, we can partition it differently. So we divide the twenty-two by eleven, resulting in two, and similarly for the other division. We add them, giving four, which is the result of dividing forty-four by eleven.

120.10 An example of the second: suppose someone said, “Divide ninety-six by twelve”. We decompose the divisor twelve into what it is composed of, which is six and two, or three and four, which is similar. Suppose we decompose it the first way. We divide the ninety-six by the two first, resulting in forty-eight. We likewise divide it by the six, the other denominator, resulting in eight, which is the required number. And if we divide the dividend first by the six, then what results by the two, it still gives eight. So know it.

120.16 An example of the third: suppose someone said, “Divide thirty-five by fifteen”. You rework each of them by a fifth, which reconciles them. For the dividend, its fifth is seven, and for the divisor, its fifth is three. We divide seven by three, resulting in two and a third, which is equal to the result of dividing the thirty-five by the fifteen. So know it.

link to parallel text 120.20 [Another] type of division is specifically called apportionment. The way you work it out is that you add the apportioned parts and you take it to be a denominator. Then you multiply each of the apportioned parts by the dividend, and you divide the results by the denominator, giving the required number.

121.4 For example, a man is bankrupt. Some people give him ten dinars, which he accepts. Suppose there are three donors, and they divide it according to their own wealth. One of them has four dinars, the second five, and the third six. We add up these parts, which surpass them, that is, the apportioned parts, and their sum is fifteen dinars. We make it a denominator. Then we multiply what the first has in hand by the dividend, ten, giving forty. We divide it by the denominator, resulting in two dinars and two-thirds of a dinar, which is his obligation of the ten. Likewise we multiply what the second has in hand by the ten, giving fifty, and we divide it by the denominator, resulting in three dinars and a third of a dinar, which is his obligation of the ten. Then we likewise multiply what the third has in hand by the ten, giving sixty, and we divide it by the denominator, resulting in four dinars, which is his obligation of the ten. So if we add up these three results it gives ten.

link to parallel text 121.14 There are four other ways to do this. One of them is that we denominate what each of them has in hand with the denominator, and we multiply the result by the dividend. This gives the required number. The second is that we denominate the dividend with the denominator, and what results is called the part of the share. We multiply it by what each one has in hand, giving the required number. In the third, we divide the denominator by what each one has in hand, and we divide the dividend by the results, giving the required number. In the fourth we divide the denominator by the dividend and we divide what each one has in hand by the results. This gives the required number.

121.23 There are other ways [of solving this problem] which involve combining the proportion, switching it, or some of the other conditions of the proportion, as will be seen later, almighty God willing.[63] What we have described is sufficient for anyone to understand.

link to parallel text 122.2 In the case that the apportioned parts are fractions, multiply everything in the problem by the smallest number divisible by the denominators. And if all of the parts have a common divisor, remove it by exchanging the parts or reconciling them.

122.5 For example: a man is bankrupt, and he receives twelve dinars. He has three donors, the first of whom has four dinars and a third of a dinar. The second has five dinars and a fourth of a dinar, and the third has six dinars and a sixth of a dinar. We multiply each of the parts that each of them has in hand by the smallest number divisible by their denominators. Knowing the smallest number divisible by the denominators in this example and in others is reached by decomposition.

122.10 We decompose each of the denominators into the numbers of which it is composed, and we drop numbers from the second that are repeats of numbers in the first, and from numbers of the third that are repeats of numbers in the first and remaining in the second, and from numbers in the fourth that are repeats in the previous numbers, and so on to the last. Then we compose the remaining numbers by multiplication. If there is no repeated number, then they are all different, so multiplying them together gives the smallest number divisible by the denominators.[64]

122.15 If we want to apply this in the present example, we find that the denominator of a third, which is three, cannot be decomposed. The denominator of a fourth, which is four, decomposes into two and two, and there is no repetition, so we keep them with the three. The denominator of a sixth decomposes into three and two. Both are repeats, so we discard them and we compose the rest, and that is three by two, giving six, and six by two, giving twelve, which is the least number divisible by the denominators.

122.19 We multiply everything in the problem by it. We multiply it first by what the first has in hand, resulting in fifty-two. We exchange it with what he has in hand. Then we likewise multiply it by what the second has in hand, resulting in sixty-three, and we also exchange it with what he has. Then we likewise multiply it by what the third has in hand, resulting in seventy-four, and we also exchange it with what he has in hand. Then we see if these parts have a common divisor, and we remove it and replace each one with its reconciled number, as mentioned.

123.3 To know how to remove common divisors in this example and others, we decompose the numbers according to how they are composed, and we drop the repeated numbers in each of them from the total. Then we compose what is left of each of them by multiplication to get the reconciled amounts. Whenever no number remains, exchange it with one, since multiplying a number that has vanished involves no duplication. Thus the exchanged one is the reconciled amount. The common divisor of these numbers can always be obtained by means of the denominated part of the repeated number that was dropped.[65]

123.9 If it is two, the common divisor is a half, and if it is five, then it is a fifth, and if it is ten, then it is a tenth, and if it is eleven, then it is a part of eleven, and so on.

123.11 After testing the parts mentioned in the example above by this procedure, we find them to be different, so we add them. This is the denominator, and it is one hundred eighty-nine. So we multiply what each of them has in hand by the twelve and we divide by the denominator. The result for the first [donor] is three dinars and two ninths of a dinar and five sevenths of a ninth of a dinar, which is his obligation of the twelve. For the second it is four dinars, which is his obligation of the twelve. And for the third it is four dinars and six ninths of a dinar and two sevenths of a ninth of a dinar, which is his obligation of the twelve dinars. If we add these, they likewise result in twelve. And, if we wish, we can also work this out by the other methods mentioned above.

link to parallel text 123.18 For denomination, the most well-known way is that you decompose the denominating number into the numbers of which it is composed, and you take them as denominators. Then you divide what you want to denominate by them, resulting in the answer. One obtains its value by means of the ratio of these parts to the denominators serving as divisors.

link to parallel text 123.22 For example, suppose someone said, “Denominate eleven with fifteen”. We decompose the fifteen, the denominating number, into the numbers from which it is composed, which are five and three. We put them below a line and we divide the eleven first by the three, and we put the remainder above it. And we divide the quotient by the five, which is smaller than it. We put it above it, so it yields this figure: 2~\,3\over 3~\,5. Thus we form the ratio of the three to the five that is below it, and the two to the three that is below it, and we attach the ratio to the five. This gives the size of eleven with respect to fifteen, which is three fifths and two thirds of a fifth. So know it.

link to parallel text 124.8 A lesser known way is that you divide the denominating number by the denominated number, and you denominate one with the result. Or you denominate one with the denominating number and you take this ratio of the denominated number. Or you multiply the denominated number by some [convenient] number and you divide the result by the denominating number, and [you divide] the result by that multiplied number.[66]

124.12 An example of the first [way]: suppose someone said, “Denominate four with twelve”. We divide the twelve by the four, and we denominate one with the result. This gives the required number, which is a third.

124.14 An example of the second: suppose someone said, “Denominate nine with fifteen”. We denominate one with the fifteen, to get a third of a fifth. We take from the nine a third of its fifth. The result is the required number, which is three fifths.

124.17 And an example of the third: suppose someone said, “Denominate ten with sixteen”. We multiply the ten by whatever number we wish, such as eight, to get eighty. We divide it by the sixteen, and the result of that by the eight, to get the required number, which is five-eighths. So know it.

link to parallel text 124.20 To decompose numbers requires that some preliminary remarks be learned. They are:

125.1 Every number that does not begin with units has a tenth and a fifth and a half, [the latter] being a characteristic of every even number. For example, fifty and similar numbers. Its half is twenty-five, its fifth is ten, and its tenth is five.

link to parallel text 125.5 If it begins with a five, it has a fifth. If it begins with units, and if it is even, then one casts out numbers using one of the three moduli, in the sense of what is said above on casting out. If it is cast out entirely by nines, then it has a ninth and a sixth and a third. For example, thirty-six is cast out entirely by nines. Its ninth is four, its sixth is six, and its third is twelve.

125.11 If the remainder is three or six, then it has a sixth and a third. An example with a remainder of three is sixty-six, and similar numbers. Its sixth is eleven and its third is twenty-two. An example with a remainder of six is forty-two, and similar numbers. Its sixth is seven and its third is fourteen.

125.16 If the remainder is something else, then cast out eights. If it is cast out entirely, then it has an eighth and a fourth. For example, sixty-four is cast out entirely by eights. Its eighth is eight and its fourth is sixteen.

125.18 If the remainder is four, then it has a fourth. For example, sixty-eight. Its fourth is seventeen.

125.20 And if the remainder is something else, then cast out sevens. If it is cast out entirely, then it has a seventh. For example, fourteen is cast out entirely by sevens, and its seventh is two.

126.2 If it is not cast out entirely, then it only has a half and its half is odd, [and you then] look for deaf parts. For example, twenty-six is not cast out entirely by any of the three moduli, so it only has a half, and its half is thirteen, which is an odd number. So know it.

126.5 If [the number] is odd, cast it out by two numbers, nine and seven. If it is cast out entirely by nines, then it has a ninth and a third. For example, eighty-one is cast out entirely by nines. Its ninth is nine and its third is twenty-seven.

126.9 If the remainder is three or six, then it has a third. An example with a remainder of three is thirty-nine. Its third is thirteen. An example with a remainder of six is one hundred twenty-three. Its third is forty-one.

126.12 And if the remainder is some other number, then cast out sevens. If it is cast out entirely, then it has a seventh. For example, seventy-seven is cast out entirely by sevens, and its seventh is eleven.

link to parallel text 126.14 If it is not cast out entirely, then look for deaf parts by dividing by them; and you continue dividing the required number to be decomposed by deaf parts until you arrive at a number that divides into it, or you get a number whose square is greater than your given number, or [equivalently, that] the result from the division is equal to or less than the divisor and leaves a remainder after the division. At this point you know that it is one of the deaf parts, and the denomination is formed from it.

127.1 For example, suppose we want to decompose two hundred twenty-one. We find that it is not cast out entirely by the two moduli, so we need to divide it by deaf parts, as mentioned. We find that it is not divisible by eleven, which is the first of them. But it is divisible by thirteen, and the result of the division is seventeen. So it is composed from thirteen by seventeen. If a square of the thirteen had been greater than the dividend,[67] we would have also known that it is deaf. If the result had been equal to the divisor thirteen, or if it had been less, with a remainder from the dividend, we would also have known that it is deaf. We denominate it with separate pieces, saying something like “three parts of this” or “a hundred twenty parts of that” or something similar.

link to parallel text 127.9 Subsection on finding deaf parts.

127.10 This process is called the sieve. Here you write the odd numbers beginning with three. Then, for each of these numbers, you count off its successors by the number of units in it. When this is done, the next number is composite, and it counts that [first] number. You continue to do this until you reach a number whose square is greater than the last number in the sieve, at which point you know that the work is finished. Each marked number is composite and each unmarked number is deaf.

127.15 For example, we write down the odd numbers from three consecutively, as mentioned, in a table, as in this figure:

link to parallel text 128.1 If we want to know which among them is composite with three, we count it off from its cell, ending at the cell of the seven, so the nine that is next is composite with the three. So we put a mark above it. Similarly, we count beginning with the cell of the nine, ending with the cell of the thirteen. So the fifteen that is next is also composite with the three, and we put a mark above it. Do this until the end of the sieve. Likewise, do this with the five and the seven, but not with the nine, since it is composite, and not with any other composite number. This ends in our example with counting by thirteen. We know that the work is finished because its square is one hundred sixty-nine, which is greater than one hundred forty-five, which is the last number in the prescribed sieve.

128.9 And if we wish, we can work with a sieve greater or smaller than this one, since the method is the same for all of them. Every number in this sieve that is marked is composite, and every unmarked number is deaf, as we noted. These deaf parts are counted only by one, as we recall from the first part of the book.[68] We cannot find a number which, when multiplied by a number gives, for example, thirteen or one hundred fifty-one and the like. So if someone said to you, “Of which numbers is thirteen composed”, you would say thirteen by one. Multiplication by one is not duplication, as mentioned before. And this is the answer to similar [questions], so know it.

link to parallel text 129.1 Section Six, on restoration and reduction.

129.2 Restoration is reconstitution (iṣlāḥ), and reduction is its opposite. The purpose of restoration and reduction is to know what to multiply by a number to produce the required outcome. Restoration is only for taking the smaller to the greater, and reduction is for the opposite, that is, from the greater to the smaller.

129.6 An example of restoration: suppose someone said, “By how much must one restore eight, for example, so that it gives nineteen?” An example of reduction: suppose someone said, “By how much must one reduce fifty, for example, so that it gives six?”

129.8 To work out restoration, you divide the restored number by the number to be restored, which gives the required number. For example, suppose someone said, “By how much must one restore three so that it gives six?” You divide the six, the restored number, by the three, the number to be restored, resulting in two. If you multiply that by the three it becomes six, which is the required number.

129.12 To work out reduction, denominate the reduced number with the number to be reduced. This results in the answer. For example, suppose someone said, “By how much must one reduce eight so that it yields three?” You denominate the three, the reduced number, with the eight, the number to be reduced. This gives three eighths. If you multiply the three eighths by the eight, it becomes three, which is the answer. This is the end of the first chapter, with God’s benediction and His good guidance.

link to parallel text 131.1 Chapter Two, on fractions.

133.1 The fraction is the ratio between two numbers that are a part or parts. The ratio between the part and its name is called a fraction.

link to parallel text 133.3 For example, three and six. The relation that arises with the ratio of the smaller to the greater is called the fraction. It cannot be named in terms of the three alone because of the separation, and likewise it cannot be named in terms of the six alone; nor in terms of [the two words] taken together, since the ratio is not a sensible object but is the idea of a specific intelligible object. It is called a fraction, like terrain with fractures that ascends and descends; or also like the surface, the solid, and the line, and there is nothing of this in discrete quantities except similar abstractions. Furthermore, the fraction consists of names, like “the half” and others that will be presented below, almighty God willing.

133.10 We intend to present calculation with fractions in six sections.

link to parallel text 134.1 Section One, on the names of fractions and numerating them.

link to parallel text 134.2 Ten fractions have simple names. The first is the half, which is the greatest, whose figure is 1\over 2. Then the third, whose figure is 1\over 3. Then the fourth, whose figure is 1\over 4. Then the fifth, whose figure is 1\over 5. Then the sixth, whose figure is 1\over 6. Then the seventh, whose figure is 1\over 7. Then the eighth, whose figure is 1\over 8. Then the ninth, whose figure is 1\over 9. Then the tenth, whose figure is 1\over 10. Then the “part”, of which there are many types. You can say a part of eleven, whose figure is 1\over 11, and a part of seventeen, whose figure is 1\over 17, and so on.

134.8 You can form the dual and plural of these fractions. Adding terminates for each fraction when you have [one] less than the name of the part. The name is the greater of the two numbers in the ratio of one of them to the other. For example, we say a fourth and two fourths and three fourths, but we do not say four fourths. Likewise we say a seventh and two sevenths and three sevenths and four sevenths and five sevenths and six sevenths, but we do not say seven sevenths. So know it.

link to parallel text 135.1 You can attach these simple names to one another so that the resulting name is a combination of two or more names. For example, we say two eighths and a seventh of an eighth, whose figure is 1~\,2\over 7~\,8, and likewise for similar examples. Another example is eight ninths and four sevenths of a ninth and two sixths of a seventh of a ninth and a third of a sixth of a seventh of a ninth, whose figure is 1~\,2~\,4~\,8\over 3~\,6~\,7~\,9, and likewise for similar examples.

link to parallel text 135.8 Numeration is that you reduce all of what is given to you in a particular problem to a finer fraction in it…[69] Know that the finer fraction in it is the part named with respect to all the denominators of the problem.

link to parallel text 135.10 When the fraction is related, like five-sixths and four-fifths of a sixth and two-thirds of a fifth of a sixth, whose figure is 2~\,4~\,5\over 3~\,5~\,6, we multiply what is above the first denominator by the denominator that follows it. That means to relinquish its name, which is sixths, in favor of fifths of sixths, since the second denominator is a number of what is in the one of the first denominator in units. So we add it with the four that is above it, since they are fifths of sixths, then we multiply that by the three, the third denominator, which is the number of what is in the one of the second denominator in units. This results in thirds of fifths of sixths. We add it with the two, since it is thirds of a fifth of a sixth. This gives the numerator of the problem, which is how many thirds of fifths of sixths there are in it, and that is eighty-nine. The multiplication of the denominators one by the other, which is ninety, is what is in the whole one of these parts, and a part of one of them is a finer fraction in the problem.

link to parallel text 136.8 If the fraction is distinct, like five-sixths and four-fifths, whose figure is {4\over 5}\,{5\over 6}, we multiply the five-sixths by five, the denominator of the fifths, and that is the number of what is in the one sixth of fifths, so it becomes fifths of sixths. And we multiply the four-fifths by six, the denominator of the sixths, and that is the number of what is in the one fifth of sixths. So it becomes fifths of sixths. The sum of these two numerators are parts of thirty, which is sixths of fifths or fifths of sixths, both being equal.

link to parallel text 137.1 The fraction may be portioned, in which the fractions are taken one of another, and which is how they are described in the expression. An example is three fourths of five sixths, whose figure is 5~\bullet ~3\over 6 ~\bullet ~4. We multiply the three by the five, giving fifteen fourths of a sixth, or sixths of a fourth, which are parts of twenty-four parts of the unit. This is because five of the sixths, of which we want three of its fourths, [is found] by taking its fourth, which is five fourths of a sixth, and multiplying it by three, giving fifteen fourths of a sixth, or by taking a fourth of three times it. And three times it is its multiplication by three, giving fifteen sixths, and a fourth of that is fifteen fourths of a sixth, which is also fifteen sixths of a fourth. So three fourths of five sixths is five sixths of three fourths, since multiplying three by five is like multiplying five by three.[70] So know it.

link to parallel text 137.11 …and this is different for different [kinds] of fractions, that is, the numeration. There are five kinds [of fractions]: simple, related, distinct, portioned, and excluded.

link to parallel text 137.13 The numerator of a simple fraction is what is above it. For example, we said “a seventh”: its numerator is the one that is above the line. Likewise for combined fractions, like if we said, “a third of a seventh”: its numerator is also the one that is above the line.

link to parallel text 138.1 The numerator of a related fraction is what is above the first denominator multiplied by the next denominator, then adding to the end of the line; or, it is what is above the first denominator multiplied by the denominators after its denominator, and what is above the second denominator also multiplied by the denominators after its denominator, and so forth until the line is completed, then adding them together.

138.5 For example, suppose someone said, “Numerate five eighths and four sevenths of an eighth and three fifths of a seventh of an eighth and two thirds of a fifth of a seventh of an eighth”, whose figure is 2~\,3~\,4~\,5\over 3~\,5~\,7~\,8. We multiply the five that is above the first denominator by the seven, the second denominator, and we add the four that is above it, giving thirty-nine. We also multiply that by the five, the third denominator, and we add the three that is above it, giving one hundred ninety-eight. We likewise multiply it by the three, the fourth denominator, and we add the two that is above it to it, giving five hundred ninety-six, which are eighths of sevenths of fifths of thirds. This is the numerator, and its figure is 596.

138.12 For the second way, we multiply the five that is above the first denominator by the other denominators except its denominator, as mentioned. This results in five hundred twenty-five. Keep it in mind. Then we multiply the four that is above the second denominator by the denominators after it, resulting in sixty. Keep this in mind as well. Then we multiply the three that is above the third denominator by the three, the fourth denominator, and we add what is above it, since there is nothing left after that denominator, giving eleven. We add it with the two remembered numbers, which gives five hundred ninety-six, which is the numerator, as before.

link to parallel text 139.1 The numerator of a distinct fraction is found by multiplying the numerator of each part by the other denominators, then adding them together.

139.2 For example, suppose someone said, “Numerate five sevenths and half a seventh and four sixths”. We put them below two lines as in this figure: {4\over 6}~{1~\,5\over 2~\,7}. We find that the first part is related, so we take its numerator in the manner shown above, to get eleven. We multiply it by the six, the denominator of the second part, resulting in sixty-six. Keep it in mind. We find that the second part is simple, so its numerator is the four that is above the denominator. We likewise multiply it by the denominators of the first part, giving fifty-six. We add this with the remembered number to get a total of one hundred twenty-two, which is sevenths of halves of sixths. This is the numerator, and its figure is 122.

link to parallel text 139.10 The numerator of a portioned fraction is found by multiplying what is above the line, each one by the next.

139.11 For example, suppose someone said, “Numerate seven-ninths of five-sixths of three-tenths”. We put them on a line and we always separate them with marks, as in this figure: ~3~\bullet ~5~\bullet ~7\over 10~\bullet ~6~\bullet ~9.We multiply the seven that is above the first denominator by the five that is above the second denominator, and their product by the three that is above the third denominator, to get the numerator. That comes to one hundred five sevenths of sixths of tenths, and its figure is 105.

link to parallel text 140.1 The numerator of an excluded fraction of disconnected [type] is found the same way as for distinct fractions, by subtracting the smaller from the greater.

140.2 For the disconnected [type], what follows the “less” is not taken from what precedes it, but rather is taken from one, and then it is removed. For example, if we said “a half less a third”, we mean less a third of one. So, after taking a third of the one, it is removed from the half. And when he said “the same way as for distinct fractions” he meant that you multiply what is before the “less” with what is after it in two different steps. We multiply the numerator of each part by the other denominator, and then we subtract the smaller from the greater.

140.8 For example, suppose someone said, “Numerate six eighths less a ninth of one”. We put them on a line as in this figure: {1\over 9}~\scalebox {-1}[1]{\ell }~{6\over 8}. We multiply the six, the numerator of what is before the “less”, by the nine, the denominator after it, giving fifty-four. From that we drop the product of the one, the numerator after the “less”, by the eight, the denominator before it, which is eight. The remainder is forty-six eighths of ninths, which is the numerator. Work out similar problems the same way, as explained above.

link to parallel text 140.14 For the connected [type], multiply the numerator of the diminished fraction by the numerator of the excluded fraction, and likewise multiply by their denominators, and then subtract the smaller from the greater.

140.16 The connected [type] occurs when what follows the “less” is taken from what precedes it, without mediation. For example, suppose we said “a half less its third”. A third of the half is a sixth, so it is as if someone had said “a third”, or “a half less a sixth”, so it becomes disconnected.

141.1 Suppose someone said, “Numerate six-sevenths and half a seventh less its third”. We put them on a line, as in this figure: {1\over 3}~\scalebox {-1}[1]{\ell }~{1~\,6\over 2~\,7}. We take the numerator of what is before the “less”, which is the diminished fraction. We multiply it by the denominators of what is after the “less”, which is the excluded fraction, to get thirty-nine. Keep it in mind. Then we also multiply the numerator of the diminished fraction by the numerator of the excluded fraction, similarly giving the complement thirteen. We drop it from the remembered number. The remainder is twenty-six sevenths of halves of thirds, and it is the numerator.

link to parallel text 141.7 Supplementary remark. If there are repeated deletions in which each of them has the conjunction “and”, it is with respect to the first [term]. And in relation to the first [term], whether they are all connected to or detached from the diminished fraction, [treat them as] distinct fractions excluded from the diminished fraction.[71]

141.11 A first example: suppose someone said, “Numerate five and a third less its fourth and less its seventh and less its fifth”. We put it down on a line like this figure: {1\over 5}~\scalebox {-1}[1]{\ell }~{1\over 7}~\scalebox {-1}[1]{\ell }~{1\over 4}~\scalebox {-1}[1]{\ell }~{1\over 3}\,5. The fourth and the seventh and the fifth are distinct fractions excluded from the five and a third, which is the diminished part, and they are all connected to it. Removing the second and third particles of exclusion, the figure in this problem becomes {1\over 5}\,{1\over 7}\,{1\over 4}\,\scalebox {-1}[1]{\ell }\,{1\over 3}\,5. We work it out as before, with the result that the numerator is nine hundred twelve thirds of fourths of sevenths of fifths, and its figure is 912.

link to parallel text 142.6 A second example: suppose someone said in the previous problem, “less a fourth of one and less a seventh of one and less a fifth of one”. Then these are distinct fractions removed from the diminished part, which are all detached. Write them also as if they were connected and work it out as before. The result is that the numerator is one thousand nine hundred ninety-one thirds of fourths of sevenths of fifths, and its figure is 1991.

link to parallel text 142.10 If the exclusions are repeated without a coordinating conjunction,[72] in such a way that each [term] is excluded from the one before it, whether they are detached or connected, then we take the [last] excluded fraction and the diminished fraction [before it], as in the other problem, and we work it out as shown above, whether they are connected or detached. The numerator we get is excluded from the one before it. Then we work that out similarly. The result is the numerator excluded from the one before it. Continue like this to the first [fraction].[73] If we wish, we can work it out for disconnected fractions as in the two preceding cases.

link to parallel text 143.1 Whenever some of them are disconnected and others are connected, then the connected fractions must be transformed to their disconnected form when writing it down, so that all of them become disconnected. For instance, five sixths less three of its fourths is five sixths less three fourths of five sixths. Now what is written after the “less” is a disconnected fraction.[74]

link to parallel text 143.5 And if there is a whole number before the fractions in a problem, it is multiplied by the denominators and then added with the numerator to become fractions. For example, suppose someone said, “Numerate five and five sixths and three fourths of a sixth”. So we put it down on a line like this: {3~5\over 4~6}\,5. We multiply the whole number five by the six, the first denominator, and what results by the four, the second denominator. This gives the result one hundred twenty. We add it with the numerator of the fraction, which is twenty-three. This gives a total of one hundred forty-three sixths of fourths, and its figure is 143, which is the numerator.

link to parallel text 143.13 And if it is after them, multiply the numerator by it, since the fractions are portions of it. For example, suppose someone said, “Numerate four sevenths and six eighths of ten”. We put it down on a line like this: 10\,{6\over 8}\,{4\over 7}. The numerator of the fractions, as shown before, is seventy-four. We multiply it by the whole number ten to get seven hundred forty sevenths of eighths, which is the numerator, and its figure is 740.

link to parallel text 144.4 And if it is in the middle, it is either attached to what is before it, so it is on the end, or it is attached to what is after it, so it precedes it. You numerate it according to one of the two cases, with the remaining [fractions] as distinct in the latter case, and in the former case one multiplies by the numerator of the remaining [fractions].

144.7 “Attached to what is before it” means that only the first fraction is taken of the whole number, so it is one part. The remaining fraction is the distinct part. You multiply the numerator of each part by the denominator of the other and you add the results.

144.10 For example, suppose someone said, “Numerate four ninths of five, and three sixths”. We put down the problem on a line like this: {3\over 6}\,5\,{4\over 9}. We multiply the four that is above the nine by the whole number five, giving twenty, which is the numerator of the first part. Then we multiply it by the six, the denominator of the second part, giving one hundred twenty. Keep it in mind. Then we multiply the three, the numerator of the second part, by the nine, the denominator of the first part, giving twenty-seven. We add it with the remembered number to get one hundred forty-seven sixths of ninths, which is the numerator. Its figure is 147.

link to parallel text 145.1 And the “attached to what is after it” [means] that the first fraction is taken of the whole number and of the fraction after it. So the whole number is attached to what is after it, and it precedes it. So you numerate it with it and you multiply that by the numerator of the remaining [fraction], which is the first fraction, since it is a portion of it.

145.4 For example, suppose someone said, “Numerate two thirds of seven and four sevenths”. We put it down like this: {4\over 7}\,7\,{2\over 3}. We multiply the whole number seven by the seven, the denominator of the fraction, and we add the four that is above it and we multiply that by two, the numerator of the remaining [fraction], which is the first fraction. This gives one hundred six thirds of sevenths, which is the numerator, and its figure is 106.

145.8 One understands from this that for every fraction, or more than one fraction, that is taken only of a whole number, the whole number comes after it. So we numerate it with it as one part. And the other fraction which is not taken of it we regard as distinct. And for every fraction, or more than one fraction, together with the whole number, if the whole number precedes it, then we numerate it with it as one part, and one multiplies its numerator by the numerator of the fraction taken of that whole number and the fraction that is with it.

145.13 Any remaining fractions in the problem other than those taken of the whole number and what is with it are distinct fractions. So the whole number and what is with it and what is taken of that is one part, and every fraction among the distinct fractions is a part. So one multiplies the numerator of each fraction by the denominator of the other and we add the results.[75]

link to parallel text 146.3 Common divisors between the numerator and denominators should be removed. We mentioned the way to do this by decomposition in the section on division.[76] For portioned fractions in particular one should remove the common divisors before the numeration, since the numbers ascribed to the numerator are above the line and the numbers ascribed to the denominators are below the line. So we drop the repeated [factors] from both of them.[77]

link to parallel text 147.1 Section Two, on adding and subtracting fractions.

link to parallel text 147.2 To work out the addition, you multiply the numerator of each line by the denominators of the other, and you divide the sum by the denominators. And for subtraction you drop the smaller from the greater before dividing by the denominators.

147.4 An example of addition: suppose someone said, “Add three and four fifths and six eighths to four tenths and three eighths of a tenth and half an eighth of a tenth”. We put the addend on a line and the augend on another line below it, like in this figure:

We numerate the upper fraction as before. Its numerator is one hundred eighty-two. We multiply it by the denominators of the lower [fraction], to get twenty-nine thousand one hundred twenty. Keep it in mind. Then we multiply the numerator of the lower [fraction], which is seventy-one, by the denominators of the one above, resulting in two thousand eight hundred forty. We add it with the remembered number to get thirty-one thousand nine hundred sixty, and its figure is 31960. We divide it by the denominators of the two lines. The result is the required number, and that is four and nine tenths and seven eighths of a tenth and half an eighth of a tenth, and its figure is {1~7~\,\,9\,\over 2~8~10}4. And its answer is given by five.

link to parallel text 147.15 Another example, of subtraction: suppose someone said, “Subtract seven tenths of two less a third of one from four and three fourths of five sixths”. We put the minuend down on a line and the subtrahend on another line below it, like with the addend, as in this figure:

148.3 We numerate the upper [fraction] as before. Its numerator is one hundred eleven. We multiply it by the denominators of the lower [fraction], resulting in three thousand three hundred thirty. Keep it in mind. Then we multiply the numerator of the lower [fraction], which is thirty-two, by the denominators of the upper [fraction], resulting in seven hundred sixty-eight. We drop it from the remembered number, leaving the required number, which is two thousand five hundred sixty-two, and its figure is 2562. We divide it by the denominators of the two lines. The result is the answer, which is three and five tenths and three sixths of a tenth and half a sixth of a tenth, and its figure is {1~3~\,\,5\,\over 2~6~10}3. And its answer is given by subtraction.

link to parallel text 149.1 Section Three, on multiplying fractions.

link to parallel text 149.2 This is the portioning of one of two fractions by the amount of the other. This contrasts with whole numbers or the duplication of fractions by the amount of a whole number. If the multiplier is a whole number and the multiplicand is a fraction, or vice versa, then either one extracts the whole number by the amount of the fraction, or one duplicates the fraction by the amount of the whole number.

link to parallel text 149.6 To work it out, that is the multiplication of fractions, you multiply the numerator of each line by the numerator of the other, and you divide the result by the denominators.

149.8 For example, suppose someone said, “Multiply three fourths and a third by three ninths and four sixths of a ninth and a fifth of a sixth of a ninth”. We put the multiplier on a line and the multiplicand below it on another line, as in this figure:

We multiply the numerator of the upper [fraction], which is thirteen, by the numerator of the lower [fraction], which is one hundred eleven. It results in one thousand four hundred forty-three, and its figure is 1443. We divide it by the denominators, resulting in the required number, which is four ninths and a fourth of a fifth of a sixth of a ninth, and its figure is 1~\,0~\,0~\,4\over 4~\,5~\,6~\,9. And its answer is given by one.

link to parallel text 150.2 Another example: suppose someone said, “Multiply a third of four and an eighth by a fifth of two thirds of ten”. We put the problem down on two lines as before, like in this figure:

We multiply the numerator of the upper [fraction], which is thirty-three, by the numerator of the lower [fraction], which is twenty, resulting in six hundred sixty, and its figure is 660. We divide it by the denominators, resulting in the required number, which is one and five sixths, and its figure is {5\over 6}1. And its answer is given by two.

link to parallel text 151.1 Section Four, on division and denomination.

link to parallel text 151.2 To work these out you multiply the numerator of each line by the denominators of the other, and you divide the result for the dividend by the result for the divisor, or you denominate.

151.4 An example of division: suppose someone said, “Divide six and a third by four fifths of seven eighths of three”. We put the dividend on a line and the divisor on another line below it, as in this figure:

We multiply the numerator of the dividend, which is nineteen, by the denominators of the divisor, resulting in seven hundred sixty. Its figure is 760, which is the result for the dividend. Keep it in mind. Then we multiply the numerator of the divisor, which is eighty-four, by the denominators of the dividend, resulting in two hundred fifty-two. Its figure is 252, which is the result for the divisor. We divide the remembered number by it. The result is the required number, which is three and a seventh of a ninth, and its figure is {1~\,0\over 7~\,9}\,3. And its answer is given by four.

link to parallel text 151.14 Another example, of denomination: suppose someone said, “Denominate three and a fourth less two ninths of it with six and two eighths and three fifths”. We put down the problem as in this figure:

We multiply the numerator of the denominated number, which is ninety-one, by the denominators of the denominating number to get the result of three thousand six hundred forty, and its figure is 3640. Keep it in mind. Then we multiply the numerator of the denominating number, which is two hundred seventy-four, by the denominators of the denominated number to get the result of nine thousand eight hundred sixty-four, and its figure is 9864. We denominate the remembered number with it to get the required number, which is fifty parts of one hundred thirty-seven parts and five ninths of a part of one hundred thirty-seven parts, and its figure is 5~\,\,50\,\over 9~137. And its answer is given by subtraction.

link to parallel text 152.9 When the denominators of the two lines are equal, you divide the numerator by the numerator or you denominate with it, without multiplying by the denominators.

152.11 An example of division: suppose someone said, “Divide eight and nine tenths and two thirds of a tenth by five tenths and a third of a tenth”. We put down the problem like this:

We divide the numerator of the dividend, which is two hundred sixty-nine, by the numerator of the divisor, which is sixteen. The result is the required number, which is sixteen and six eighths and half an eighth, as in this figure: {1~\,6\over 2~\,8}\,16.

153.3 Another example, with denomination: suppose someone said, “Denominate two and a third with six and two thirds”. We put down the problem like this:

We denominate the numerator of the denominated number, which is seven, with the numerator of the denominating number, which is twenty. This results in the required number, which is three tenths and half a tenth, and its figure is \,1~~3~\over 2~\,10.

link to parallel text 153.7 When the two numerators are equal, you divide the denominators of the divisor by the denominators of the dividend, or you denominate, without multiplying by the numerator, since, if we multiply by the denominators, the dividend will be composed from its numerator and the denominators of the divisor, and the divisor will be composed from its numerator and the denominators of the dividend. The two numerators vanish when the common divisors are removed. This is also the cause for the previous method.

153.12 For example, suppose someone said, “Divide five by five sixths”. We divide the six, the denominator of the divisor, by one.[78] It gives six, since whenever the dividend or the divisor is a whole number it is its numerator, and its denominator is one.

153.15 Similarly, suppose someone said, “Denominate five sixths with five”. You denominate one with six, giving a sixth. So know it.

link to parallel text 154.1 Section Five, on restoration and reduction.

154.2 To work these out you divide the restored number by the number to be restored, or you denominate the reduced number with the number to be reduced, to get the required number. When he said “required number” he meant what is multiplied by the number to be restored in order to restore, or by the number to be reduced in order to reduce.

154.6 There are six problems of restoration. One of them is restoration of a fraction to a fraction. For example, suppose someone said, “By how much must we restore a half so that it gives nine tenths?” We divide the nine tenths, the restored number, by the half, the number to be restored. The result is the required number, which is one and eight tenths.

154.10 The second is restoration of a fraction to a whole number and a fraction. For example, suppose someone said, “By how much must we restore two sevenths and half a seventh so that it gives five and a half?” We work this out as shown above to get the required number, which is fifteen and four tenths.

154.13 The third is restoration of a fraction to a whole number. For example, suppose someone said, “By how much must we restore two thirds of five sevenths so that it gives ten?” We work it out as shown above to get the required number, which is twenty-one.

154.16 The fourth is restoration of a whole number to a whole number and a fraction. For example, suppose someone said, “By how much must we restore five so that it gives ten and four sixths?” We work it out as shown above to get the required number, which is two and a tenth and a third of a tenth.

155.2 The fifth is restoration of a whole number and a fraction to a whole number. For example, suppose someone said, “By how much must we restore four and three tenths and half a tenth so that it gives eight?” We work it out as shown above to get the required number, which is one and seventy-three parts of eighty-seven parts.

155.6 The sixth is restoration of a whole number and a fraction to a whole number and a fraction. For example, suppose someone said, “By how much must we restore three and three fifths less a third of one so that it gives twelve and three fifths?” We work it out as shown above to get the required number, which is three and six sevenths. So understand it and you will succeed, almighty God willing.

155.11 And for reduction there are also six problems. One of them is reduction of a fraction to a fraction. For example, suppose someone said, “By how much must we reduce seven tenths so that it becomes a third?” We work it out as described. You denominate the third, the reduced number, with the seven tenths, the number to be reduced. This results in the required number, which is three sevenths and a third of a seventh.

155.16 The second is reduction of a whole number to a whole number and a fraction. For example, suppose someone said, “By how much must we reduce eight so that it becomes two and a half?” We work it out as shown above to get the required number, which is two eighths and half an eighth.

155.19 The third is reduction of a whole number to a fraction. For example, suppose someone said, “By how much must we reduce ten so that it becomes three fourths?” We work it out as shown above to get the required number, which is three fourths of a tenth.

156.3 The fourth is reduction of a whole number and a fraction to a whole number and a fraction. For example, suppose someone said, “By how much must we reduce seven and a fourth so that it gives three and four sixths?” We work it out as shown above to get the required number, which is fourteen parts of twenty-nine parts and four sixths of a part of twenty-nine parts.

156.7 The fifth is reduction of a whole number and a fraction to a whole number. For example, suppose someone said, “By how much must we reduce eleven and nine tenths and four sevenths of a tenth so that it becomes five?” We work it out as shown above to get the required number, which is thirty-eight parts of ninety-three parts and eight ninths of a part of ninety-three parts.

156.12 The sixth is reduction of a whole number and a fraction to a fraction. For example, suppose someone said, “By how much must we reduce two and a third so that it becomes a ninth?” We work it out as shown above to get the required number, which is three sevenths of a ninth. So know it and manage it.

link to parallel text 157.1 Section Six, on converting.[79]

link to parallel text 157.2 This section covers two kinds. The intent of one kind concerns only the name, such as when it is said “five sixths and three fourths: how many tenths is it?” We want to denominate these two fractions by naming the fraction in tenths. So we work it out as described, resulting in one and five tenths and five sixths of a tenth, which is what we get from gathering these two fractions together after turning them into tenths. So we shifted the problem from naming in terms of sixths and fourths to naming in terms of tenths and its fractions. This is shifting one kind of fraction to another kind, and it is this kind that is intended in the book.

link to parallel text 157.9 The intent of the second kind is: how many of that name, taken as units, are in the whole [fraction]? Work out this kind the same way as for whole numbers. Whenever we want to convert it, we look back to see how it was described in the second type of multiplication.[80]

157.12 Suppose someone said, “Five sixths and three fourths: how many tenths are in it?” We multiply them by the whole number ten, resulting in one and five tenths and five sixths of a tenth, which is the answer. This is the amount that comes about from the tenth, which is fifteen tenths and five sixths of a tenth.[81]

158.1 Unlike for the first kind, this kind [of conversion] does not require division by the denominator of the converted fraction, since it is similar to the case in which someone said, “Five dirhams, how many tenths are in it?” We multiply the five by the ten to get the result of fifty, which is the answer. The same [rule] can be applied to fractions. Since this kind belongs to the section on multiplication, it is not mentioned by the author in the book. He mentioned only the kind specific to the section [on fractions].[82]

link to parallel text 158.7 To work this out, that is, the intended kind, you multiply the numerator of the fraction to be converted by the denominator of the converted fraction. One divides the result first by the denominators of the fraction to be converted and then the result by the denominator of the converted fraction. The advantage of this kind is that one can convert a fraction into a finer fraction.

158.11 For example, suppose someone said, “Six eighths and four tenths, how many ninths are in it?” We put the fraction to be converted on a line, and the denominator of the converted fraction on a line below it, as in this figure:

We multiply the numerator of the fraction to be converted, which is ninety-two, by the nine, the denominator of the converted fraction, resulting in eight hundred twenty-eight, and its figure is 828. We divide it first by the denominators of the fraction to be converted, then what is left by the denominator of the converted fraction, resulting in the required number. This is one and a ninth and three tenths of a ninth and four eighths of a tenth of a ninth, and its figure is {4~\,\,\,3\,~\,1\over 8~\,10~\,9}\,1. The same works for similar [examples].

159.5 This completes the chapter, with God’s blessing.

link to parallel text 161.1 Chapter Three, on roots. Related to this, we cover what we intend on this topic in four sections.

link to parallel text 163.1 Section One, on taking a root of a whole number and a root of a fraction.

link to parallel text 163.2 These are divided into two varieties, rational and surd. A rational [root] is any number whose ratio to one is known. This can be a whole number, a fraction, or a whole number and a fraction.

link to parallel text 163.4 A surd [root] is one whose ratio to one is unknown. For example, a root of ten, a root of a half, and a root of ten and a half. Surds come in two varieties: those that are expressed with [the word] “root” once, like those just mentioned, and which are called rational in square, and those expressed by “root” more than once, like a root of a root of ten, and which are called medial.[83]

163.9 And the root is any number, which multiplied by itself, results in the number whose root is sought. Examples have been given above.

link to parallel text 163.11 In language, it is the origin of everything. He said “root” (jadhr) with a “dh” and either an “a” or an “i” as the vowel for the “j”. Our professor al-shaykh Abū l-ʿAbbās [Ibn al-Bannāʾ], God be satisfied with him, told me that the “a” is more appropriate.

link to parallel text 163.14 The ranks in a whole number “have a root” and “do not have a root” for each successive place. This is evident by examination for the units and tens. The hundreds have a root because they come from multiplying the tens by themselves, and the thousands do not have a root because they are in relation to the hundreds in the position that the tens are in relation to the units, and similarly for what comes after that. A place is said to have a root if there is a number in it that has a root.[84]

link to parallel text 164.3 Some conditions for a number may indicate that it does not have a root. But if they do not hold it only implies that it may have a root. These are:

164.5 Any number that begins with a two or a three or a seven or an eight does not have a root.

164.6 Any number that begins with a one, and half of its tens is different from the number of hundreds in the even and the odd,[85] does not have a root, like three hundred forty-one and four hundred sixty-one and the like.

164.10 Any number that begins with a five, and its tens is not twenty, does not have a root, like seventy-five and one hundred eighty-five and the like.

164.12 Any number that begins with a six, and its tens are even, does not have a root, like forty-six and three hundred twenty-six and the like.

164.14 Any number that does not begin with six, and its tens are odd, does not have a root.[86]

164.15 Indications for non-square numbers, including those given above, mean the same as knowing that the number does not have a root.

164.17 Indications for [the first digit of] a square are five numbers: a one, with the condition that its tens [place] has a half; a five, with the condition that its tens are twenty; and a six, with the condition that its tens are odd. This leaves nine and four. If one of these is the first digit, as mentioned, and its tens are odd, then it does not have a root, and if it is even then it might.

165.3 Any number beginning with an odd number of zeros does not have a root, like ten, twenty-one thousand, three thousand, and the like.

165.5 Any number beginning with an even number of zeros, and for which the number [remaining after the zeros are deleted] does not have a root, [also] does not have a root, like five hundred, thirty thousand, and the like.

165.8 Any number that is not exhausted after casting out nines, and the remainder is not a one or a four or a seven, does not have a root,[87] like four hundred twenty-five and the like.

165.11 Any number that is not exhausted after casting out eights, and the remainder is not a one or a four, does not have a root, like two hundred sixty-six and the like.

165.14 Any number that is not exhausted after casting out sevens, and the remainder is not a one or a two or a four, does not have a root, like three hundred forty-nine and the like.

link to parallel text 166.1 The way to take a root of a whole number is to count off the ranks with “root”, “no root” to the end of the line. Once you arrive at the last “root”, put a number below it so that if you multiply it by itself, it cancels what is above it or it leaves the smallest possible whole number as remainder. Then you back up, double it, and put it below the place [marked] “no root”. Now you look for a number to put below the “root” preceding it, so that when you multiply it by the doubled number one step back and then by itself, it either cancels what is above it or it leaves the smallest possible remainder. Then if it did not cancel, you do it again. You continue doubling the backed-up number and shifting until you have covered all the line. What you then have in the second line, before doubling, is the root.

link to parallel text 166.9 If something remains, denominate it with double the whole part of the root if it is equal to or smaller than the root; and if it is greater than the root, always add one to it, and two to double the root, then you denominate it with it and you add the denomination to the whole part. This gives the root that you multiply by itself, and it is an approximation of a root of the given number.

link to parallel text 166.13 For example, suppose someone said, “How much is a root of six hundred twenty-five?” We put it on a line like this: 625. The first place has a root, the second does not have a root, and the third has a root, as mentioned above. We then look for a number to put below the six, which is in a rank that has a root, such that when we multiply it by itself it either cancels the six or it leaves the smallest possible whole number as remainder. We find that it is two. We multiply it by itself, giving four. We drop it from the six, leaving two, which we put in its place. We back up the doubled two and put it below the place [marked] “no root”, which is below the two. Then we look for a number to put below the [rank that] has a root that is before the backed up number below it, and that is below the five. We find it is five. It cannot be anything else.

166.21 If it were a six, you could only put a four or a six below it, and if it were a one, you could only put a one or a nine below it, and if it were a four, you could only put a two or an eight below it, and if it were a nine, you could only put a three or a seven below it. So know it.

link to parallel text 167.3 So we multiply the five by the doubled four, giving twenty. The twenty-two above its head cancels the twenty, leaving two in its place. Then we likewise multiply it by itself, giving twenty-five, and above its head is twenty-five, which cancels it. The meaning of the five and the doubled two after it is twenty-five, which is the required number. Work it out the same way if there are more digits.

link to parallel text 167.8 Another example: suppose someone said, “How much is a root of twenty?” We know that it does not have a rational root since it begins with one zero. We work it out as explained above, and we divide the remainder according to the calculation described above. We look for a number such that when we multiply it by itself, it cancels it or it leaves the smallest possible remainder. We find it is four. We multiply it by itself, giving sixteen. The remainder is four, which is equal to the root. We denominate it with its double, which gives a half. We add it to the whole part of the root, giving four and a half, which is a root of twenty by approximation.

link to parallel text 167.14 Another example: suppose someone said, “How much is a root of fifty-four?” We work it out as before, to get seven for the whole part of the root and five for the remainder, which is smaller than the root. We denominate it with its double, giving two sevenths and half a seventh. We add it to the root, to get the sum of seven and two sevenths and half a seventh, which is the required root by approximation.

link to parallel text 168.1 Another example: suppose someone said, “How much is a root of ninety-two?” We work it out as before to get nine for the whole part of the root and eleven for the remainder, which is greater than the root. So we add one to it, and two to double the nine, which is the root. We denominate the smaller with the greater and we add it to the root. The result is the required root, which is nine and three fifths. If we multiply the nine and three fifths, which is the approximated root, by itself, it results in ninety-two and four fifths of a fifth. The approximation is found with the added fraction.

link to parallel text 168.8 And if you want to refine the approximation, denominate it with double the root. Drop the result from the root, leaving a root whose square is closer to the number whose root is required than the first square.

link to parallel text 168.10 He said “denominate it”, by which he meant the fraction added to the square that was found by approximation; because taking the root by approximation can be obtained from a previous close smaller square as seen above, and it can also be obtained from a previous close greater square, which is what he meant when he said “denominate it, etc.”. The description of how to work it out is that we drop the number whose root is required from the square, and we denominate the remainder with double a root of the square, and we also subtract the result from a root of the square. The remainder is a root of the number by approximation.

link to parallel text 168.16 If we want to take a root of ninety-two using the previous greater square, we make the greater square ninety-two and four-fifths of a fifth. If we drop the number from it, the remainder will be the fraction found by approximation. So we denominate it with double the root, as mentioned, to get half a sixth of a tenth. We drop it from the root, leaving nine and five-tenths and five-sixths of a tenth and half a sixth of a tenth. A square of this remainder is closer than the first square. So know it.

link to parallel text 169.1 There is another method of approximation, which is that you multiply the number whose root is required by a greater square number, and you take a root of the result by approximation, and you divide by a root of the multiplied square. The result is the approximated root.

169.4 For example, suppose someone said, “How much is a root of twelve?” We multiply it by sixteen, for instance, resulting in one hundred ninety-two. We take its root to get thirteen and six-sevenths. We divide it by a root of sixteen. The result is a root of twelve by approximation, and that is three and three-sevenths and a fourth of a seventh.

link to parallel text 169.8 The condition “the smallest possible whole number”[88] [is needed] because if one diverges from the well-known method by working with fractions, then the remainder will be smaller than the remainder with whole numbers.[89]

link to parallel text 169.10 And suppose someone said, “How much is a root of six hundred twenty-five?” This example is from the earlier problem. We put two and a half below the six, so its square is six and a fourth. This exceeds [the six]. So you take away the six with the six, and take away the fourth with the twenty-five, which is a fourth of a hundred. So all the numbers vanish, and the remainder begins with the last two zeros. We take one of them. Thus the two and a half are tens, which is twenty-five. We double the two and a half to get five, and we place it below the tens. And we look for something to multiply by the doubled number. We find it to be nothing, since there are zeros above it. We put down a zero and we halve what we doubled, so it yields half of fifty.[90]

link to parallel text 169.17 Another example: suppose someone said, “How much is a root of seven hundred twenty-nine?” If we work it out with whole numbers, then the remainder from the seven, which is seven hundred, in its rank is three. And if we work it out with fractions, then the remainder is smaller. So if we put two and a half below it, three fourths of one remains in that rank, and three fourths of a hundred is seventy-five. We add it to the twenty-nine that is with it, to get one hundred four. Then we back up the doubled two and a half, which is five, and we look for a number to multiply by the five and by itself. We find it is two, and nothing remains of the number. So we halve what we doubled, which is fifty. Its half is twenty-five, so the total root is twenty-seven. Or we double the two, so it becomes fifty-four, and we take half of it.[91]

link to parallel text 170.6 Another example: suppose someone said, “How much is a root of a hundred?” The basic rule for this and similar problems is that we always take half of the number of zeros, and we add them to a root of the remaining number to get the root. The hundred begins with two zeros. We take one of them and we add it to a root of the remaining one, giving ten, which is a root of the hundred. So know it.

link to parallel text 170.10 To take a root of a fraction, you multiply the numerator by the denominator and you divide a root of the result by the denominator. If the numerator has a rational root and the denominator does too, then divide a root of the numerator by a root of the denominator.

link to parallel text 170.14 With regard to taking a root, I mean of a fraction, there are four types. In one of them the numerator has a rational root and the denominator does too. Work it out as described.

170.16 For example, suppose someone said, “How much is a root of four-sixths and a sixth of a sixth?” Its figure is 1~\,4\over 6~\,6. We take a root of the numerator, giving five. We divide it by a root of the denominator, which is six. The result is the required root, and that is five-sixths.

171.1 Another example: suppose someone said, “How much is a root of twelve and a fourth?” We take a root of the denominator, giving two. We divide seven, a root of the numerator, by it. The result is the required number, which is three and a half. And if we wish, we can work it out by the first method, since it is general, while this one is particular. So know it.

171.5 In the second [type] neither of them has a rational root, so work it out by the first method.

171.6 For example, suppose someone said, “How much is a root of four-ninths and three sixths of a ninth?” Its figure is 3~\,4\over 6~\,9. We multiply the numerator by the denominator to get one thousand four hundred fifty-eight. We take its root, which is thirty-eight and three parts of nineteen parts and half a part of nineteen parts. We divide it by the denominator. The result is the required number, which is thirteen parts of nineteen parts and three ninths of a part of nineteen parts and five-sixths of a ninth of a part of nineteen parts and half a sixth of a ninth of a part of nineteen parts, and its figure is 1~\,5~\,3~\,13\over 2~\,6~\,9~\,19.

171.13 In the third [type] the denominator has a rational root but the numerator does not have a rational root. For this type, if we wish, we can work it out by the first [method] or by the second.

171.15 For example, suppose someone said, “How much is a root of ten and seven-eighths and half an eighth?” Its figure is {1~\,7\over 2~\,8}10. If we want, we can work it out by the first method. We multiply the numerator by the denominator, resulting in two thousand eight hundred. We take its root, which is fifty-two and forty-eight parts of fifty-three parts and half a part of fifty-three parts. We divide it by the denominator to get the required number, which is three and sixteen parts of fifty-three parts and two eighths of a part and a fourth of an eighth of a part of fifty-three parts, and its figure is {1~\,2~\,16\over 4~\,8~\,53}\,3.

172.6 By the second method, we take a root of the numerator, which is thirteen and three parts of thirteen parts. We divide it by a root of the denominator. The result is the required number, which is three and four parts of thirteen parts, and its figure is {4\over 13}\,3. This method is closer than the first [method].

172.10 In the fourth [type] the numerator has a rational root and the denominator does not have a rational root, so work it out by the first method.

172.12 For example, suppose someone said, “How much is a root of four sevenths and half a seventh?” Its figure is 1~\,4\over 2~\,7. We multiply the numerator by the denominator and we take a root of the result. This is eleven and two parts of eleven parts and half a part of eleven parts. We divide it by the denominator. The result is the required number, which is eight parts of eleven parts and five sevenths of a part and three fourths of a seventh of a part of eleven parts, and its figure is 3~\,5~\,\,8\,\,\over 4\,~7~\,11.

173.2 Of the four types, a root of the first one is found exactly, and the other three are found by approximation. If we wish to approximate [more accurately] the root like we did for whole numbers, then follow the same procedure.

link to parallel text 173.4 To take roots of binomials and apotomes, you drop a fourth of a square of the smaller of the two terms from a fourth of a square of the greater one; you take a root of the remainder, you add it to half of the greater term, and you also subtract it from half of the greater term; and you drop a root on each of them. If the number whose root is required is a binomial, then its root is the sum of these two roots, and if it is an apotome, then its root is the difference between these two roots.

link to parallel text 173.10 I say we need to begin with an introduction to clarify binomials and apotomes and how to find them. After that we will take their roots, almighty God willing.

173.12 We say that there are six binomials and six apotomes.

173.13 A binomial is a number and a root of a number, or a root of a number and a root of a number in which the two can only be joined with the coordinating conjunction.[92] For example, five and a root of three, and a root of five and a root of three.

173.16 An apotome consists of two terms in which the smaller term is removed from the greater by the particle of exclusion.[93] For example, five less a root of three, and a root of five less a root of three.

link to parallel text 174.1 Roots of the first three binomials and apotomes are closer to being rational in rank than roots of the other three. The first three can be distinguished from the others by multiplying the difference between the squares of the terms by a square of the greater one. If the result is a square, then it is one of the first three, and if it is not a square, then it is one of the second three.

174.5 The greater term is rational in the first and the fourth. An example of the first is five and a root of twenty-one, and an example of the fourth is two and a root of two.

174.8 The smaller is rational in the second and the fifth. An example of the second is five and a root of forty-five, and an example of the fifth is five and a root of seventy-two.

174.11 Neither of them is rational in the third and the sixth. An example of the third is a root of ten and a root of eighteen, and an example of the sixth is a root of seven and a root of eight.

link to parallel text 174.14 It is necessary to recall their characteristics if we want to find them. If we subtract a square from a square and the remainder is not a square, and we join a root of the remainder with a root of the greater square, this gives the first binomial. Or we make whatever number we wish the greater term, and [we make] the smaller a root of a surface whose two sides are in numerical ratio,[94] with the condition that it is not rational. However, this procedure is rare.

175.1 We subtract a non-square number from a square such that the remainder is not a square, and we join a root of the remainder with a root of the square to get the fourth binomial.

175.3 We multiply two squares by their difference, which gives non-squares, and we join a root of the greater of the two results with a root of their difference to get the second binomial.

175.5 We multiply two squares by something other than their difference, which gives non-squares, and we join a root of the greater of the two results with a root of their difference to get the third binomial.

175.7 We add a square to a square so that the sum is not a square, and we join a root of the sum with a root of one of the two squares to get the fifth binomial.

175.9 We add a non-square number to a square so that the sum is not a square, and we join a root of the sum with a root of the added number to get the sixth binomial.[95]

link to parallel text 175.11 These are the six binomials. Now let us return to examples of their roots. Example: suppose someone said, “Eight and a root of sixty: how much is a root of that?” We work this out as described. We drop a fourth of a square of a root of the sixty, since it is the smaller, and that is fifteen, from a fourth of a square of the eight, since it is the greater, and that is sixteen. The remainder is one. We take its root, giving one. We add it to half of the eight, since it is the greater, giving five, and we drop it likewise from its half. The remainder is three. We drop a root on the five and the three. This gives a root of five and a root of three, which is the required number. The same rule applies to other [examples].

175.19 And suppose someone said, “Eight less a root of sixty: how much is its root?” We work this out as before. We drop a root of the three from a root of the five. A root of their difference is the required root, and that is a root of the five less a root of the three.

link to parallel text 176.1 Another way to do it is that we drop a square of the smaller term from a square of the greater one, and we take a root of the remainder. We add it to the greater term and we take a root of half of the sum. And likewise, we subtract it from the greater term and we take a root of half of the sum. If the number whose root is required is a binomial, then its root is the sum of these two roots. If it is an apotome, then its root is the difference between the two roots.

176.6 Suppose someone said, “Eight and a root of fifty-five: how much is its root?” This is the first binomial. We take its root as before, to get a root of five and a half and a root of two and a half. It is called “one of the binomials”. Its apotome is the first apotome, and an apotome of its root is a root of its apotome, and it is called “one of the six apotomes”.

link to parallel text 176.10 Suppose someone said, “Seven and a root of one hundred twelve: how much is its root?” This is the second binomial. We take its root as before to get, after the addition and subtraction, a root of a root of eighty-five and three fourths and a root of a root of one and three fourths. This is called “the first bimedial”, and its apotome is the second apotome. An apotome of its root is a root of its apotome, and it is called “a first apotome of a medial”.

link to parallel text 176.15 Suppose someone said, “A root of thirty-two and a root of fourteen: how much is its root?” This is the third binomial. We take its root as before. This gives, after the addition and subtraction, a root of a root of twenty-four and a half and a root of a root of a half. This is called “the binomial of the second bimedial”, and its apotome is the third apotome. An apotome of its root is a root of its apotome, and it is called “a second apotome of a medial”.

link to parallel text 176.20 Suppose someone said, “Seven and a root of thirty: how much is its root?” This is the fourth binomial. We take its root as before, to get three and a half and a root of four and three fourths, taking its root, and three and a half less a root of four and three fourths, taking its root. It is called “the major”, and its apotome is the fourth apotome. And an apotome of its root is a root of its apotome, and it is called “the minor”.

link to parallel text 177.5 Suppose someone said to you, “Three and a root of twenty: how much is its root?” This is the fifth binomial. We take its root as before, to get a root of five and a root of two and a half and a fourth, taking its root, and a root of five less a root of two and a half and a fourth, taking its root. It is called “[the number whose] power is a rational and a medial”. Its apotome is the fifth apotome, and an apotome of its root is a root of its apotome. It is called “the joining with a rational to become a whole medial”.

link to parallel text 177.11 Suppose someone said, “A root of ten and a root of eleven: how much is its root?” This is the sixth binomial. We take its root as before, to get a half and a root of two and three fourths, taking its root, and a root of two and three fourths less a half, taking its root. It is called “[the number whose] power is a bimedial”. Its apotome is the sixth apotome, and an apotome of its root is a root of its apotome. It is called “the joining with a medial to become a whole medial”.

link to parallel text 179.1 Section Two, on adding and subtracting roots of numbers.

link to parallel text 179.2 You multiply the two numbers, one of them by the other, when you want to add or to subtract their roots. If the result is a square, then the roots of the two numbers can be added and subtracted. If it is not a square, then they cannot be added or subtracted. If you know they can be added, take two roots of the result and add it to the sum of the two numbers. Take a root of this sum to get the required number.

link to parallel text 179.7 For example, suppose someone said, “Add a root of three to a root of twenty-seven”. We multiply the three by the twenty-seven, giving eighty-one, which is a square. We take two of its roots, giving eighteen. We add it to the sum of the two numbers, and a root of that sum is the required number, which is a root of forty-eight.

link to parallel text 179.11 Another way is that we divide one of the two addends by the other, we add one to the result, and we multiply the sum by the divisor. The result is their sum.[96] So here we divide a root of the twenty-seven by a root of the three, resulting in three. We add one to it, and we multiply the sum by a root of the three, the divisor, as explained [in the section on] multiplying roots.[97] This results in a root of forty-eight, which is the required number, as shown above.

link to parallel text 179.16 Another example: suppose someone said, “Add a root of two to a root of eight”. We multiply the two by the eight, giving sixteen, and we take two of their roots, giving eight. We add it to the sum of the two numbers, and a root of that sum is the required number, and that is a root of eighteen. If we wish, we can work it out the second way to get the required number.

link to parallel text 179.20 Another example: suppose someone said, “Add half a root of twenty to two roots of five”. Half a root of twenty is less than one root, so we transform it to one root, as described in the section on division.[98] This gives, according to what was explained on working out the multiplication, a root of five. And two roots of five are more than one root, so we transform them to one root to again get a root of twenty. It is as if someone had said, “Add a root of five to a root of twenty”. We work it out as before to get the required number, which is a root of forty-five.

link to parallel text 180.6 Likewise, whenever the roots are of different ranks, we transform them to one rank. For example, suppose the addend is a root of a number rational in square and the augend is a root of a root of a number, that is, a medial. We transform [the number] rational in square to a medial in ratio[99] with the other [addend], at which point we add them.[100]

link to parallel text 180.10 Another example: suppose someone said, “Add a root of three to a root of fifteen”. We find that their surface is not a square, so they are incommensurable. We thus add them with the coordinating conjunction to get a root of three and a root of fifteen. Any examples like this, that can only be added with the coordinating conjunction, are called binomials.

link to parallel text 180.15 Another example: suppose someone said, “Add half a root of a root of eighty to a third of a fourth of a root of six hundred eighty-four”. It is known, according to what is explained in the section on multiplication,[101] that half a root of a root of eighty is a root of a root of five, and that a third of a fourth of a root of six hundred eighty-four is a root of four and three fourths. It is as if someone had said, “Add a root of a root of five to a root of four and three fourths”. We transform them to the same rank, as mentioned before. So the problem becomes as if someone had said, “Add a root of a root of five to a root of a root of twenty-two and four eighths and half an eighth”. Their surface is likewise not a square, so we add them with the coordinating conjunction, and that is a root of a root of five and a root of a root of twenty-two and four eighths and half an eighth. So know it.

link to parallel text 181.4 For subtraction, you subtract two of the roots that result from multiplying the two numbers from the sum of the two numbers, and you take a root of the remainder to get the required number.

181.6 For example, suppose someone said, “Subtract a root of eight from a root of thirty-two”. We multiply the eight by the thirty-two, giving two hundred fifty-six. We drop two of its roots, which are thirty-two, from the sum of the two numbers, and we take a root of the remainder. It is the required number, which is a root of eight.

link to parallel text 181.10 Another way to do this is to divide one of the subtrahends by the other. Then take the difference between the result and one, and multiply by the divisor to get the required number.

181.12 For example: suppose someone said, “Subtract a root of twelve from a root of twenty-seven”. We divide a root of the twenty-seven by a root of the twelve, resulting in one and a half. We take the difference between it and the one, which is a half. We multiply it by a root of the twelve, which is the divisor, resulting in a root of three, which is the required number.

181.16 If we denominate a root of the twelve with a root of twenty-seven, it results in two thirds. We take the difference between it and the one, and that is a third. We multiply it by a root of the twenty-seven, the denominating number, and the result is the required number, which is a root of three. So know it.

link to parallel text 182.1 If the subtrahend or the minuend is more than one root or less, or if the ranks of their roots are different, then it is necessary to transform them to one root or the same rank, as with addition.

link to parallel text 182.4 Another example: suppose someone said, “Subtract a root of eight from a root of ten”. We find that their surface is not a square, so they are incommensurable. So subtract it using the particle of exclusion, and that is a root of ten less a root of eight. Work out [other problems] similarly. So know it. Here, too, any examples like this one that can only be subtracted using the particle of exclusion are called apotomes. So understand it.

link to parallel text 183.1 Section Three, on multiplying roots.

183.2 To work this out you multiply one of the numbers by the other and you take a root of the result. This is the result of multiplying a root of one of them by a root of the other.

link to parallel text 183.4 For example, suppose someone said, “Multiply a root of eight by a root of nine”. We multiply the eight by the nine, and a root of the outcome is the required number, which is a root of seventy-two.

link to parallel text 183.7 Another example: suppose someone said, “Multiply a root of a root of five by a root of a root of seven”. We multiply the five by the seven, and we drop a root of the root, which the two multiplicands had, on the result. This gives the required number, which is a root of a root of thirty-five. The rule is similar for other medials, no matter how far from rational they are.

183.11 Another example: suppose someone said, “Multiply a root of a root of a root of three by a root of a root of a root of eight”. We multiply the eight by the three, and we drop a root of a root of the root on the result to get the required number, and that is a root of a root of a root of twenty-four.

link to parallel text 183.15 Another example: suppose someone said, “Multiply three by a root of seven less two”. To work this out we multiply the three by a root of the seven, and we subtract from the outcome its multiplication also by the excluded two. The remainder is the required number, which is a root of sixty-three less six. The principle behind multiplying appended and deleted terms will be covered in [the chapter on] algebra,[102] with the power of almighty God.

link to parallel text 183.20 If you want to multiply a number by a root of a number, square the number and work with the two numbers as shown above.

184.1 For example, suppose someone said, “Multiply three by a root of seven”. We multiply the three by itself, giving nine. Then it is as if someone had said, multiply a root of nine by a root of seven. We work it out as before, resulting in the required number, which is a root of sixty-three.

link to parallel text 184.4 Another example: suppose someone said, “Multiply two by a root of a root of three”. We multiply the two by itself, and the result by itself. This is multiplied by the three. A root of a root of the outcome is the required number, which is a root of a root of forty-eight. The rule is similar if there are more [roots] than these. So know it.

184.8 From this principle, I mean his saying “if you want to multiply a number by a root of a number, etc.”, one knows how to work it out by transforming the problem to one root when there is a term with more than one root or less than one root, and by duplicating a root of a number or partitioning it.

link to parallel text 184.11 Here is an example with a root that is more than one root. Suppose someone said, “Multiply two by two roots of seven”. It is necessary that we focus on what [makes] this number of a root of the seven a [single] root. To do this we multiply the two, which is the number of the root, by itself, and the outcome by the seven. We take a root of the result, which is the required number, and that is a root of twenty-eight. So it is as if someone had said, “Multiply two by a root of twenty-eight”. We work it out as before, resulting in the required number, which is a root of one hundred twelve.

link to parallel text 184.18 Another example: suppose someone said, “Multiply five by three roots of a root of two”. So we likewise focus on what number makes three roots of a root of two a [single] root of a root. To do this we multiply the three, the number of roots, by itself, and the outcome by itself, and the outcome of that by the two, and we take a root of a root of the result. This is the required number, which is a root of a root of one hundred sixty-two. So it is as if someone had said, “Multiply five by a root of a root of one hundred sixty-two”. We work it out as before to get the required number, which is a root of a root of one hundred one thousand two hundred fifty. Work it out similarly if there are more [roots] than these.

link to parallel text 185.6 Here is an example with less than one root: suppose someone said, “Multiply two thirds by half a root of twenty”. So we focus on half a root of twenty, and we change it into what [single] root it is, as before. To do this we multiply the half by a root of the twenty, so it gives, as explained, a root of the five. It is then as if someone had said, “Multiply two thirds by a root of five”. We again work it out as before. The result is the required number, which is a root of two and two ninths.

link to parallel text 185.12 Another example: suppose someone said, “Multiply a root of five by half a root of a root of forty”. We focus on half a root of a root of forty, and we change it into what root of a root it is, as before. To do this we multiply the half by itself, and the outcome by itself, and the outcome of that by the forty. We get, as we have explained, a root of a root of two and a half. It is then as if someone had said, “Multiply a root of five by a root of a root of two and a half”. We work it out again as before, resulting in the required number, which is a root of a root of sixty-two and a half. Work out similar [examples] the same way.

link to parallel text 186.1 Here is an example of duplicating roots: suppose someone said, “Duplicate a root of three twice”. It is as if someone had said, “Multiply two by a root of three”. Work it out as before to get the required number, which is a root of twelve.

186.4 Another example: suppose someone said, “Duplicate a root of seven five times”. It is as if someone had said, “Multiply five by a root of seven”. Work it out as before to get the required number, which is a root of one hundred seventy-five. Work out similar [examples] the same way.

link to parallel text 186.8 An example of partitioning roots: suppose someone said, “How much is half of a root of ten?” It is as if someone had said, “Multiply a half by a root of ten”. Work it out as before to get the required number, which is a root of two and a half.

186.11 Another example: suppose someone said, “How much is a third of four eighths of a root of a root of sixty?” It is as if someone had said, “Multiply a third of four eighths by a root of a root of sixty. Work it out as before to get the required number, which is a root of a root of two sixths of a ninth and half a sixth of a ninth. So know it.

link to parallel text 187.1 Section Four, on dividing and denominating roots of numbers.

link to parallel text 187.2 You divide the number by the number or you denominate it with it, and you take a root of the result. What you get is the result of dividing a root of the dividend by a root of the divisor.

187.4 For example, suppose someone said, “Divide a root of twenty by a root of three”. We divide the twenty by the three, and we drop the root on the result to get the required number, which is a root of six and two thirds.

187.7 Another example: suppose someone said, “Divide a root of three by a root of eight”. We denominate the three with the eight, and we drop the root on the result to get the required number, which is a root of three eighths.

link to parallel text 187.10 Work it out similarly for medials. For example, suppose someone said, “Divide a root of a root of six by a root of a root of two”. We divide the six by the two, and we drop a root of the root on the result to get the required number, which is a root of a root of three.

187.14 Another example: suppose someone said, “Divide a root of a root of eighteen by a root of a root of thirty-two”. We denominate the eighteen with the thirty-two, and we drop a root of the root on the result to get the required number. This is a root of a root of four eighths and half an eighth. The same rule applies to other medials, no matter how far from rational they are.

link to parallel text 188.1 In these [preceding] three sections, on addition, multiplication, and division, whenever you encounter a term that is more than one root or less than one root, or roots of different ranks, transform them to one root or make them the same rank.

188.4 I say I have already introduced examples of these and how to work them out in the sections on addition and multiplication, but let us also present some examples in this section.

link to parallel text 188.6 Along these lines, suppose someone said, “Divide a root of a root of fourteen by a root of two”. The divisor is rational in square and the dividend is medial. So we transform a root of the two so that it is medial like the divisor, and then we divide. We get a root of a root of four. If we divide the dividend by it and we drop a root of the root which the two dividends had on the result, it gives the required number, and that is a root of a root of three and a half.

link to parallel text 188.11 Another example: suppose someone said, “Divide two roots of fifteen by two”. We already knew that two roots of fifteen are the same as a root of sixty, and that the two is a root of four. So it is as if someone had said, “Divide a root of sixty by a root of four”. We work it out as before to get the required number, which is a root of fifteen.

188.15 Another example: suppose someone said, “Divide half a root of twenty-four by a root of two”. We already knew that half a root of twenty-four is the same as a root of six. So it is as if someone had said, “Divide a root of six by a root of two”. We work it out as before to get the required number, which is a root of three.

link to parallel text 188.18 For division by binomials and apotomes, you multiply the dividend and the divisor by an apotome of the divisor if it is a binomial, or by its binomial if it is an apotome. Then you divide the result of the dividend by the result of the divisor.

189.1 For example, suppose someone said, “Divide twelve by five and a root of three”. We multiply the twelve, the dividend, by the five and a root of three, an apotome of the divisor, and we distribute across the appended and the deleted terms, as will be made clear in [the chapter on] algebra,[103] to get sixty less a root of four hundred thirty-two. This is the result for the dividend. Then we divide it by the result of multiplying the five and a root of three, the divisor, by the five less a root of three, its apotome, and that gives twenty-two, since any binomial multiplied by its apotome, or vice versa, results in the difference between the squares of the terms. The result from the division is the required number, which is two and eight parts of eleven parts less a root of nine parts of eleven and nine parts of eleven parts of eleven.

189.11 Another example: suppose someone said, “Divide ten by three less a root of seven”. We multiply the ten, the dividend, by the three and a root of seven, the binomial of the divisor, to get thirty and a root of seven hundred. We divide it by the result of multiplying the three less a root of seven, the divisor, by its binomial, and that is two. The result of the division is the required number, which is fifteen and a root of one hundred seventy-five. So know it.

189.16 This completes Chapter Three, with the blessing and help of God.

link to parallel text 191.1 Part Two, on the basic rules by which one arrives at knowledge of the required unknown from the posited known.

191.4 This is divided into two chapters: a chapter on working out [problems] with proportion, and a chapter on algebra.

link to parallel text 193.1 Chapter One, on working out [problems] with proportion. This can be done in two ways, with four proportional numbers and with scales.

link to parallel text 195.2 I say there are different kinds of proportions. Among them are arithmetical, harmonic, harmony, and ex-aequali proportion. For arithmetical [proportion], its principles have already been covered,[104] and for the other three, they are not described.[105] As mentioned, they are dispensable, since they originate from it, and it is the foundation of calculation. All three derive from it and it does not derive from them, as was shown in Lifting the Veil.[106]

link to parallel text 195.7 Four proportional numbers are those for which the ratio of the first to the second is as the ratio of the third to the fourth, and the product of the first by the fourth equals the product of the second by the third.

link to parallel text 195.9 Whenever you multiply the first by the fourth and you divide by the second, it results in the third; or by the third, it results in the second. And whenever you multiply the second by the third and you divide by the first, it results in the fourth; or by the fourth, it results in the first. Whichever is unknown can be worked out from the other three known numbers.

link to parallel text 195.14 The way to do this is that you multiply the isolated number different in kind from the others by the number in ratio with the unknown, and then you divide by the third number, resulting in the unknown.

link to parallel text 195.16 For example, the ratio of three to six is as the ratio of four to eight. So the three with respect to the six is a half, and the four with respect to the eight is a half. The product of the first, which is the three, by the fourth, which is the eight, equals the product of the six, which is the second, by the four, which is the third.

195.19 If we multiply the three by the eight and we divide it by the six, it results in the four; or by the four, it results in the six. And similarly, if we multiply the six by the four and we divide by the three, it results in the eight; or by the eight, it results in the three.

196.3 Suppose in this example that the fourth is unknown, which is the eight, and we want to find it. So we work it out as mentioned. We multiply the isolated number – different in kind from the other two, namely in its classification, which is what is intended, which is discussed in this book,[107] and which in this example is six, since it is related to it, and thus the remaining two are related – by the number in ratio with the unknown, which is the four, since the fourth proportional is unknown. This gives twenty-four. We divide it by the third of the three remaining [numbers], which is the first, resulting in eight, which is the unknown fourth [number]. If the first is unknown, we divide the twenty-four by the eight to get the three.

196.10 If the second is unknown, which is the six, then we multiply the eight – which is the isolated number, since it is the one related to it, thus the remaining two are related – by the number in ratio with the unknown, which is the first, since the second is the unknown in this proportion, giving twenty-four. We divide it by the four, which is the third of the remaining three. The result is six, which is the unknown second. If the third is unknown, which is the four, we divide the twenty-four by the six, resulting in the four. So know it.

link to parallel text 196.16 For the four proportional numbers, if one switches to get the ratio of the first to the third and the second to the fourth; or reverses them to get the ratio of the second to the first and the fourth to the third; or combines to get the ratio of the sum of the first and the second to one of them is as the ratio of the sum of the third and the fourth to one of them; or separates to get the ratio of the difference between the first and the second to one of them is as the ratio of the difference between the third and the fourth to one of them; or combines after switching them; or separates after switching them; or switches after combining them; or switches after separating them; or reverses after any one of these, they will remain proportional.[108]

link to parallel text 197.4 There are four ways to find the unknown other than those mentioned. First, if the fourth, for example, is unknown, we divide the second by the first, and we multiply the result by the third, to get the fourth. Second, we divide the third by the first, and we multiply the result by the second, to get the fourth. Third, we divide the first by the second, and we divide the third by the result, to get the fourth. Fourth, we divide the first by the third, and we divide the second by the result, to get the fourth. These are not presented by the author because what he presented is the source of all of them. They derive from it, and it does not derive from them. As the jurist Abū Muḥammad ʿAbd al-Ḥaqq ibn Ṭāhir reported, the first way is called the procedure, and the other four are called deductions by analogy.

link to parallel text 198.2 The [method of] scales comes from the art of geometry. It comes from the art of geometry because the ratio of the error of each scale to the difference between the scale and the unknown number is as the ratio of the assigned number to the unknown.[109]

link to parallel text 198.4 You switch, you separate, you switch. The ratio of the portion to its scale is as the ratio of the assigned number to the unknown, as was made clear in Lifting the Veil.[110]

link to parallel text 198.7 You represent this by drawing a balance, as in this figure:[111]

198.8 You place the given assigned number above the dome. You choose any number you wish for one of the scales, and you perform the prescribed additions, reductions, and so forth from among the operations. Then you confront it with what is above the dome. If you got it right, then this scale is the unknown number. If you got it wrong, then write the error above the scale if it exceeds, or below if it falls short.

198.13 Then you choose for the other scale any number you wish other than the first [number], and you work it through as you did with the first [scale]. Then multiply the error of each scale by the posited number[112] in the other [scale]. Then look. If the two errors both exceed or both fall short, then subtract the smaller from the greater, and the smaller of the two products from the greater. Then divide the remainder from the two products by the remainder from the two errors. And if one of them exceeds and the other falls short, you divide the sum of the products by the sum of the errors, resulting in the required number.

link to parallel text 199.1 For example, suppose someone said, “A quantity: taking away its third and its fourth leaves ten. How much is the quantity?”

199.2 We draw the balance as described, and we place the given ten above the dome. Then we choose fifteen for the first scale. We place it between the lines of the balance. Then we take its third and its fourth, which are eight and three fourths, and we drop it from it, leaving six and a fourth, which is the portion that we confront with what is above the dome. So we place it inside the balance, next to the scale. The remainder from the ten above the dome, after the confrontation, is three and three fourths. This is the error for the scale, which falls short. We place it below the scale as mentioned. If the portion confronted with it were, for instance, ten, then the scale would be the unknown.

199.9 We choose twelve for the second scale. We also place it inside the balance, on the other side, and we take its third and its fourth, giving seven. We subtract it from it, leaving five, which is also the portion we confront it with. We place it inside the balance, next to the scale. The remainder from the ten after the confrontation is likewise five, which is the error for the second scale, and it falls short. We place it below the scale, as in this figure:

199.14 Then we multiply the three and three fourths, the first error, by the twelve, the second posited number, to get forty-five, which is one of the two products. Then we multiply the five, the second error, by the fifteen, the first posited number, to get seventy-five, which is the second product. From this we drop the first product, which is smaller, since the two errors fall short, leaving thirty. We keep it in mind. Then we drop the first error, since it is smaller, from the second error, since it is greater. The remainder is one and a fourth. We divide the remembered number by it. The result is the unknown quantity, which is twenty-four.

link to parallel text 200.1 Another example: suppose someone said, “A quantity: we take the sum of its third and its fifth, and we add to it half of the remainder, so it comes to twenty-three. How much is the quantity?”

200.3 We draw the balance similarly and we place the twenty-three above its dome. We choose forty for one of the scales. We take its third and its fifth, and we add half of the remainder to it, as mentioned, to get thirty and two-thirds. This is the portion we confront with what is above the dome. The error is seven and two-thirds, which is an excess. We choose forty-five for the second scale. We take its third and its fifth and half of the remainder to get thirty-four and a half, which is the portion we also confront. The error is eleven and a half, which exceeds. So we place it and the first [error] above the scales, to get this figure:

200.10 Then we multiply the seven and two-thirds, the error of the first scale, by the forty-five, the second posited number, to get three hundred forty-five, which is the first product. Doing the reverse[113] gives four hundred sixty, which is the second product. From this we drop the first product, which is the smaller, since the two errors exceed. The remainder is one hundred fifteen. We divide it by the difference between the two errors, which is three and five sixths. The result is the unknown quantity, which is thirty.

link to parallel text 201.1 Another example: suppose someone said, “A quantity: we add its tenth to the difference between its fourth and three of its fifths, so it comes to nine. How much is the quantity?”

201.3 We draw the balance similarly, and we choose twelve for one of the scales, and we drop its fourth from three of its fifths. The remainder is four and a fifth. To this we add a tenth of the scale, to get the sum five and two-fifths, which is the portion that we confront with what is above the dome. The error is three and three fifths, falling short. We put it below the scale. Then we choose twenty-five for the second scale. We again drop its fourth from three of its fifths, leaving eight and three fourths. To this we add a tenth of the scale to get the sum eleven and a fourth, which is the portion that we also confront. The error is then two and a fourth, which exceeds. We put it above the scale to get this figure:

201.10 Then we multiply the first error by the second posited number, giving ninety. Doing the reverse gives twenty-seven. We add these products, since one of the errors exceeds and the other falls short, to get the sum one hundred seventeen. We divide it by the sum of the two errors, which is five and four fifths and a fourth of a fifth. The result is the unknown quantity, which is twenty. So know it.

link to parallel text 201.14 If you wish, you can choose the first number or any other number for the second scale, then figure its portion, which you confront with what is above the dome. Multiply it by the first posited number, and multiply the first error by the second posited number. If the first error falls short, you add the two products. If it exceeds, you take their difference. You divide the result by the portion of the second scale to get the required number.

link to parallel text 202.3 This second method is used only when there is a proportional relation. For example, suppose someone said, “Ten: you divide it into two parts so that a third of one of them is a fourth of the other”.[114]

202.5 We draw the balance similarly, and we put the ten above the dome. Then we choose two numbers for the scale so that a third of one of them is a fourth of the other. They are, for example, three and four. We confront their sum, since it is the portion, with the ten. The error is three, which falls short. Then we choose [numbers for] the second scale the same way, such as six and eight. We figure its portion, which is the sum of the numbers, to get fourteen. This gives the following figure:

202.10 Regardless of which part we want to find, we multiply the first error by the upper number of the second, and the portion of the second by the upper number of the first. Then we continue with the procedure. The result is the required part, and the remainder from the ten is the other part. Suppose in this example that we want the smaller part. We multiply the three, the first error, by the six, the upper number of the second, to get eighteen. Then we multiply the fourteen, the second portion, by the three, the upper number of the first, to get forty-two. We add it with the first product, since the error falls short, to get sixty. We divide it by the fourteen, the second portion. The result is the required number, which is four and two-sevenths. The remainder from the ten is the greater part, which is five and five sevenths.

link to parallel text 203.1 Another example: suppose someone said, “Ten: you divide it into two parts. You divide the greater by the smaller, so the result is four. How much is one of them?”

203.3 One of the parts is necessarily four times the other. So we draw the balance similarly, and we place the ten above the dome. Then we choose for one scale two numbers so that one of them is a fourth of the other. They are, for example, three and twelve. We confront their sum with the ten, since it is the portion. The error is five, which exceeds. Then we make our choice for the second scale. Here we choose the same as the first. We figure its portion, which is fifteen. This gives the following figure:

203.8 We work it out as described in the previous example, to find that one of the parts is two, and the other is the remainder from the ten, which is eight. So know it. If we wish, we can work out these two problems by the first method.

link to parallel text 203.11 We follow the second method in any problem where the assigned numbers are one of the scales and its error. For example, I say, “A quantity: we subtract its third and its fourth from a third of sixty and its fourth, leaving fourteen. How much is the quantity?”

203.15 The sixty is one of the scales, and the fourteen is its error, which exceeds. We choose whatever number we wish for the other scale, and we find its third and its fourth, which is the portion that we would confront with what is above the dome, if there were a number there. We work it out as mentioned, so the quantity is thirty-six.

204.1 Thus, the method of scales derives from the four proportional numbers. You know that approaches for working it out come from relating the two errors and from the differences of the two scales when combined and separated.[115]

link to parallel text 204.3 Regarding this, my professor, the jurist and great erudite Abū l-ʿAbbās [Ibn al-Bannāʾ], God bless him, dictated to me while I was studying with him on Wednesday, the twenty-eighth of the month of rajab in this year:

link to parallel text 204.6 “There are three approaches. One of them is that you multiply the difference between the scales by one of the errors. If the two errors both exceed or both fall short, you divide the product by their difference. If one of them exceeds and the other falls short, you divide the product by their sum. You add the result of this to the scale whose error you multiplied if it falls short, and you subtract it from it if it exceeds. This gives the required number.

link to parallel text 204.12 “The second approach is that you multiply the difference between the scales by the sum of the errors if they both exceed or both fall short, and you divide by their difference. If one of them exceeds and the other falls short, you multiply the difference between the scales by the difference between the errors, and you divide by the sum of the errors. Keep this result in mind. If you wish, you add this remembered number to the difference between the scales and you take half of the sum. You add it to the scale whose error is greater if it falls short, and you subtract it from it if it exceeds, to get the required number.

204.18 “And, if you wish, take the remembered number and the difference between the scales, subtract the smaller from the greater, and take half of the remainder. Add it to the scale whose error is smaller if it falls short, and subtract it from it if it exceeds, to get the required number.

link to parallel text 205.1 “The third approach is that you multiply the difference between the scales by your assigned number. If the errors both exceed or both fall short, you divide the product by their difference, and if one of them exceeds and the other falls short, you divide the product by their sum, to get the required number. So know it.”

link to parallel text 205.5 Scales can also be set up to find the unknown in cases without a proportional relation, as in this problem: Three men want to buy a horse.[116] The first says to the second, give me half of what you have, and, together with what I have, I will have the price of the horse. The second says to the third, give me a third of what you have, and, together with what I have, I will have the price of the horse. And the third says to the first, give me a fourth of what you have, and, together with what I have, I will have the price of the horse.

205.11 We choose one scale for the three men. We assign to the first man whatever we wish, so we make it four. The second also has whatever we wish, so we make it two. Then the price of the horse is calculated to be five. We put it above the dome, and it is what we will confront. The third man has, also by calculation, nine. If we add a fourth of what the first has, it amounts to ten. The error of the first scale with these three numbers is five, which exceeds. Then we choose [the numbers in] the other scale. We assign four for the first man, the same as we assigned for him in the first [scale], and we make the second whatever we wish. We see that it should not be eight or more, since it would lead to the third having nothing. So know it. So we make it six. This makes the price of the horse, which we confront, seven. We put it above the dome also. The third necessarily has three, and adding this to a fourth of what the first has gives four. The error of the second scale is then three, which falls short. This is the figure:

206.1 We multiply the error of each scale by what each one has in hand in the other scale, and we divide the sum of the two products as mentioned, to get what each one has in hand and the price of the horse.[117] This gives us what the first has in hand, which is the four, the second four and a half, the third five and a fourth, and the price of the horse is six and a fourth.

link to parallel text 206.5 If we wish the result to be without fractions, we multiply all of the problem by the smallest number divisible by their denominators. We knew [how to do] this before,[118] and it is four. By this calculation the first has sixteen, the second has eighteen, the third has twenty-one, and the price of the horse is twenty-five. So understand it.

206.9 If we wish, we can change what is assigned for the first [man] in the first scale, and leave unchanged what is assigned for the second, since the condition is that one of them should have the same number in both scales.

link to parallel text 207.1 If we [choose to] assign the price of the horse, we put it above the dome, and we put some portion of it for the first [man], and twice its remainder for the second. Then we drop a fourth of what we put for the first from the price of the horse, leaving what the third has. Then we take what the second has and a third of what the third has, and we confront it with the assigned price. We then do this likewise for the other scale: we first make the price of the horse whatever we wish, as long as it is not the first number. Then we work it out as before.

link to parallel text 207.6 Another problem: forty birds, among which are geese, chickens, and starlings, [all] for forty dirhams. The starlings are eight for a dirham, the chickens are one for two dirhams, and the geese are one for three dirhams. How many were chosen of each kind of bird?

link to parallel text 207.9 Not all examples of this type can be worked out. Two conditions must be satisfied. One is that the numbers must be whole numbers and not fractions. Second, the price of one of the cheapest [birds], if multiplied by the number of birds, must be smaller than the total price; and the price of one of the most expensive [birds], if multiplied likewise, must be greater than the [total] price.

207.13 It is obvious in this problem that the number of starlings must be eight or sixteen or twenty-four or thirty-two, and no other [number]. If there are eight, then there remain thirty-two birds and thirty-nine dirhams. If we check the second condition, the product of the remaining individual birds by the price of the cheapest one is greater than the number of the [total] price, which is not valid. If we make the starlings sixteen, and we again check the remaining birds, then the remaining price is also not valid.

208.1 If we make the starlings twenty-four, and we again check what remains, the two conditions are met. So we suppose there are twenty-four starlings, and we assign the chickens to be whatever we wish. Suppose they are eight. Then the geese are eight, the remaining number. The error in the price is three dirhams, which exceeds. Then we choose [the numbers in] the other scale. We make the starlings twenty-four, as in the first [scale], since it is a condition for it to work out that a number be repeated in the two scales. And we make the chickens be whatever we wish that is different from the first, so let them be fourteen. Then the number of geese is two. The error is three dirhams falling short. Here is the figure:

208.8 We work it out as before to get the required number of each kind of bird and the price of each kind, whichever we want to find first.[119]

208.10 The price of the starlings is three, and their number is twenty-four. The geese are five and their price is fifteen, and the chickens are eleven and their price is twenty-two.

link to parallel text 208.12 If we make the starlings thirty-two, it is not valid since the condition for the remaining [birds] is not met. Thus this problem has only one answer. For similar problems, look to these two problems. Problems like this cannot be worked out by the second method of solving by scales, since that is specific to a proportional relation, as said above. Problems involving multiplication are not based in proportion, so they cannot be solved by the scales. So know it.[120]

208.17 Chapter One is completed, with the praise of God and His help.

link to parallel text 209.1 Chapter Two, on algebra. We present this work in five sections.

link to parallel text 211.1 Section One, on the meaning of algebra (restoration and confrontation) and an explanation of its types.[121]

link to parallel text 211.2 Restoration is the reconstitution we mentioned in Part One of this book.[122] Confrontation is subtracting each species from like [species] until there are no longer two species of the same kind in the two parts.[123] And equalization is the restoration of the deleted to the appended, and the subtraction of the appended from the appended and the deleted from the deleted of things of the same kind. This will be covered with clear examples in the section on addition and subtraction,[124] almighty God willing.

link to parallel text 211.8 Algebra depends upon three species: the number, the things, and the māls. The things are roots, since every unknown in numbers is a thing, and a root of its square when it is known. The māl is what one gets from multiplying the root by itself. The term māl serves to distinguish it from the other [species].

link to parallel text 211.13 These three species are equated to one another, in either a simple or composite way. This results in six types, three simple and three composite.

211.15 The first of the simple [types], following convention, is: some māls equal some roots. For example, three māls equal seven things. The second is: some māls equal a number. For example, five māls equal twenty. The third is: some roots equal a number. For example, three roots equal twelve.

212.6 In the first of the three composite [types], which is the fourth type, the number is isolated. For example, a māl and ten roots equal twenty-four. In the fifth, the roots are isolated. For example, a māl and four equal five roots. And in the sixth, the māl is isolated. For example, a māl equals four roots and five.

link to parallel text 213.1 Section Two, on working out the six types.

link to parallel text 213.2 For the three simple [types], you divide by the māls what they are equal to, and by the roots if there are none. The result of the division in the first and third types is the root, and in the second it is the māl. If the root is known, then the māl is known by multiplying the root by itself. And if the māl is known, then the root is known from it.

link to parallel text 213.7 For example, suppose someone said, “Three māls equal fifteen things”. The meaning of this problem is: what quantity (māl), if we take its root fifteen times, gives a sum equal to three times the quantity? This is of the first type, and it is worked out as explained above. If we divide the fifteen, the number of things, by the three, the number of māls, it results in five, which is a root of the unknown māl, the latter being twenty-five.

link to parallel text 213.13 Another example: suppose someone said, “Two māls equal eighteen”. The meaning of this problem, likewise, is: what quantity (māl), if we add itself to it, becomes equal to eighteen? This is of the second type. To work it out we divide the eighteen by the two, the number of māls, as mentioned, to get nine. This is the unknown māl, and its root is three.

link to parallel text 214.1 Another example: suppose someone said, “Five things equal twenty”. The meaning of this problem, likewise, is: what quantity (māl), if we take five of its roots, gives a result equal to the twenty? This is of the third type. To work it out we divide the twenty by the five, the number of things, since there are no māls. It results in four, which is a root of the unknown māl, and that is sixteen.

link to parallel text 214.7 To work out the fourth type, you halve the number of roots, you square the half, you add it to the number, you take a root of the result, and you drop the half from it, leaving the root.

link to parallel text 214.9 For example, suppose someone said, “A māl and two things equal fifteen”. Its meaning is, what quantity (māl), if we add two of its roots to it, becomes equal to fifteen? We take half of the things, giving one, and we square it, giving one, and we add the number to it, giving sixteen. We take its root, giving four. From this we drop the one, which is the half, leaving three. This is the unknown thing, and the māl is nine.

link to parallel text 214.14 If we wish to find the māl first, before the root, we add half a square of the number of roots to the number, then keep the sum in mind. Then we subtract a square of the number from a square of the remembered number, and we take a root of the remainder. We subtract this from the remembered number. The remainder is the māl.

214.17 If we want to work it out in the example, we add half a square of the number of things, which is two, to the fifteen, giving seventeen. We keep it in mind. Then we square the number to get two hundred twenty-five. We subtract it from a square of the remembered number, which is two hundred eighty-nine, leaving sixty-four. We take its root, giving eight. We subtract it from the seventeen, leaving nine, which is the required māl.

link to parallel text 215.3 For the fifth [type], you subtract the number from a square of half of the number of roots, and you take a root of the remainder. If you add it to the half, it gives a root of the greater māl, and if you subtract it, it gives a root of the smaller māl.

215.6 For example, suppose someone said, “A māl and eight equal six things”. Its meaning is, what quantity (māl), if we add eight to it, gives a result equal to six of its roots? We take half of the things and we square it, giving nine. We drop the number from it, leaving one. We take its root, giving one. If we add it to the half, it gives four, which is a root of the greater māl, so the greater māl is sixteen. If we subtract it from it, it leaves two, which is a root of the smaller māl, and the māl is four.

link to parallel text 215.12 Know that whenever a square of the half is equal to the number, then the half is the root, and the māl is the number.

link to parallel text 215.14 For example, suppose someone said, “A māl and nine equal six things”. We square half of the things, giving nine, and this is equal to the number. So the number is the māl, and the half is the root. And if we continue the solution, we subtract the number from the square, leaving nothing. We take its root, giving nothing. We add nothing to the half, or we subtract it from it, leaving the half. It is the root, and its square is the number. So know it.

link to parallel text 216.1 If we wish to find the māl first, before the root, we subtract the number from half of a square of the number of roots, and we keep the remainder in mind. Then we subtract a square of the number from a square of the remembered number. If we add a root of the remainder to the remembered number, it gives the greater māl; and if we subtract it from it, then the remainder is the smaller māl. Note that this is what it is when the number is smaller than a square of half of the number of roots.

216.6 If we want to work this out in the preceding example, we subtract the number from half of a square of the number of roots, which is eighteen, leaving ten. We keep it in mind. Then we subtract a square of the number, which is sixty-four, from a square of the remembered number. The remainder is thirty-six. We take its root, giving six. If we add it to the remembered number, we get sixteen, the greater māl. And if we subtract it from the remembered number, then the remainder is four, the smaller māl. So know it.

link to parallel text 216.11 The sixth [type] is solved like the fourth, except that you add the half at the end to a root of the sum to get the root.[125]

216.13 For example, suppose someone said, “A māl equals two of its roots and three”. Its meaning is, what quantity (māl) is equal to two of its roots and three? So we square half of the things and we add it to the number, giving four. We take its root, giving two. We add the half to it to get three, which is the unknown thing, and the māl is nine.

link to parallel text 216.17 If we wish to find the māl first, before the root, we add double the number to a square of the roots, and we keep half of the sum in mind. Then we drop a square of the number from a square of the remembered number, and we add a root of the remainder to the remembered number. This gives the māl.

216.20 If we want to work it out in the example, we add double the number, which is six, to a square of the roots, giving ten. We keep in mind its half, which is five. Then we drop a square of the number from a square of the remembered number. The remainder is sixteen. We add its root, which is four, to the remembered number. This gives the unknown māl, which is nine.

link to parallel text 217.1 For each of the three composite types, whenever there is more than one māl, reduce it to one māl, and reduce by that term all of the equation. And for each of them, whenever there is less than one māl, restore it to one māl, and restore by that term all of the equation. The way to work out restoration and reduction was described earlier.[126] And if you wish, divide [each of] the names[127] in the problem by the number of māls. This reduces the problem.[128] Then confront the two sides.

217.8 He indicated this only for the composite [equations], since the simple [equations] do not require restoration if there is less than one māl, nor reduction if there is more than one māl.

link to parallel text 217.10 For example, suppose someone said, “Two māls and six things equal thirty-six”. We reduce the two māls to one māl by multiplying them by a half, as mentioned above. And we reduce by this [multiplication] the things and the number. So the problem becomes: a māl and three things equal eighteen, which is the fourth type. This is also how to work out the fifth and sixth [types].

link to parallel text 217.15 And if we work it out the other way, we divide the two, the number of māls, by itself, resulting in one. And we also divide the number of things by it, resulting in three. And we also divide the number by it. So the problem becomes: a māl and three things equal eighteen, just as before.

link to parallel text 218.1 Another example: suppose someone said, “Half a māl and two things equal six”. We restore the half to one māl by multiplying it by two. We multiply it also by the things and the number. So the problem becomes: a māl and four things equal twelve, which is also the fourth type. This is also how to work out the fifth and sixth [types].

218.6 And if we work it out the other way, we divide the half by itself, resulting in one. And we also divide the number of things by it, resulting in four. We also divide the number by it, resulting in twelve. So the problem becomes: one māl and four things equal twelve, just as before. So understand it.

link to parallel text 219.1 Section Three, on addition and subtraction.

link to parallel text 219.2 Adding different species is done with the conjunction “and”, like: a māl and six things and ten dirhams.[129]

link to parallel text 219.4 …and the different excluded [species] cannot be subtracted.

link to parallel text 219.5 For example, suppose someone said, “Add a māl less a thing to ten dirhams”. The sum is: a māl and ten dirhams less a thing. So the exclusion remains as it was, since [the thing] cannot be subtracted from anything.[130]

link to parallel text 219.7 Whenever appropriate, subtract the smaller from the greater, that is, in the problem. For example, suppose someone said, “Add two māls less a māl to ten dirhams”. So we subtract the excluded māl from the two māls, so the remaining sum is a māl and ten dirhams.

219.10 Another example: suppose someone said, “Add a māl less two things to ten things”. We subtract the two things from the ten things. The remaining sum is a māl and eight things.[131]

219.12 Another example: suppose someone said, “Add a māl less five things to ten things”. We subtract the five things from the ten things. The remaining sum is a māl and five things.[132]

link to parallel text 220.1 Subtracting different species is done using the particle of exclusion. For example, suppose someone said, “Subtract a thing from a māl”, then the remainder is a māl less a thing.

220.3 Another example: suppose someone said, “Subtract ten dirhams from two māls and three things”. The remainder is two māls and three things less ten dirhams.

link to parallel text 220.5 The exclusion can be in both parts[133] or in one of them, and it might be one species or two different species. To work it out, you add the exclusion in each part to both parts at once, and then you subtract. Work out the two sides of an equation similarly whenever they have exceptions.

link to parallel text 220.9 For example, suppose someone said, “Subtract two and a thing from a māl less three things”. We find the things on the side of the minuend to be deleted, and they are three. We add them to the two parts at once. This is what is meant by equalization with restoration, since we added to the māl what it is associated with, which are the three things. Thus we restore it from deleted to appended. It becomes greater, as is desired for the minuend. We [also] equalize by restoring the subtrahend by the amount we added in the minuend, which is by the three things, since subtracting two numbers is the same as subtracting them after having added a number to both of them, or after subtracting a number from both of them. The problem is as if someone had said: “Subtract two and four things from a māl”. Then we work it out as before. The remainder after that is the required number, which is a māl less four things and less two. So know it.

link to parallel text 221.1 Another example of this: suppose someone said, “Subtract fifty-two dirhams less five things from two cubes and thirty dirhams”. We find the things on the side of the subtrahend to be deleted, and they are five. We add them to the two sides at once, as before. The problem becomes as if someone had said: “Subtract fifty-two dirhams from two cubes and five things and thirty dirhams”. Then we find dirhams on both sides. We remove the smaller quantity from the two sides at once. The problem becomes as if someone had said, “Subtract twenty-two dirhams from two cubes and five things”.

link to parallel text 221.8 This is confrontation with equalization, since when we removed the thirty from the subtrahend it becomes smaller than the original subtrahend. So we equalize, by which we remove from the minuend the same amount we removed from the subtrahend. It is worked out like before. We work the remainder as before. What remains after that is the required amount, which is two cubes and five things less twenty-two dirhams.

link to parallel text 221.13 Another example: suppose someone said, “Subtract twelve dirhams less four things from three māls less two things”. We find two deleted things in the minuend, and also four deleted things in the subtrahend. We add them. Since they are of the same species, they amount to six things. We add them to the two parts at once, or we restore both the subtrahend and the minuend by what is excluded from them, and we add their equals to the other [part], as before. The problem becomes as if someone had said, “Subtract twelve dirhams and two things from three māls and four things”. So we confront and subtract. The remainder after that is the required number, and that is three māls and two things less twelve dirhams.

link to parallel text 222.1 Another example: suppose someone said, “Subtract a cube less two māls from thirty dirhams less four things”. We restore and subtract. The remainder is the required number, which is thirty dirhams and two māls less a cube and less four things.

link to parallel text 223.1 Subsection concerning examples with the two sides of an equation.

link to parallel text 223.2 Along these lines, suppose someone said, “A māl less three things equal two and a thing”. We find the things on the side of the māl to be deleted, and they are three. We restore the māl by them, and we also add them to the two sides of the equation[134] as we did before, since whenever one adds equals to equals or subtracts equals from equals, the outcomes are equal. So the problem becomes: a māl equals two and four things, which is the sixth type.

link to parallel text 223.7 Another example: suppose someone said, “A māl less three things equal twenty-four dirhams less five things”. We restore and confront as before, so the problem becomes: a māl and two things equal twenty-four dirhams, which is the fourth type. If we wish, we can first subtract the three things which are deleted on the side of the māl from the five things deleted on the side of the number. This leaves two things deleted on the side of the number. And we finish working it out, again arriving at the fourth type.

link to parallel text 223.14 Another example: suppose someone said, “A māl less ten dirhams equal a māl less two things and a half”. We restore and confront, so the problem becomes: ten dirhams equal two things and half a thing, which is the third type.

link to parallel text 223.17 Another example: suppose someone said, “A māl and ten dirhams equal fifty-one dirhams less four things”. We restore, and the problem becomes: a māl and four things equal forty-one dirhams, which is the fourth type.

224.1 Another example: suppose someone said, “A māl and five things equal ten dirhams and two māls less a thing”. We restore and confront, so the problem becomes: six things equal a māl and ten dirhams, which is the fifth type.

224.4 Another example: suppose someone said, “Two māls and ten dirhams less two things equal three māls and seven dirhams less seven things”. We restore and confront, so the problem becomes: a māl equals three dirhams and five things, which is the sixth type. So know it. Work out similar problems the same way, with the help of God.

link to parallel text 225.1 Section Four, on multiplication and knowing the power and the term.

225.2 For the power, know that the power of the things is one, the power of the māls is two, and the power of the cubes is three. As for the term, the term for one is “things”, the term for two is “māls”, and the term for three is “cubes”. After that, there are three for each cube and two for each māl. The cube is the surface of the thing by the māl. It is called this way because it is a cube, although you do not know its amount.

225.8 Suppose someone said, “What is the power of a māl māl ?” We say four. And suppose someone said, “What is the power of a māl cube?” We say five. Suppose someone said, “What is the power of a māl māl māl ?” We say six. And suppose someone said, “What is the power of a māl cube māl cube?” We say ten. To work these out, we always count each māl as two, since that is its power, as mentioned. Likewise, each cube is always three, since that is its power, as mentioned. The result is the associated power.

225.13 And suppose someone said, “What is the power of a māl cube māl māl ?” We would say nine. Suppose someone said, “What is the power of a cube māl cube cube māl māl ?” We would say fifteen. Work it out this way for other combinations, too.

link to parallel text 226.1 Conversely, suppose someone said, “What is a term for four?” We would say a māl māl. And if someone said, “What is a term for seven?” We would say a cube māl māl. And if someone said, “What is a term for six?” We would say a māl māl māl, or a cube cube.

226.3 We can also work this out by separating the powers by twos or threes or by grouping them: first the māl, and then the cube, and then the māl and the cube. Suppose someone said, for example, “What is a term for eight?” We would say a māl māl māl māl. Or if we wish, we can say a cube māl cube, or a cube cube māl, or a māl cube cube, or anything else that is allowable.

226.7 Suppose someone said, “What is a term for nine?” We would say a cube cube cube, or if we wish, we can say a cube māl māl māl, or if we wish, we could continue or stop. [It can be] anything that is allowable. So know it.

link to parallel text 226.9 To multiply these species, add the power of the multiplicand and the power of the multiplier. Then the sum of the powers is the power of the result. To multiply a number by one of these species, the result is the same species.

link to parallel text 226.12 For example, suppose someone said, “Multiply five things by seven things”. We multiply the number of the multiplicand, which is five, by the number of the multiplier, which is seven, giving thirty-five. Then we add the powers of the two multiplicands, giving two. This two is the power of the thirty-five resulting from the multiplication, so it is māls. So the result of the multiplication is thirty-five māls, which is the required number.

226.17 Another example: suppose someone said, “Multiply ten things by six māls”. We multiply the number of the multiplicand by the number of the multiplier, which gives sixty. Then we add the powers of the two multiplicands, giving three. We make it the power of the result, which you recall is sixty. This is the required number, which is sixty cubes.

227.4 Another example: suppose someone said, “Multiply a thing by a cube”. The result of multiplying the number of the multiplicand by the number of the multiplier is one. The sum of their powers is four, which is the power of the one, the result of the multiplication. So it is a māl māl.

227.7 Another example: suppose someone said, “Multiply six by four māls”. We multiply the multiplicand by the number of the multiplier, giving twenty-four. The outcome of the multiplication in this example is māls, so it is then twenty-four māls.

link to parallel text 227.10 Another example: suppose someone said, “Multiply seven by three māls cube”. We multiply the multiplicand by the number of the multiplier, giving twenty-one. The outcome of the multiplication in this example is likewise māl cube, so it is then twenty-one māls cube. So know it.

link to parallel text 227.14 Whenever you equate between the māls māls and the cubes and the māls, or the cubes and the māls and the things, or any combination in which you do not have a number, subtract the smaller of the powers from the power of each of them. Then you equate what remains, one with the others, in the same way as the equation.

227.17 For example, suppose someone said, “Three māls māl equal four cubes and ten māls”. We find the power of the māls to be the smallest power in the equation, and it is two. If we drop it from the power of the māls māls, which is four, there remains two, and its term is māls. We also drop it from the power of the cubes, which is three, leaving one, and its term is a thing. And we reduce the māls to a number. The problem becomes: three māls equal four things and ten dirhams, which is the sixth type.

228.1 Another example: suppose someone said, “Three cubes equal ten māls and twenty things”. We find the power of the things to be the smallest power in the equation, and we work it out as before. The problem becomes: three māls equal ten things and twenty dirhams, which is also the sixth type.

228.4 Another example: suppose someone said, “A cube and ten māls equal thirty-nine things”. We work it out as before. The problem reduces to: a māl and ten things equal thirty-nine dirhams, which is the fourth type. Work out similar examples the same way.

link to parallel text 228.8 If it is not possible to transform it to [one of] the six types, then do not work it out since it would be of no use for anything.

link to parallel text 228.9 The multiplication of two appended or deleted [terms], one of them by the other, is appended, and the multiplication of an appended [term] by a deleted [term] is deleted.

228.11 For example, suppose someone said, “Multiply five things by thirteen less four things”. We multiply the five things by the appended thirteen, and we drop from the result the product of the five things again by the deleted four things. The remainder, then, is the required number, which is sixty-five things less twenty māls.

link to parallel text 228.15 Another example: suppose someone said, “Multiply eight less two things by seven less four māls”. We multiply the appended eight by the appended seven, to get an appended [amount]. We add to it the product of the two deleted things by the four deleted māls, since it is appended. And we drop from the sum the product of the appended eight by the deleted four māls, since it is deleted, and the product of the deleted two things by the appended seven, since it is deleted. The remainder, then, is the required [number], and that is eight cubes and fifty-six less fourteen things and less thirty-two māls. So know it.

link to parallel text 229.1 Section Five, on division.

link to parallel text 229.2 To divide one of these species by a lower species, drop the power of the divisor from the power of the dividend. What remains is the power of the species of the result of the division.

229.4 For example, suppose someone said, “Divide ten māls by two things”. We divide the number of māls by the number of things, and we call the resulting five by the name for the difference between the power of the things and the māls, and that is a thing. The result of the division is five things, which is the required number.

link to parallel text 229.8 Another example: suppose someone said, “Divide fifteen cubes by three things”. We divide the number of cubes by the number of things, and we call the resulting five by the name for the difference between the power of the things and the cubes, and that is a māl. The resulting five is then five māls, which is the required amount. So know it.

link to parallel text 229.12 And whenever you divide a species by itself, the result is a number.

229.13 For example, suppose someone said, “Divide twelve māls by three māls”. We divide the number of the divisor by the number of the dividend, resulting in four. There is no difference between the powers,[135] so we call the term[136] of the resulting four “number”, most certainly.

229.16 Whenever you divide one of these species by a number, the result is the same species.

link to parallel text 229.17 For example, suppose someone said, “Divide twelve things by four dirhams”. We divide the number of the dividend by the number of the divisor, resulting in three. And the number of the divisor has no power, so that dropping from the power of the dividend leaves the power of the dividend as the power of the result, so we call it with its term. The result is three things, which is the required number.

link to parallel text 230.1 If there is an exclusion in the dividend, divide both the excluded [term] and the diminished [term] by the divisor, and exclude the result of the excluded [term] from the result of the diminished [term]. The outcome of this is the result of the division.

230.4 For example, suppose someone said, “Divide twelve cubes less three māls by two things”. We divide the twelve cubes, which is the diminished [term], by the two things, which is the divisor. And the exclusion in this division is the three māls, which is the excluded [term that we] likewise [divide] by the two things. The remainder is the required number, which is six māls less a thing and half a thing.

230.8 Another example: suppose someone said, “Divide ten māls less three things by two dirhams”. We work it out as before. The result is five māls less a thing and half a thing, which is the required number.

link to parallel text 230.11 One cannot divide the lower of two species by the higher except by cancelling their common divisor, by which you subtract the power of the smaller from the powers of both of them.

230.13 For example, suppose someone said, “Divide six māls by three cubes”. We divide the number by the number. Then the result is divided by the difference between the two ranks, which is a thing. The result is two divided by a thing.

link to parallel text 230.16 And one cannot divide by a diminished [term]. So understand it and figure it out with the help of God.

230.17 For example, suppose someone said, “Divide ten māls by three less a thing”. So we say the result is ten māls divided by three less a thing. The answer is the same as the question. So understand it.

link to parallel text 231.1 Section [on secret numbers].

231.2 Let us conclude this work with three problems of witty reckoning, like those with which arithmeticians still conclude their compositions.

link to parallel text 231.4 One of them is that we tell someone to drop his [secret] number from ten, then to drop a square of the remainder from a square of his number. If [a square of] the remainder is smaller, we ask for the remainder. We divide it by ten, and we add ten to the result and then take half of the sum[137] to get the secret number. And if a square of the remainder is greater, he drops a square of the secret number from it, and we ask for the remainder. We divide it by the ten, and we drop the result from ten. Half of the remainder is the secret number. If we wish, we can ask that he drop his secret number from something other than the ten. Following the procedure will give the required number.

link to parallel text 231.13 The second problem. We tell someone to divide the ten into two parts secretly. Then we tell him to divide a square of one of them by their surface, and we ask for the outcome. If we knew this, it is the ratio of one of the two parts to the other. So we divide the ten according to this ratio. And similarly, we can work with whatever number we wish other than the ten, dividing it secretly. This results in the two secret parts.

link to parallel text 232.1 The third problem. A secret number is divided into two secret numbers. How much is it? And how much are each of its parts? We tell him to multiply one of the parts by the other and to square each of them, then to drop a square of the smaller from a square of the surface, and we ask for the remainder; and to drop the surface from a square of the greater, and we ask for the remainder. We take a root of the difference between the two asked numbers, which is the difference between the parts. Then we divide the sum of the asked numbers by it to get the secret number, which is the sum of the two parts. If we add to it a root of the difference between them, it gives double the greater, and if we subtract it from their sum, it leaves double the smaller. So know it.

232.9 These three problems were dictated to me by my professor, the great jurist Abū l-ʿAbbās [Ibn al-Bannāʾ], may God be pleased with him.

link to parallel text 233.2 As the servant [of God], the named, the tempted, ʿAbd al-ʿAzīz ibn ʿAlī ibn Dāwud al-Hawārī al-Miṣrātī, may God have mercy on him, said: With the praise of God for His excellent assistance, I have achieved what I set out to accomplish to the extent of my ability, and not absolving myself of my errors and of anything leading to wrong ideas, I beg Almighty God’s protection and to preserve me from loosing His favors. He is sufficient for us, and He is the best disposer of our affairs.

This work was completed on Saturday, the eighteenth of Dhū al-Qaʿda, 704. May God’s generosity benefit the author. Full prayers upon His prophet and servant, our Master and Lord Muḥammad, and upon his family and companions and followers.

Footnotes

[1]  The work on the sieve begins at 127.10.

[2]  I.e., calculating a cube root.

[3]  Ibn al-Bannāʾ.

[4]  Copied from (Ibn al-Bannāʾ 1994, 212.9-10).

[5]  Literally, “Number has twelve simple names from which all of its names are formed”.

[6]  From here to the end of the section we write the numbers in the order they are written in Arabic, and with the stated plurals of “thousand”. Unlike thousands, hundreds are not made plural when there is more than one because the numbers “two hundred”, “three hundred”, up to “nine hundred” are single compound words in Arabic. After that, we switch back to the English order, and without the plural.

[7]  They are only units, since there is only a 9.

[8]  I.e., a number expressed with a repetition of the word “thousands”.

[9]  Literally, “tens thousands thousands thousand” and “hundreds thousands thousands thousand”.

[10]  Copied from Ibn al-Yāsamīn, (Zemouli 1993, 131.2-5).

[11]  Copied from (Ibn al-Bannāʾ 1994, 214.3-5).

[12]  From 76.7 below.

[13]  From 78.1 below.

[14]  Copied from (Ibn al-Bannāʾ 1994, 214.13-15).

[15]  From 79.1 below.

[16]  Copied from (Ibn al-Bannāʾ 1994, 214.18-215.1).

[17]  Copied from Ibn al-Yāsamīn, (Zemouli 1993, 131.5-8).

[18]  Copied from (Ibn al-Bannāʾ 1994, 226.5-7).

[19]  Copied from Ibn al-Yāsamīn, (Zemouli 1993, 131.9-14).

[20]  Copied from Ibn al-Yāsamīn, (Zemouli 1993, 131.15-16).

[21]  I.e., the units in a particular rank are the tens for the rank before it.

[22]  Copied from Ibn al-Yāsamīn, (Zemouli 1993, 131.17).

[23]  Ibn al-Bannāʾ copied 76.7-12 from Ibn al-Yāsamīn, (Zemouli 1993, 132.1-6).

[24]  I.e., double the number of squares, like from the fourth square to the eighth square.

[25]  I.e., the sum of their indexes.

[26]  At 104.1 below.

[27]  This sentence is copied from Ibn al-Yāsamīn, (Zemouli 1993, 136.6).

[28]  Copied from Ibn al-Yāsamīn, (Zemouli 1993, 136.7-9).

[29]  Copied from (Ibn al-Bannāʾ 1994, 220.8-12).

[30]  This passage belongs in the Condensed Book, but it is not in Souissi’s edition. It is copied from Ibn al-Yāsamīn, (Zemouli 1993, 136.9-12).

[31]  Copied from Ibn al-Yāsamīn, (Zemouli 1993, 136.13-14).

[32]  Copied from (Ibn al-Bannāʾ 1994, 228.13-15).

[33]  The Medina and Oxford manuscripts show \begin{matrix}{3074}\\[-.5mm]6543\\[-.5mm]3469\end{matrix}.

[34]  Literally, “working out”.

[35]  Copied from (Ibn al-Bannāʾ 1994, 245.8-15).

[36]  Copied from (Ibn al-Bannāʾ 1994, 245.16-247.2).

[37]  I.e., the nine, called “ninety” here because it is positioned one place forward.

[38]  I.e., the section on multiplying fractions, at 149.1.

[39]  Copied from (Ibn al-Bannāʾ 1994, 248.11-17).

[40]  Copied from (Ibn al-Bannāʾ 1994, 248.18-249.4).

[41]  Copied from (Ibn al-Bannāʾ 1994, 249.5-11). Ibn al-Bannā gives a section on numeration in Lifting the Veil at p. 272.14. Al-Hawārī covers numeration beginning at 135.8.

[42]  Copied from (Ibn al-Bannāʾ 1994, 255.3-13).

[43]  We translate as “last digit” what is more literally “what is in the last rank”.

[44]  This is misstated. The multiplier, not the three, is shifted.

[45]  Copied from Ibn al-Yāsamīn, (Zemouli n.d., 15.2-9).

[46]  The reference is to (Ibn al-Bannāʾ 1994, 215.12).

[47]  Above at 71.6.

[48]  The three is already there, so there is no need to write it.

[49]  I.e., divide it by 10.

[50]  Copied from Ibn al-Yāsamīn, (Zemouli n.d., 16.5-7).

[51]  Copied from Ibn al-Yāsamīn, (Zemouli n.d., 16.8-13).

[52]  Copied from Ibn al-Yāsamīn, (Zemouli n.d., 16.14-17.2).

[53]  Copied from Ibn al-Yāsamīn, (Zemouli n.d., 17.3-7).

[54]  Copied from Ibn al-Yāsamīn, (Zemouli n.d., 17.8-10).

[55]  Copied from Ibn al-Yāsamīn, (Zemouli n.d., 17.11-13).

[56]  Copied from Ibn al-Yāsamīn, (Zemouli n.d., 17.13-15).

[57]  Copied from Ibn al-Yāsamīn, (Zemouli 1993, 119.13-14).

[58]  Copied from (Ibn al-Bannāʾ 1994, 263.3).

[59]  Copied from Ibn al-Yāsamīn, (Zemouli 1993, 119.16).

[60]  Copied from (Ibn al-Bannāʾ 1994, 263.4).

[61]  Copied from (Ibn al-Bannāʾ 1994, 263.4-9).

[62]  Copied from (Ibn al-Bannāʾ 1994, 263.10-264.8).

[63]  Manipulations of proportions are covered starting at 196.16.

[64]  Copied from (Ibn al-Bannāʾ 1994, 266.15-20).

[65]  Copied from (Ibn al-Bannāʾ 1994, 267.1-5).

[66]  Copied from (Ibn al-Bannāʾ 1994, 267.11-14).

[67]  The text mistakenly has “a square of the dividend”.

[68]  On prime numbers, at 66.7.

[69]  This sentence continues at 137.11.

[70]  Copied from (Ibn al-Bannāʾ 1994, 272.14-273.19).

[71]  Copied from (Ibn al-Bannāʾ 1994, 275.8-9).

[72]  The coordinating conjunction is the word “and” (wa).

[73]  Copied from (Ibn al-Bannāʾ 1994, 275.14-16).

[74]  Copied from (Ibn al-Bannāʾ 1994, 276.3-5).

[75]  Copied from (Ibn al-Bannāʾ 1994, 276.16-277.2).

[76]  At 123.3 above.

[77]  Copied from (Ibn al-Bannāʾ 1994, 277.4-6).

[78]  Copied from (Ibn al-Bannāʾ 1994, 279.15).

[79]  Everything from 157.2 to 158.6 is slightly condensed from (Ibn al-Bannāʾ 1994, 279.18-280.20).

[80]  At 95.6 above.

[81]  Copied from (Ibn al-Bannāʾ 1994, 279.18-280.12).

[82]  Copied from (Ibn al-Bannāʾ 1994, 280.16-20). Al-Hawārī added “by the author” and changed Ibn al-Bannāʾ’s “I mentioned” to “he mentioned”.

[83]  Copied from (Ibn al-Bannāʾ 1994, 283.3-7).

[84]  Copied from (Ibn al-Bannāʾ 1994, 283.8-11).

[85]  I.e., the parity of half of the tens digit is different from that of the hundreds digit.

[86]  Copied from (Ibn al-Bannāʾ 1994, 284.14-285.1).

[87]  Copied from (Ibn al-Bannāʾ 1994, 285.2-285.6).

[88]  From 166.3 and 166.16 above.

[89]  Copied from (Ibn al-Bannāʾ 1994, 283.11-12).

[90]  Copied from (Ibn al-Bannāʾ 1994, 284.4-9).

[91]  Copied from (Ibn al-Bannāʾ 1994, 283.13-284.3).

[92]  The coordinating conjunction is the word “and” (wa).

[93]  The particle of exclusion is most often the word “less” (illā).

[94]  I.e., they are commensurable.

[95]  Copied from (Ibn al-Bannāʾ 1994, 287.19-288.19).

[96]  Judging by the corresponding passage at 181.10, this sentence should have been attributed to Ibn al-Bannāʾ. It is not in the Condensed Book, so it should be in small caps.

[97]  At 183.1 below.

[98]  At 187.1 below.

[99]  I.e., commensurable.

[100]  Judging by the passage at 182.1, this passage should have been attributed to Ibn al-Bannāʾ. It is not in the Condensed Book, so it should be in small caps.

[101]  At 186.11 below.

[102]  At 228.9 below.

[103]  As in the example at 228.11.

[104]  At 73.17 above.

[105]  I.e., in Ibn al-Bannāʾ’s Condensed Book.

[106]  (Ibn al-Bannāʾ 1994, 293.17). By “as mentioned” he means “as mentioned in Lifting the Veil”. See our commentary at 195.2.

[107]  Just above, at 195.14. Ibn al-Bannāʾ does not discuss it in Lifting the Veil.

[108]  Copied from (Ibn al-Bannāʾ 1994, 294.3-9).

[109]  Copied from (Ibn al-Bannāʾ 1994, 297.17-18).

[110]  (Ibn al-Bannāʾ 1994, 301.4ff).

[111]  This is the only figure in Ibn al-Bannāʾ’s Condensed Book.

[112]  Literally “the whole [number]”, i.e., the number in its entirety before performing operations and taking the difference with the number above the dome. We also translate this as “posited number” in the passages at 199.14, 200.10, 201.10, and 201.14.

[113]  That is, the same calculation with the scales reversed.

[114]  Copied from (Saidan 1986, 557.7).

[115]  Copied from (Ibn al-Bannāʾ 1994, 298.13-19).

[116]  The word dābba means a “riding animal”, such as a horse, mule, or donkey. We translate it as “horse”.

[117]  Copied from (Ibn al-Bannāʾ 1994, 299.1-14).

[118]  At 122.2 above.

[119]  Copied from (Ibn al-Bannāʾ 1994, 299.15-300.17).

[120]  Copied from (Ibn al-Bannāʾ 1994, 300.18-22).

[121]  The “types” are the six types of equations, described beginning at 211.13.

[122]  At 129.1 for whole numbers, and at 154.1 for fractions.

[123]  I.e., in the two parts of a subtraction, or in the two sides of an equation.

[124]  At 219.1 below.

[125]  In the Condensed Book this is placed just after the passage at 214.7.

[126]  At 129.1 and 154.1 above.

[127]  By “names” (here alqāb) he means the terms of the equation.

[128]  Literally, “what results is a returning to the problem”.

[129]  Copied from (Ibn al-Bannāʾ 1994, 313.3).

[130]  Copied from (Ibn al-Bannāʾ 1994, 313.3-5).

[131]  Copied from (Ibn al-Bannāʾ 1994, 313.6-9).

[132]  Copied from (Ibn al-Bannāʾ 1994, 313.10-11), except that Ibn al-Bannāʾ solves “add a māl less ten things to five things”.

[133]  That is, in the minuend or the subtrahend.

[134]  He should add them only to the side with two and a thing, because the act of restoration already adds them to the māl.

[135]  Literally, “there is nothing between the powers”.

[136]  Literally, “we name the name”, with ism.

[137]  This sentence is our reconstruction of what appears to be a corrupt passage. Literally, it reads “We divide it by ten, and what results, we add to it half of its remainder to the ten”. Only the Oxford manuscript attempts to correct it, by writing “We divide it by ten, and what results, we add to the remainder from the ten”. This is mathematically correct, but we cannot add the remainder from the ten because that is not revealed to us. This MS also has “square” scratched in above “remainder”.

Commentary on al-Hawārī’s Commentary

link to parallel text 59.1 [Introduction]

A copyist wrote the introductory words “In the name of God…said:”. Al-Hawārī first praises God and Muḥammad, and then Abū Yaʿqūb, Marinid ruler of Morocco from 1286 to 1307. Last, praise is given to Abū Muḥammad ʿAbdallah Ibn Abū Madyan, an intellectual and an important government minister from 1302 to 1307.

link to parallel text 59.15 Al-Hawārī asked permission from Ibn al-Bannāʾ to write this commentary, which Ibn al-Bannāʾ granted. The latter had already written a commentary on his own book, titled Lifting the Veil from the Face of the Operations of Arithmetic. What the Condensed Book and Lifting the Veil lacked, according to al-Hawārī, were ample numerical examples of the rules.

link to parallel text 61.1 Four of the manuscripts we consulted show the phrase “may God forgive him”, which agrees with manuscripts of the Condensed Book. The Medina manuscript has instead “may God preserve his splendor, his reputation, and keep his memory whole”. This version implies that Ibn al-Bannāʾ is already deceased.

link to parallel text 61.9 Part I. Known numbers.

link to parallel text 63.1 Chapter I.1. Whole numbers.

link to parallel text 65.1 Section I.1.1. The divisions of numbers and their ranks.

link to parallel text 65.2 Euclid defined “number” in the beginning of Book VII of the Elements as “a multitude composed of units”.[1] Because these units are indivisible, Euclidean numbers are restricted to positive integers. This is in contrast with the numbers of practical Arabic arithmetic, which include fractions and irrational roots. Al-Fārisī, a Persian contemporary of Ibn al-Bannāʾ, gives the definition of number of “the arithmeticians” as “a quantity you obtain from one by repetition or partition or both, and it is clear by this meaning that the type is divided into whole numbers and their fractions”.[2] Ibn a-Bannāʾ condenses and combines the Greek and Arabic definitions: “A number is a collection of units, and it is divided according to how it is produced into two kinds: whole and fractional”. Later, he made a philosophical apology for fractional numbers in Lifting the Veil in which he claimed that Euclid’s definition is not really a definition at all, but merely an expression of “what is in the soul”.[3] Later commentators entered into the discussion, including al-Mawāḥidī (ca. 1382), Ibn Qunfudh (1370) and Ibn Ghāzī (1483).[4]

Al-Hawārī will explain the various ways of representing fractions in Chapter 2, beginning at 131.1. The examples he gives here are, in the notation described later in the book, 1\over 2, 3~\,1\over 8~\,2 (equivalent to our 7\over 16), {1\over 4}~{1\over 9} (our 13\over 36), 6~\bullet ~7\over 7~\bullet ~8 (our 3\over 4), and {1\over 9}~\scalebox {-1}[1]{\ell }~{5\over 6} (our 13\over 18).[5]

link to parallel text 65.6 Reading right to left, the first place is the units place. Zero was not considered to be a digit, but was instead a sign indicating an empty place in the representation of a number.

link to parallel text 65.10 Evenly-even numbers are the powers of two starting with 2, and evenly-odd numbers are the double of an odd number \ge 3. Evenly-evenly-odd numbers are numbers in between: they are the product of some power of two \ge 4 by an odd number \ge 3. The classification of even numbers into evenly-even, evenly-odd, and evenly-evenly-odd comes from Greek number theory. These definitions are taken from Nicomachus’s Arithmetical Introduction, translated into Arabic by Thābit ibn Qurra in the late ninth century.[6]

Euclid’s definitions of these terms in Elements Book VII are different, and do not correspond to the standard Greek definitions. He defines a number to be “even-times even” if it is the product of two even numbers, and “even-times odd” if it is the product of an even number by an odd number. This way a number can be both, like 28=14\cdot 2=7\cdot 4. Also, Euclid has nothing corresponding to the “evenly-evenly-odd”.

Note the wording “each of the sixteens”. When we take half of 32, the result is a single number, 16. For al-Hawārī, halving 32 means to partition it into a pair of 16s. Numbers in medieval Arabic arithmetic admit multiplicity. See our comments at 163.2 below.

link to parallel text 66.7 The word we translate as “prime” is awwal. This word also means “first, foremost”, etc., and so is close to our “prime” or “primal”. Ibn al-Bannāʾ and al-Hawārī also use the word aṣamm (“deaf”) to mean “prime”, though other arithmeticians, like al-Uqlīdisī and al-Baghdādī, use that word to mean “irrational”. The use of the word “deaf” for “prime” has to do with Arabic ways of expressing fractions, which we explain below at 134.2. We describe the words for “irrational” at 163.2. A prime number can also be called basīṭ (“simple”), in contrast to composite numbers, which are formed from more than one prime.

An “oddly-odd” number is the product of two odd numbers, both \ge 3. This time Euclid, not Nicomachus, is probably the source. Definition VII.10 in Elements reads “An odd-times odd number is that which is measured by an odd number according to an odd number”, i.e., it is the product of two odd numbers. Nicomachus has no such definition. He calls composite odd numbers simply “composite” (sýnthetós, translated as murakkab by Thābit).[7]

The word for “composite” in Ibn al-Bannāʾ and al-Hawārī is similarly murakkab, or some related form. (Sometimes we translate these words as “composed” or “composition”.) The associated verb rakiba serves to multiply the factors together to produce the number composed of those factors. See below at 196.16 and 211.13 for related forms of this word in the context of proportions and algebra.

link to parallel text 66.13 Al-Hawārī presents another classification of numbers. All numbers are either prime or composite, and composite numbers come in three types: (a) perfect squares, (b) products of two or more different numbers, and (c) perfect cubes. He does not mention that some numbers fall into more than one category. The number 36, for example, is a square ( 6^2) and a product of two different numbers ( 4\cdot 9). The classification of composite numbers is explained for even numbers starting at 66.17, then again for odd numbers at 67.17.

We have translated the adjective majdhūr as “has a root”. A number “has a root” if its square root is rational, like 9 or 16\over 25. We could have translated it as a single word like “rootable”, but that seemed too awkward. Later, first at 163.14, we will encounter the same concept for ranks. A rank is a majdhūra if a number of that rank can have a root. This is true for the units, the hundreds, ten thousands, and every other rank after that.

link to parallel text 66.17 The words ḍilʿ (“side”), musaṭṭaḥ (“surface” or “plane”), and murabbaʿ (“square”) are geometric terms used in arithmetic in a metaphorical sense. They derive from the corresponding Greek words in the number theory books VII to IX in Euclid’s Elements, and are first encountered in Definition 16 at the beginning of Book VII. They do not imply any underlying geometric conception of number.

link to parallel text 67.12 The word mukaʿʿab can mean a geometrical cube, or, as here, it can be an arithmetical term meaning “[perfect] cube”, like 27 or 8\over 125. The related word kaʿb in the present passage means “cube root”, but it is also the name given to the third power of the unknown in algebra (first encountered at 221.1), which we translate as “cube”. The meanings are clear by the context.

link to parallel text 68.11 Perhaps al-Hawārī is thinking of the fact that “root” and “cube root” are particular to dimensions 2 and 3 respectively, while “side” is the term for the n-th root for any particular n.

To “decompose” (ḥalla) a number means to express it as the product of two or more numbers. The opposite operation is to “compose” (rakiba) two or more numbers into their product.

Ibn al-Bannāʾ and al-Hawārī explain a rule for extracting square roots numerically beginning at 166.1. Similar rules for extracting cube roots were well known in their time, but al-Hawārī remarks here that going through the work is “of little benefit”. Instead, he briefly explains how to find the cube root of a perfect cube by factoring. For example, to find the cube root of 216, one can break it down as 3\times 72, then to 3\times 9\times 8, and then to 2\times 2\times 2\times 3\times 3\times 3. Then one can piece it together as 6\times 6\times 6, so 6 is the cube root.

link to parallel text 68.18 Ibn al-Bannāʾ uses a few different words to explain the base ten system for writing numbers with the Indian figures:

Rank (martaba). This word indicates the position of a digit in a number. The Arabic word suggests a ranking of the digits, as al-Hawārī explains at 68.18. Ranks are sometimes designated by ordinal numbers, like first, second, third, etc., or by the names “units”, “tens”, “hundreds”, etc. For example, the “7” in 17,285 is the fourth rank, or the rank of thousands.

Place, position (manzila). Like martaba, this word indicates the position of a digit in a number. But here the Arabic word evokes the image of a place or a home, “because the numbers reside in them”, as remarked at 68.18. Again, the “7” in 17,285 is in the fourth place, or the thousands place. Sometimes the word mawdhiʿ, also meaning “place”, is used with the same meaning (at 108.12, 109.4).

Index (uss). This is the number indicating the position of the digit. The index of units is 1, of tens is 2, of hundreds is 3, of thousands is 4, etc. The Arabic word uss is used today to indicate the exponent in mathematics. We could have translated it as “power”, but the numbers would be off by one. For us 10 is the first power of ten, while the corresponding index is 2. Later, in the chapter on algebra, the word uss is used to mean the power of the unknown, and there it matches our exponents. The uss of the second degree unknown is 2, for instance. Thus we translate it as “power” there.

Name (ism). The name of a digit is “units”, “tens”, “hundreds”, “thousands”, “ten thousands”, etc., depending on its place. For instance, the name of the 7 in 17,285 is “thousands”.

Species (nawʿ or jins). There are three species of number: units, tens, and hundreds. These are repeated for the thousands, for the thousand thousands (i.e., millions), etc. So the species of the “7” in 17,285 is units, and the species of the “1” is tens. The same two words nawʿ and jins are used for the different “species”, or what we would call the powers of the unknown, in algebra.

At 68.18 Ibn al-Bannāʾ limits “rank” and “place” to units, tens, and hundreds, these being repeated for the thousands, thousand thousands, etc. But at 70.2, 70.23, 72.4, and 105.1 al-Hawārī regards these terms as progressing indefinitely instead of repeating.

One word absent in the book is ḥarf, meaning “digit”. Other medieval books, such as Principles of Indian Reckoning of Kūshyār ibn Labbān, use ḥarf to mean “digit”. The word is also used by our authors, but with the meanings of “letter”, “particle”, or “conjunction”. In many instances, beginning in the section on addition, we translate martaba as “digit”. This is not technically correct since the rank is a location and the digit is a number that is placed there, but it makes the reading easier. We also often translate manzila as “digit”, too, first at 74.9; and in the passages at 111.7 and 112.5, we render ʿadad (“number”) as “digit” where appropriate.

There were several Arabic words that played the roles of our words “type”, “kind”, “species”, “division”, or “variety”. The ones appearing in our book are nawʿ, jins, ḍarb, qism, and ṣanf. These words were more or less interchangeable in Arabic mathematics, whether for types of number, of fraction, of addition (and other operations), of proportion, of equation, etc. It is partly from the ways numbers were classified, and from the descriptions of the two types each of multiplication and division (95.3, 117.9), that we can recognize numbers in Arabic mathematics as being numbers of something. See our commentary at 95.3 below.

link to parallel text 69.2 By the thirteenth century CE, and perhaps much earlier, two distinct styles of writing the Indian numerals had developed in the Islamic world.[8] The Western forms, written in the Maghreb and al-Andalus, ultimately led to the European forms 123456789 and 0, while the Eastern forms led to the current Arabic forms of the numerals, ١٢٣٤٥٦٧٨٩ and ٠. There are of course variations within each style.

Naturally, al-Hawārī wrote the Western forms, and it is these that are described by the poet. The poem makes use of the similarities between the shapes of the numerals and the shapes of letters of the Arabic alphabet to teach the student how to write the numerals. The copyist of the Medina manuscript was not familiar with the Western way of writing the numerals, so he followed the instructions in the poem to the letter (pun intended). Figure 1 below shows the numerals 987654321, with the Eastern forms written underneath.

Fig. 1: Digits in the Medina MS.

Starting from the right, the “1” looks like the letter alif, which is just a vertical line. The “2” is written as the letter ḥā, and the “3” is the ligature of the letters ḥā and jīm, pronounced ḥajja as if it were a word. The “4” should be the ligature of ʿayn and wāw, pronounced ʿuw, but the copyist forgot to add the waw. See below in Figure 2 for the “4” in the Istanbul manuscript. The “5” is shown as an ʿayn alone, so it looks like the “4” in this figure. The “6” is shaped like the letter hā (the copyist failed to close the loop), the “7” like an anchor, and the “8” is said to be a couple zeros (small circles), one above the other, connected by a vertical line (the alif). The line is not evident in the figure, and is generally not written at all. The Tunis manuscript writes the “8” the normal way, as Too late, but just below it follows the instructions with the vertical line: Too late (and afterward the copyist wrote the Eastern forms of the digits). Finally, the “9” is written as the letter wāw.

The copyist of the Istanbul manuscript was already familiar with the Western forms. These are shown in Figure 2.

Fig. 2: Digits in the Istanbul MS.

For comparison, Figure 3 shows the number 9367184225 from the same manuscript, with the Eastern forms above the Western forms. The Eastern “5”, on the right in black, is mistakenly written as a “4”.

Fig. 3: More digits, from the Istanbul MS.

link to parallel text 70.2 There were no words for “million”, “billion”, etc. in Arabic arithmetic. They wrote instead “thousand thousand”, “thousand thousand thousand”, etc.

link to parallel text 70.8 The places/ranks of a number were spoken in Arabic in a different order than they are in English. We preserve the Arabic order in our translation from here to the end of the chapter (through page 72 of our edition) to give the reader a feel for how the numbers were expressed.

Arabic numbers less than one hundred were spoken with the units first, like “four and sixty” instead of “sixty-four”. So “four and sixty thousands” is 64,000, and “four and sixty thousands and three hundred thousands” is 364,000, though this was often spoken as “three hundred thousands and four and sixty thousands”. In Arabic, the hundreds (like “three hundred” and “five hundred”) are compound words, so we translate “three hundred” instead of “three hundreds”. The word for "Thousands" is stated separately from its number, and is made plural when there is more than one of them. They wrote “a thousand” for one, and “two thousands”, “three thousands”, etc. for more than one.

The plural becomes more complicated when we get to the millions. When there are more than ten of something, the plural form of an Arabic noun is written the same as the singular form. So we read “four trees”, but fourteen of them reads like “fourteen tree”. Thus, what appears to be “ten thousands thousand” should be understood as “ten thousands thousands”. If there are 10,000 of the second word “thousand”, it must be plural.

Also, where we say “one hundred”, “one thousand”, etc., in Arabic the “one” is not written. They expressed these numbers with the implied indefinite article, as “[a] hundred”, “[a] thousand”, etc.

link to parallel text 71.12 Subsection on knowing the index of the repeated number.

link to parallel text 72.1 A 1, 2, or 3 is needed for the rule, so if the number is divisible by three, then three is regarded as the remainder.

link to parallel text 73.1 Section I.1.2. Addition.

link to parallel text 73.2 The act of adding two numbers, according to al-Hawārī’s definition, should result in a single expression. Sometimes, though, the sum can only be expressed with more than one expression. For example, adding 978 to 456 yields one expression, 1434 (at 75.9), as does adding “a root of two to a root of eight” ( \sqrt{2}+\sqrt{8}), which gives “a root of 18” ( \sqrt{18}, at 179.16). The example at 180.10 shows an addition that results in two expressions: adding a root of three ( \sqrt{3}) to a root of fifteen ( \sqrt{15}) can only be expressed as “a root of three and a root of fifteen”. We write this as \sqrt{3}+\sqrt{15}, but as we explain below at 219.1, the operation of addition is not inherent in this composite expression. The current section covers addition of whole numbers, which always results in one expression.

Four different verbs are used in the book to mean “to add”: jamaʿa, ḥamala, zāda, and ḍāfa. Lane’s definitions of jamaʿa begin “to collect; bring, or gather together”.[9] He starts off his definitions of zāda with “to increase, or augment, or grow”, while the various meanings of ḥamala begin with “to bear it, carry it, take it up and carry it, convey it, or carry it off or away”. Lane gives no definition of ḍāfa that relates to addition, but Wehr has “to be added, be annexed, be subjoined, be attached”.[10] There are similar variations for words meaning “exceed”/“surpass” (zāda, faḍala, ʿalā) and “sum” (majmū and related forms, jumla).

Ibn al-Bannāʾ covers five types of addition in the Condensed Book:

Adding numbers with no known relation. He covers the basic process of adding numbers in Indian notation beginning at 74.9.

Adding sequences of numbers with a known disparity, at 76.7.

2a. In one kind of disparity, the ratio between consecutive terms is constant.

2b. In the second kind, the difference between consecutive terms is constant.

Adding consecutive numbers, their squares, and their cubes, at 79.13.

Adding consecutive odd numbers, their squares, and their cubes, at 80.5.

Adding consecutive even numbers, their squares, and their cubes, at 80.20.

The rules for summing finite series, extending from 76.7 to 82.4, are (mostly) originally Greek in origin, but were probably borrowed from some intermediate Arabic source.[11]

link to parallel text 73.7, 73.17 Ibn al-Bannāʾ’s distinction between a disparity in quantity (kamm), in which the difference between consecutive terms is constant, and a disparity in quality (kayf), in which the ratios of consecutive terms is constant, comes from Nicomachus.[12] The terms “quantity” and “quality” appear again in the passage at 92.17, but with different meanings.

The Arabic for “geometric progression” is nisba handasiyya, literally “geometric relation”. Ibn al-Bannāʾ writes “known disparity” instead of “known relation” because the word nisba (“relation”) might imply the geometric progression. See below at 193.1 for more on the word nisba.

link to parallel text 74.17 Al-Hawārī gives two examples of the first type of addition, starting with the addition of 4043 to 2685. First, the two numbers are written on two lines, like this:

Then the units 3 and 5 are added, and the result is put above (we put changes from the previous figure in red):

The tens are next. Because 4+8=12 has two digits, 1 is added to the 6 in the hundreds place of the lower number, and a 2 is put above the tens place:

There is nothing to add to the 7, so 7 is put above the hundreds place:

Finally, the thousands place is 6, from the sum of 2 and 4. The answer is 6,728:

Because this method requires erasing and replacing, it was intended to be worked out on a dust-board or wax tablet, and not with ink.

Operating on zero

The rules for operating on numbers expressed in Indian notation call for the addition, subtraction and multiplication of digits, and sometimes one or both of these digits is a zero. The zero signifies a place where there is no number at all, so we should ask what it meant to operate on it. For this we need to understand that the operations themselves were thought of in a more material sense than our binary operations on abstract sets. Even the notion of a set is a modern one – there was no word for “set” in Greek, Latin, Arabic, Sanskrit, or medieval Italian. Premodern mathematicians, Europeans and Indians included, had no concept of a set as an object.

Addition for al-Hawārī was not an operation on R⁺ satisfying the commutative and associative axioms. It was simply the appending of a number with another number, or the gathering of numbers together, which were all regarded as amounts of something (dirhams, men, hours, etc.). To add five to three was like combining the five silver dirhams in one purse with the three silver dirhams in another purse, or like extending a length of five adhruʿ by three more adhruʿ, or like adding three mathāqīl of grain to five mathāqīl. Not even Euclid found it necessary to provide a definition for addition conceived like this, though some Arabic arithmetic books, al-Hawārī’s included (at 73.2), characterize the operation.

For the operation of addition, Ibn al-Bannāʾ provides a special instruction when there is a zero present (74.9): “Then you add each digit of one of the addends to its counterpart in the other. If there is no counterpart, then the answer is the addend, as if it had a counterpart”. Al-Hawārī follows this rule in the present calculation: “Nothing corresponds to the seven in the upper line, so it is considered to be the sum of that rank and that of its counterpart as if it had something”. Adding nothing to 7 to get 7 does not mean that 0 assumes the role of an operable quantity. Instead, no addition takes place at all. Think of it like combining the money in two purses: one with 7 dirhams and the other empty. There is no act of combining to perform. Subtraction works similarly. In the passage at 83.19 al-Hawārī is faced with the subtraction of 0 from 9: “So we subtract this nothing of the minuend from the nine of the subtrahend, leaving nine”. No subtraction takes place when taking nothing away from an amount, so it leaves the amount unchanged.

Multiplication by zero is explained in the passage at 114.4: “multiplying the number by the zero or the zero by the number is identical. It comes from voiding the number or duplicating zero. Neither of these gives a number, so its sign is always a zero”. The word behind “voiding” (taṣfīr) is related to the word for “zero” (ṣifr). The former could have been translated as “emptying” or “zeroing”, and “zero” could be replaced with “nothing”. This duplication conforms to the standard definition of multiplication, given by Ibn al-Bannāʾ at 95.2: “Multiplication consists of the duplication of one of two numbers by however many units are in the other”. Duplicating nothing a number of times surely gives nothing, so the multiplication makes sense even if zero, being nothing, is not a number.

See the passages in the translation at 83.19,84.13, 90.7, 90.18, 123.3, and 215.14 for other operations with zero and/or nothing.

link to parallel text 75.9 Al-Hawārī’s second example shows addition starting from the highest power term. He adds 978 to 456, first writing one above the other as before:

Working from the hundreds place, 9+4=13, so 13 is placed above:

Next, 7 + 5 = 12, so a 2 is put above the 7, and 1 is added to the 3 next to it:

Finally, 8 + 6 = 14, so the 4 is put above the 8, and 1 is added to the 2 next to it to get the answer, 1,434:

link to parallel text 76.7 For the second type of addition, Ibn al-Bannāʾ works with the famous chessboard problem. In some books, a grain of wheat is placed in the first square, two grains in the second, four in the third, etc. Ibn al-Bannāʾ simply places numbers in the squares, as did Abū Kāmil when he wrote about it in the late ninth century CE at the end of his Book on Algebra.[13] A 1 is placed in the first square, a 2 in the second, a 4 in the third, and continuing so that each square has double the previous square, like this:

Ibn al-Bannāʾ gives the rule for finding the sum of the numbers from the first square up to the 2^nth square. Al-Hawārī gives the example for the 2^4th = 16th square. The following iteration is performed:

Take the 1 in the first square. Add 1 to get 2.

Square it to get 4. This is 1 more than what is in the first two squares ( 1 + 2), and it is also what is in the third square (4).

Then square the 4 to get 16. This is 1 more than what is in the first four squares ( 1+2+4+8), and it is also what is in the fifth square (16).

Square the 16 to get 256. This is one more than what is in the first 8 squares ( 1+2+4+8+16+32+64+128), and it is also what is in the 9th square (256).

Square the 256 to get 65,536. This is 1 more than what is in the first 16 squares, and it is also what is in the 17th square.

So, the sum of the numbers in the first 16 squares is 65,535.

The figure shown in the translation is the one found in the Medina, Tehran, and Tunis manuscripts. The Istanbul and Oxford manuscripts show this figure instead (only Istanbul has the 65536 written on the left):

link to parallel text 77.9 Ibn al-Bannāʾ then gives a variation in which the first square has a number other than 1, and the rule for filling out the remaining squares in the chessboard is the same: each square is double the one before it. For example, if the first square has a 3, then the succeeding squares are 6, 12, 24, etc. The rule to find the sum of the first 2^n squares is to follow the procedure as if a 1 were in the first square, then one multiplies the result by the number that is in the first square. Al-Hawārī gives the example of adding 4+8+16+32+64+128+256+512. The sum of the first eight squares starting with 1 is 1+2+4+8+16+32+64+128=255. Multiplying this by 4 gives 1020, which is the required sum.

link to parallel text 78.1 Another variation is when the ratio of consecutive terms is some number other than 1\over 2. The example given by al-Hawārī starts with 16, and each square is 2/3 of the succeeding square. He gives the first five numbers: 16, 24, 36, 54, 81. Putting Ibn al-Bannāʾ’s rhetorical rule into modern form, the sum will be {16\cdot (81-16)\over 24-16}+81=211. We leave the general rule as an exercise for the reader.

link to parallel text 79.1 Ibn al-Bannāʾ then gives a rule for summing sequences of numbers in which the difference, rather than the ratio, of consecutive terms is constant. Al-Hawārī’s example is to add the 6 numbers starting with 10, and with a difference of 3. If we were to write it all out the sum would be 10+13+16+19+22+25, but we are working only with the known numbers 6, 10, and 3. The rule begins by finding the last number, which in this case is 3\cdot (6-1)+10=25. Then (25+10)\cdot ({1\over 2} of 6)=105 is the sum. Ibn al-Bannāʾ’s rhetorical rule can be expressed in modern notation this way: if there are n numbers starting with a and with a difference of d, then the last number b is d(n-1)+a, and the sum is (b+a)\cdot {1\over 2}n.

link to parallel text 79.13 The third type of addition covers consecutive numbers, their squares, and their cubes. It is true that the first of these is a special case of the second type of addition, but here the first and last numbers are both given, and the rule is then used to find the sums of the squares and the cubes.

To add the numbers from 1 to 10, multiply half of the 10 by one more than the 10: 5\times 11=55. In modern notation, \displaystyle {\sum _{k=1}^nk=\tfrac{1}{2}k\cdot (k+1)}.

To add the squares of these numbers, or 1+4+9+16+\cdots +100, the rule is ({2\over 3}\cdot 10+{1\over 3})\cdot 55=385. In general, \displaystyle {\sum _{k=1}^nk^2=\left(\tfrac{2}{3}n+\tfrac{1}{3}\right)\sum _{k=1}^nk}.

To add consecutive cubes, square the sum of the numbers. The example is 1+8+27+\cdots +1000=55^2=3025. In general, \displaystyle {\sum _{k=1}^nk^3=\left(\sum _{k=1}^nk\right)^2}.

Medieval Arabic mathematicians wrote “[a] square” of a number, with the implied indefinite article, rather than “the square” because their numbers admit multiplicity. See below at 163.2 for a more detailed explanation for the case of roots.

link to parallel text 80.5 The fourth type of addition is to add the consecutive odd numbers, their squares, and their cubes. For the first of these, square half of one more than the last number. Al-Hawārī’s example is to find 1+3+5+7+9. The answer is ({1\over 2}(9+1))^2=25. In general, if the last number is n, then the sum 1+3+5+\cdots +n= ({1\over 2}(n+1))^2.

The sum of the squares of consecutive odd numbers up to n is {1\over 6}n(n+1)(n+2). In the example, al-Hawārī finds that 1^2+3^2+\cdots +9^2={1\over 6}9\cdot 10\cdot 11=165.

If we let a be the sum of the odd numbers up to n, then the sum of the cubes of the odd numbers to n^3 is a(2a-1). Al-Hawārī calculates that 1^3+3^3+5^3+7^3+9^3=25\cdot 49=1225.

link to parallel text 80.20 The fifth and last type of addition deals with consecutive even numbers, their squares, and their cubes. Al-Hawārī adds the even numbers from 2 to 10 by calculating {1\over 2}(2+10)\cdot {1\over 2}10=30.

For the squares 2^2+4^2+\cdots +10^2 one adds 2\over 3 of 10 to 2\over 3 of 1, and the result is multiplied by the sum, 30: 6{2\over 3}+{2\over 3}=7{1\over 3}, and 7{1\over 3}\cdot 30=220.

Al-Hawārī gives an alternative rule, using the example of adding the squares 2^2+4^2+\cdots +12^2. Here one multiplies a sixth of the last number (12) by the product of the next two numbers (13 and 14). Taking 1\over 6 of 12 gives 2, and 2 by 182 is 364, which is the required sum. This happens to be the same rule he gave for adding the odd squares.

The sum of the even cubes 2^3+4^3+\cdots +10^3 “is given by multiplying the sum by its double”. We already know that the sum 2+4+\cdots +10 is 30, so we multiply 30 by 60 to get 1800, which is the answer.

link to parallel text 83.1 Section I.1.3. Subtraction.

Just as with addition, several verbs are used for subtraction. The most common, and the one which appears in chapter titles and instructions, is ṭaraḥa. We translate it as “to subtract”. The verb ṭaraḥa is also used for what we call “casting out”, as in casting out nines to check the answer to a calculation. Two other common verbs for subtraction are saqaṭa, “to drop”, and naqaṣa, which we also translate as “to subtract”. Rarer are the verbs nazala, “to remove”, and dhahaba, “to take away”.

To announce the result of a subtraction the verb baqiya (“to remain, leave”) is used. The word for “remainder” is bāqī. Often we translate a phrase whose literal meaning is “[there] remains” or “what remains” as “the remainder”. The word for the “residue” after casting out nines or eights or sevens, covered starting at 87.15, is the related word baqiya, and we translate bāqiya as “residual”. These two words appear one time each to mean “remaining” and “remainder”, respectively, at 110.16 and 118.20.

link to parallel text 83.2 Ibn al-Bannāʾ writes of two kinds of subtraction. The first is the subtraction of one number from another with Indian numerals, and the second is “casting out” to check the answer of a calculation. Between these two kinds al-Hawārī inserts a description of repeated subtractions, starting at 86.1, that he took from Lifting the Veil.

Two examples are given for the first kind. Al-Hawārī begins with the example 5035-4968 and proceeds from the highest rank to the lowest. First, the greater number is written above the smaller:

The 4 is subtracted from the 5, and the result is written above:

For the hundreds place, there is nothing (i.e., a 0) in the minuend, so one takes nothing away from the 9 in the subtrahend, leaving 9. This is subtracted from the 1 above the 5, which is really 10 since we are now working in the hundreds place. 9 from 10 leaves 1, so the first 1 is replaced with a 0, and this new 1 is placed above the hundreds place:

In the tens place the 3 is smaller than the 6, so we subtract 3 from 6 to get 3, and this is subtracted from the 10 above, leaving 7. The figure then becomes:

The situation is similar for the units place. Since 5 is less than 8, we subtract their difference, which is 3, from the 70 above. This leaves 67 as the answer:

link to parallel text 84.13 Al-Hawārī gives a second example that starts with the units place. He begins with:

The 3 is less than 9, so we add 10 to the 3 and then subtract 9, leaving 4. This is placed above the 3. To compensate for the added 10, a 1 is added to the 6 next to the 9:

For the tens place we have a similar situation: 4 is less than 7. So add 10 to the 4 and subtract, leaving 7. Then add one to the 4 in the bottom row to get:

Next, taking 5 from 5 leaves nothing, so a 0 is placed above them:

Taking 3 from 6 leaves 3, so the figure becomes:

The answer is the 3,074 on top.

link to parallel text 85.16 Rūmī signs were used in a system of calculation practiced in Western North Africa and al-Andalus. See §3 in the Introduction.

link to parallel text 86.1 The part on repeated subtractions is taken from Ibn al-Bannāʾ’s commentary, and is not mentioned in the Condensed Book. It is not one of the two categories of subtraction he mentions at 83.2 at the start of the chapter. In modern notation the expression “ten less eight less seven less five less two” is 10-(8-(7-(5-2))). This is explained in words from the inside out: “subtract two from five, and the remainder from seven,…” and in an English version of the Arabic notation it would be 10~\ell ~8~\ell ~7~\ell ~5~\ell ~2, where the \ell stands for “less”. We are keeping the direction of the Arabic figures, so we write it in the translation as 2~\scalebox {-1}[1]{\ell }~5~\scalebox {-1}[1]{\ell }~7~\scalebox {-1}[1]{\ell }~8~\scalebox {-1}[1]{\ell }~10. See the discussion at 219.4 below for an explanation of the word “less” (illā).

In two of the five manuscripts we consulted, Istanbul and Oxford, the numbers are separated by the word illā. Here it is in the Istanbul manuscript: Too late. In the Medina manuscript only the last part of the word is drawn, so it looks like an upside-down “ \ell”: Too late. The two other manuscripts, Tunis and Tehran, put three dots in place of the word illā for this figure. Here is the figure from the Tunis manuscript: Too late.[14] Both manuscripts, curiously, write the illā in other instances.

link to parallel text 86.15 One could simply perform the operations as stated: subtract 2 from 5, then subtract the result from 7, etc. But Ibn al-Bannāʾ gives three other rules to work it out. For his first rule he distinguishes between the minuend, in this case 10, and the subtrahends, which here are 8, 7, 5, and 2. Add the even subtrahends (the 7 and 2) to the 10, and from this subtract the odd subtrahends, giving 19-13=6.

link to parallel text 86.18 Another way is to collapse the string of numbers in groups of three. We “subtract the middle from the sum of the extremes, leaving the remainder as one number”. Taking the 10, 8, and 7, calculate 10+7-8=9, and replace all three with the 9 to get 2~\scalebox {-1}[1]{\ell }~5~\scalebox {-1}[1]{\ell }~9. Repeating the process gives 9+2-5=6. One does not need to start with the first three numbers. This works starting from any three consecutive numbers.

link to parallel text 87.9 A third way is to perform alternating subtractions and additions, beginning with the 10. Subtract 8 from 10, then add 7 to the remainder, then subtract 5, and finally add 2.

link to parallel text 87.11 See our extended discussions below, at 219.1 and 219.4, for an explanation of “appended” (zāʾid) and “deleted” (nāqiṣ). They do not mean “positive” and “negative”. Because numbers were numbers of something counted or measured, negative numbers would have been meaningless to medieval arithmeticians.

link to parallel text 87.15 The second kind of subtraction is what we call in English “casting out”. In “casting out nines”, which is still taught today, the remainder from division by 9 can be found by adding the digits and removing multiples of 9. Al-Hawārī gives the example of 6435. He adds the digits one by one, casting out nines as he goes. So 6+4=10, and removing a 9 leaves 1. Then 1+3=4, and 4+5=9. “This is cast out entirely”, meaning nothing remains. Nothing is literally no number at all, and not our modern number 0.

Al-Hawārī also gives examples of the rules for casting out eights and casting out sevens. Because multiples of 200 are divisible by 8, one only needs to deal with the first three places in casting out eights. Al-Hawārī’s example is 5393. The 5000 is cast out entirely, as is the 200 from the 300. The remainder from the 100 is 4, then 2 is multiplied by the 9, and to these are added the 3, giving 25. Casting eights from this leaves 1. In general, the remainder of a number of the form 1ab (i.e., 100+10a+b) is the same as the remainder of 4+2a+b.

link to parallel text 88.10 Casting out sevens is more complicated. No multiple of 10 is divisible by 7, so all digits must be taken into account. The remainders of each power of 10 are different for the first six powers, after which they repeat. These remainders must be memorized, and al-Hawārī illustrates Ibn al-Bannāʾ’s rule of expressing this sequence of remainders in abjad form. The letters appearing there are:

Arabic | Transliteration | Value
Too late | alif | A | 1
Too late | bā | B | 2
Too late | jīm | J | 3
Too late | dāl | D | 4
Too late | hā | H | 5
Too late | wāw | W | 6

The remainders of 1, 10, 100, 1000, 10000, and 100000 are 1, 3, 2, 6, 4, and 5, respectively. At 88.14 he gives a short poem designed to help the student memorize these letters.

link to parallel text 88.17 Indian numerals were sometimes called al-ghubār, or “dust” numerals, after the dust-board on which they were commonly written.

link to parallel text 89.4 The example al-Hawārī gives is to cast out sevens from 23,786,435. He writes the digits above their corresponding letters:

For us, it is easier to use the Indian numerals:

Multiplying the digits in the corresponding places, and casting out sevens if necessary, gives:

For example, in the thousands place we multiply 6 by 6 to get 36, and casting out sevens leaves 1. The digits in the top row add up to 22, and casting out sevens again leaves 1.

link to parallel text 90.3 Ibn al-Bannāʾ gives two variations for casting out sevens that do not require the memorization of the sequence of abjad numerals. The first is an iteration: first multiply the highest power digit by three (the residue of 10), cast out sevens if necessary, and then add the result to the previous digit. Al-Hawārī’s example is 58,064. Starting with the highest power term, multiply 5 by 3 to get 15. Cast out sevens to get 1. Then add this 1 to the 8 to get 9. Now repeat: multiply 9 by 3 to get 27, leaving 6 after casting out. Add 6 to nothing (the 0) to get 6. Multiply by 3 to get 18, and cast out sevens, leaving 4. Add 4 to 6 to get 10. Thrice 10 is 30; cast out to get 2, and add it to the 4 to get 6. This is the answer.

link to parallel text 90.15 The other variation takes into account two digits at a time. Casting out sevens from the 58 in 58,064 leaves 2. Although al-Hawārī does not write it, the residue of 58,064 is the same as the residue of 2,064. Next, the residue of 20 is 6, so we now consider 664. The residue of 66 is 3, so now the number is 34. Its residue is 6, which is the answer.

link to parallel text 91.1 Subsection on the way to test [calculations] by casting-out.

Next, Ibn al-Bannāʾ turns to applications of these techniques of casting out to check the results of arithmetical operations. These include addition, subtraction, multiplication, and division/denomination. For the latter two it works even for fractions. Here are a couple of al-Hawārī’s examples.

link to parallel text 91.4 For addition, he works with the example 43+64=107. He could cast out nines, eights, or sevens, and for this and subsequent examples he chooses sevens. The residue of 43 is 1, and the residue of 64 is also 1. Add them to get 2, which should be (and is) the residue of 107. If the residues added to 7 or more, one would cast out a 7 to make it less than 7.

link to parallel text 92.3 In subtraction there is a problem if the residue of the minuend is smaller than the residue of the subtrahend. Take for example 29-13=16. The residue of the 29 is 1, and the residue of the 13 is 6. We cannot subtract 6 from 1, so we add 7 to the 1 and then subtract: 8-6=2, and this 2 is the residue of the remainder 16. For casting out nines, add nine to the residue of the minuend, and for casting out eights, add eight.

link to parallel text 92.17 An example for the multiplication of fractions is {1\over 3}\times 14{1\over 4}=4{3\over 4}. The residues of the multipliers are 1\over 3 and 1\over 4. Ibn al-Bannāʾ does not say so, but these are the residues of the numerators. If, for instance, the multiplier were 3{1\over 4}, he would have found the residue to be 6\over 4, since 3{1\over 4} as a single fraction is 13\over 4, and the residue of 13 is 6.

The product of the 1\over 3 by the 1\over 4 is 1\over 12, or as Ibn al-Bannāʾ puts it, “a third of a fourth”. The numerator of this fraction is 1. The product 4{3\over 4} is 19\over 4, and Ibn al-Bannāʾ takes the residue of the numerator, which is 5. But these are fourths, not twelfths, so he multiplies by 3 to get 15 (for 15\over 12), whose residue is 1. This agrees with the residue of the multipliers.

The numerator 1 of “a third of a fourth” is equal to the numerator of the answer, so they are equal in quantity. The kinds of fractions they are, thirds of fourths, are the same, making them equal in quality. This terminology comes from Aristotle’s Categories, probably via Ibn Sīnā.[15] See above at 73.7 for another use of the words “quantity” and “quality”.

link to parallel text 93.6 As Ibn al-Bannāʾ will explain in the section on division at 118.14, the term “division” is used for the division of a greater number by a smaller number, and “denomination” for the division of a smaller number by a greater number. For the division a\div b= c, where a>b, the a is the dividend and b the divisor. If a<b, then a is the denominated [number] and b is the denominating [number] or what we call the denominator. For both kinds, c is the quotient or result. So “the result and the divisor or denominating number” are c and b, and the “dividend or denominated number” is a.

The Arabic words for the “denominator” of a fraction (imām, and less frequently maqām) are unrelated to the verb “denominate” (sammā) and related nouns such as “denominating [number]” (musammā minhu).

link to parallel text 93.10 For both division and denomination, al-Hawārī gives an example with whole numbers and an example with fractions.

The Arabic word for the “numerator” of a fraction is basṭ, or occasionally the related word mabsūṭ. These words are unrelated to the word for “number”, which is ʿadad. We translate the related verb basaṭa as “to numerate”. Its meaning is to find the numerator of a fraction that is “a combination of two or more names” (135.1). For example, to numerate the fraction “five sixths and three fourths” (at 93.15) means to express it as “thirty-eight fourths of a sixth” ( 38\over 4\cdot 6). Finding the numerator is described at length in the first section on fractions, beginning at 135.8.

link to parallel text 93.15 The notation for distinct fractions shows one next to the other, as explained later at 136.8 and 139.1. This problem is to divide {3\over 4}~{5\over 6} by {1\over 2}. We would write {3\over 4}~{5\over 6} as {5\over 6}+{3\over 4}, and the result of the division as 3{1\over 6}. The residue of 3{1\over 6}={19\over 6} is 5, or 5\over 6. The residue of 1\over 2 is 1, or 1\over 2. Multiplying them, one gets 5\over 12, or, as Ibn al-Bannāʾ says, “five halves of a sixth”. The residue of the dividend, {3\over 4}~{5\over 6}={38\over 24}, is 3, with a denominator of 24. So the 5\over 12 must be converted to 24ths, making it 10\over 24, which is 3\over 24 after casting out again. This matches the residue of the 38\over 24.

link to parallel text 94.1 Al-Hawārī gives an example of checking the result of the denomination of whole numbers even if it might be superfluous in practice. The example is to denominate 11 with 15, which for us gives the fraction 11\over 15. The residue of the denominated number 11 is of course the same as the residue of the numerator 11 of the result, so there is nothing to check. The only aspect that makes this appear to be a problem is that al-Hawārī follows common Arabic practice by expressing the result not as 11\over 15, but as “three fifths and two thirds of a fifth”. We might write this as {3\over 5}+{2\over 3}{1\over 5}, but for al-Hawārī it would be shown as 2~\,3\over 3~\,5 (the notation for this fraction is explained at 123.18). The numerator and denominator of this fraction must be calculated in order to find its residue, but this brings us right back to the 11 and 15 we started with.

link to parallel text 94.5 Checking the result of the denomination of fractions requires some work. Writing Ibn al-Bannāʾ’s example in notation, it is to denominate 2~\,2\over 3~\,6 with 1~\,5\over 3~\,8, which results in 2\over 3. After finding the numerators, the problem remains to denominate 8\over 6\cdot 3 with 16\over 8\cdot 3. The product of the residues of 2\over 3 and 16\over 8\cdot 3 should equal the residue of 8\over 6\cdot 3, but we need to be sure the denominators are the same to make it work. The residue of the numerator of 2\over 3 is 2. The residue of the numerator of 16\over 8\cdot 3 is also 2, and multiplying the 2 (as 2\over 3) by the 2 (as 2\over 8\cdot 3) gives 4\over 3\cdot 3\cdot 8. Because the 6 in the denominator of the 8\over 6\cdot 3 is lacking in the 4\over 3\cdot 3\cdot 8, we multiply the numerator and denominator of the latter by 6 to get 24\over 3\cdot 3\cdot 6\cdot 8, and its residue is 3, which is the “answer”. We now turn our attention to 8\over 6\cdot 3. The residue of the 8 is 1, and this must be multiplied by 3 and then by 8 to make the denominators match. This gives 24\over 3\cdot 3\cdot 6\cdot 8, and its residue is 3, which agrees with the answer.

link to parallel text 95.1 Section I.1.4. Multiplication.

link to parallel text 95.2 Arabic arithmetic books often define multiplication as “the duplication of one of two numbers by however many units are in the other”, as Ibn al-Bannāʾ phrased it here in the Condensed Book.[16] This definition by “duplication” may come from Book VII of Euclid’s Elements: “A number is said to multiply a number when that which is multiplied is added to itself as many times as there are units in the other, and thus some number is produced”.[17] In fact, Abū l-Wafāʾ (tenth century) writes in his arithmetic book: “Euclid stated the meaning of multiplication in Book VII of his work The Elements, as did Nicomachus of Gerasa in the Arithmetic. They said that multiplication is the duplication of one of two numbers by the amount of what is in the other in units”.[18]

It would not matter so much that Nicomachus, in fact, gives no definition of multiplication in his Arithmetical Introduction if it were not for what Abū l-Wafāʾ says about division: “Division according to the example (qiyās) of Euclid and Nicomachus is the partitioning (tafrīq) of one of two numbers by the amount of units in the other”.[19] Neither Euclid nor Nicomachus defines division in their books, and if this statement is not intended to make such a claim, it at least implies that this definition, too, originates with the Greeks.

Contrary to what Abū l-Wafāʾ’s testimony might imply, there is evidence that the “duplication” definition of multiplication was already circulating among Arabic arithmeticians before they began reading Euclid. The definition appears in al-Khwārazmī’s algebra book, written sometime between 813 and 833 CE. This book presents the practical algebra connected with finger-reckoning, and the only hint of Greek influence is the presence of letters to label points in some diagrams.[20] Later, al-Khwārazmī copied the definition into his book on calculating with Indian numerals. This way of characterizing multiplication is rather intuitive, and it is entirely possible that arithmeticians developed or learned it independent of Euclid.

It may have been his desire to tie practical Arabic arithmetic with the Greek tradition that led Abū l-Wafāʾ to imply a false origin for both definitions. We have found another instance of this kind of “false history” in the tenth century philosopher al-Fārābī. In the third part of his Enumeration of the Sciences he classifies the various branches of mathematics. He describes algebra as being “concerned with the ways of figuring out how to find numbers that apply Euclid’s principles on the rational and irrational in the tenth book of the Elements…”[21] Yet not one of the extant books on algebra written before him, and to our knowledge none written after him, show any sign of applying principles from Book X to algebra.[22] Al-Fārābī’s wishful connection of algebra to Euclid’s Elements seems to have been a way to legitimize the prevalence of irrational numbers in algebra and to suggest a Greek inspiration for the art.

Whatever its source, the “duplication” definition of multiplication was devised with whole numbers in mind. Abū l-Wafāʾ criticizes it on this point, and shows through an example of the price of cloth that it also works for fractions. He justifies this extension by appealing to propositions about lines, reinterpreted numerically, from Book II of the Elements.[23] Ibn al-Bannāʾ gives a different definition for the multiplication of fractions, below at 149.2. Neither he nor al-Hawārī defines the multiplication of irrational numbers, even if these multiplications are performed in the book (183.2). Definitions in practical books like Ibn al-Bannāʾ’s should be read more as intuitive characterizations that introduce students to the concepts rather than precise definitions of the kind that Aristotle or David Hilbert would have approved.

link to parallel text 95.3 Ibn al-Bannāʾ distinguishes between two types of multiplication. In the first type, one puts a copy of the multiplicand in place of each unit of the multiplier. The example is “three men: each of them has five dirhams. You multiply five by three, which gives fifteen dirhams”. Each of the three men is substituted with five dirhams, to get a total of fifteen dirhams. There is a change in meaning (maʿnan) because the three (men) became fifteen (dirhams). Also, the units shift from (three) men to (fifteen) dirhams, which is a change in terms (lafẓ).

The example for the second type is “five dirhams, how many thirds does it contain?”. This type is called “conversion” because the units for the amount “five dirhams” is converted from dirhams to thirds of a dirham. There are 5 of the former and 15 of the latter. The total amount “five dirhams” remains the same, so there is no change in meaning. But the shift in units, from dirhams to thirds of a dirham, is a change in terms. “All of what is in the multiplier in units”, or three (thirds of a dirham), “is equal to the one [dirham] of the multiplicand”. Thus the “[number of] units in the multiplier”, or three, “is the number of what is in one of the multiplicand in parts”, or, more loosely translated, “is the number in each part [i.e., each dirham] of the multiplicand”. Conversion of fractions is covered later, at 157.9, and the two corresponding types of division are explained beginning at 117.9.

Ibn al-Bannāʾ’s description of the first type, at 95.3, is problematic. It reads “in putting down the multiplier, each one of them is equal to the one of the multiplicand”. Reading the example for this type at 95.10, it could be translated as: “in putting down the multiplier, each one of them stands for (mithl) the single entity (wāḥid) of the multiplicand”. One should substitute for each unit in the multiplier the entire multiplicand, not just its unit (“the one”). But this translation is at odds with the description of the second type of multiplication, which begins: “all of what is in the multiplier in units is equal to the one of the multiplicand”. The same phrase “the one of the multiplicand” in this case must mean its unit, and not the multiplicand as a single entity. Perhaps the description of the first type is misstated.

In Lifting the Veil, Ibn al-Bannāʾ completes his descriptions and examples by writing “The first type is combining (tarkīb) and the second is dissecting (tafṣīl)”, though al-Hawārī does not copy it. In the example of the first type the three copies of five dirhams are combined, and in the example of the second type each of the five dirhams is dissected into three pieces.

Modern mathematicians work with only one kind of multiplication because our numbers all belong to the same abstract set. Instead, in Arabic arithmetic a number is a number of something, whether it be men, dirhams, mithqals, or abstract units. This is why Ibn al-Bannāʾ distinguishes between a change in terms and a change in meaning. (See also our comments on division, below at 117.9.) This idea of different kinds of number is also behind the names of the powers of ten (hundreds, thousands, etc.), the names of fractions (thirds, fourths, etc.), and the names of the powers of the unknown in algebra (numbers, things, māls, cubes, etc.). So “three men” was a particular kind of 3, like “three boats”, “three thousands”, “three fifths”, and “three things”.

Arabic authors often designated numbers by their kind, and several examples are found in al-Hawārī’s book. Just above, at 95.7, we read “the units in the multiplier”; at 114.12, we find “the ranks of the result”; at 208.1; “if we make the starlings twenty-four”; and at 215.15, “we square half of the things”. These nouns mean, respectively, “the number of units”, “the number of ranks”, “the number of starlings”, and “the number of things”. If one has 24 starlings, for example, then that collection of birds is an instance of the number 24, and one can indicate their number by saying merely “the starlings”.

This idea that a number has two aspects, its meaning (value) and its term (the kind of number), breaks down with irrational roots, though our authors do not discuss it. This is because it makes no sense to have an irrational number of anything, like \sqrt{5} bricks, for example. This becomes problematic in algebra, where the “coefficient” (the “number”, or meaning) of a term had to remain rational even when multiplying by an irrational number. See the discussion below in the last paragraph of our commentary to 219.1.

link to parallel text 95.15 Rules for multiplying numbers, Ibn al-Bannāʾ tells us, fall into one of three categories: those with shifting, those with half-shifting, and those without any shifting. Methods involving shifting are designed for the dust-board or wax tablet, where it is easy to erase and rewrite the digits. Those without shifting were intended for the lawḥa with ink pen, and will also work on paper. He also describes some techniques of mental arithmetic at the end of the chapter, beginning at 108.9, that he took mainly from Ibn al-Yāsamīn’s Grafting of Opinions.

link to parallel text 95.17 The first multiplication rule with shifting is called “sleeper multiplication”, perhaps because the numbers are placed horizontally as if they are sleeping on a bed. There is another method of “sleeper multiplication”, without shifting, that features numbers written horizontally. It is described at 103.14.

link to parallel text 96.3 Al-Hawārī gives the example of multiplying 43 by 54. They are arranged on the dust-board like this:

The overall scheme will be to multiply the 4 of the 43 by the 54, then the 54 will be shifted to the right, and then the 3 from the 43 will be multiplied by the 54. First the 4 from the 43 is multiplied by the 5, and the 20 is placed above:

The 4 from the 43 is then multiplied by the 4 below, giving 16. The 1 from the 16 is added to the 0 we just wrote, and the 6 of the 16 replaces the 4:

The 54 is then shifted to the right one unit, giving this figure:

Now we multiply the 3 by the 5 below. The resulting 15 is added to the 16 above to get 31:

Now the 3 by 4 gives 12. The 1 is added to the 1 on top, and the 2 replaces the 3:

The result of the multiplication of 43 by 54 is 2,322.

It is important to know when to add a digit and when to replace a digit. One adds to digits that have been calculated, while one replaces digits from the original problem. This way, the digits of the multiplicand, 43, are replaced with the digits of the evolving calculated number.

link to parallel text 97.1 The other method of multiplication with shifting is called vertical multiplication. It works the same way as the sleeper multiplication just described. Al-Hawārī’s example is to multiply 42 by 37. The digits are arranged vertically, with units on top:

Like before, we first multiply the 4 of the 42 by the 3 of the 37 to get 12. This is placed next to the 3, with the 2 above the 1:

The 4 is then multiplied by the 7, giving 28. This will be placed to the left, too. As before, we replace the digit 4 of the multiplicand with the 8, and we add the 2 to the calculated 2 that lies below it:

Now the multiplier is shifted up one unit:

The 2 on the left will now be multiplied by the 37. First, 2\times 3=6, so we add the 6 to the calculated 8, giving 14. Because of the extra digit, the 1 is added to the 4 below:

Finally, 2\times 7=14. The 4 replaces the 2, and the 1 is added to the 4 below it:

The result of multiplying 42 by 37 is 1,554.

link to parallel text 98.4 The method by half-shifting is a technique for squaring numbers. Al-Hawārī’s example is to multiply 463 by itself. First, the number is written with dots between the digits:

The 4 is squared, and the resulting 16 is written above:

Next, double the 4 to get 8, and put this in place of the first dot:

This 8 is regarded as being shifted, and it is multiplied by the 6 to its right to get 48. This result is added above the 8, treating it like 160+48:

Next the 6 is squared, giving 36. This is added to the 2080 above:

Now the 6 is doubled to get 12, and this is shifted one place to the right so that it replaces the “6 •”. One must also shift the 8, treated as 80 since it is one place to the left, that was doubled before. This is added to the 12, making 92:

Now the last 3 must be multiplied by this shifted 92. The 3 by the 9 gives 27, which is added to the 16 above the 89, making 43:

Next 3 by 2 is 6, and this is added above the 2:

Last, we square the 3 to get 9, and add this above the 3:

The result of multiplying 463 by itself is 214,369.

link to parallel text 99.14 There are several different ways to perform the third kind of multiplication, without any shifting. Ibn al-Bannāʾ first explains “table multiplication”, which we call lattice multiplication. Al-Hawārī’s example is to multiply 435 by 287. These are drawn around a grid with diagonals:

Ibn al-Bannāʾ mentions that the 287 can be placed on the left or the right. Most manuscripts, and our translation, show it on the right, but putting it on the left makes the final addition easier. Next, in each of the nine squares we put the product of the column digit by the row digit. For example, for the upper left square 7\times 4=28, so a 2 goes under the diagonal and an 8 above the diagonal. The spaces are then filled out like this:

We then add the numbers between the diagonal lines. The upper right shows a 5, which is the units. Between the first and second diagonal lines are a 1, 3, and 0. These add to 4, which is the tens place of the answer. Adding the numbers between the next two diagonals gives 8+2+4+4=18, so the hundreds place is an 8, and the 1 is carried to the next sum. Continue like this to get the answer 124,845.

link to parallel text 101.16 For “vertical multiplication” al-Hawārī multiplies 183 by 347. These are arranged vertically, separated by some space:

The units place of the answer will be in the top row between the numbers, the tens place in the row below that, etc. First, we multiply the 3 from the 183 one by one by the digits of 347. The 21 from 3\times 7 is put on the right like this:

Next, 3\times 4=12. The 2 is added to the 2 in the tens place to become 4, and the 1 is placed below:

So far we have 141. As in the Oxford manuscript, we will cross out discarded digits. Next, the 3 is multiplied by the 3 to get 9, which is added to the 1 on the bottom to make 10:

Each digit on the left is then multiplied by the 8, bringing the diagram to this state:

Last, the 1 is multiplied by the 7, 4, and 3:

Thus 183\times 347=63{,}501.

link to parallel text 103.14 The horizontal version of “vertical multiplication” is called “sleeper multiplication”, like the method described above at 95.17. Al-Hawārī multiplies 253 by 987, first writing them on two lines like this:

First, the 3 is multiplied by the digits in 987, beginning with the 7. The 21 is written above:

Next is 3\times 8=24. The 4 is added to the 2 of the 21, making 6, which is placed above, and the 2 is put to the left:

So far the result is 261. Again, we cross out defunct digits as we go, but most manuscripts just leave them alone. Next, 3\times 9=27, and this is added the same way:

Now the 5 is multiplied by the 7, 8, and 9, giving this figure:

Finally, after multiplying the digits by 2, the final figure is:

The result of multiplying 253 by 987 is 249,711.

link to parallel text 104.1 For rūmī calculation see 85.16 above and §3 in the Introduction.

link to parallel text 106.11 This next method, called “repetition”, is just a curiosity of calculation. It works when the digits in each number are all the same and they both have the same number of digits, like 7777 by 9999, or, in al-Hawārī’s example, 444 by 333. These are written on two lines like this:

Under this we put a 1, 2, 3, etc., until we get to the end of one number and the beginning of the next. Then the numbers descend back to 1. These numbers will serves as multipliers:

Multiply the 3 of 333 by the 4 of 444 to get 12. Multiplying the 12 by the 1 on the left gives 12, which is put above like this:

Next, the 12 is multiplied by the 2 to get 24, and this is added above, one place to the right. The 2 is added to the 2 of the last 12 to make 4, so we replace it as we did in the last scheme:

The 12 is then multiplied by the 3, giving 36, and this is added above the same way:

Continuing, we multiply the 12 by the 2, and then finally the 12 by the 1. The final figure is:

So the product of 333 by 444 is 147,852.

link to parallel text 108.9 The techniques of multiplication presented so far are meant to be worked out in Indian notation on a board, and can be applied to any two (positive) integers. The remaining techniques are either shortcuts originating in finger-reckoning, or they are board techniques that work for specific kinds of numbers. Many of these rules were copied by Ibn al-Bannāʾ from Ibn al-Yāsamīn’s Grafting of Opinions, a book that is a kind of hybrid between Indian calculation and finger reckoning. Even though that book is expressly devoted to calculation with Indian numerals, it is organized like a book on finger reckoning in that it covers multiplication and division before passing to addition and subtraction; also, it still retains the rules of mental calculation that Ibn al-Bannāʾ copied into his book. In the following breakdown of Ibn al-Bannāʾ’s remaining rules, we indicate which come from finger-reckoning (FR) and which were taken from Ibn al-Yāsamīn (Y):

In addition, at 109.16 al-Hawārī gives a variation on a finger-reckoning rule that he attributes to al-Yāsamīn.

Multiplication “by excess” is a trick for doing the calculation mentally when at least one of the numbers is between 10 and 19. To multiply 12 by 15, think of the 12 as 2 more than 10. Divide the 2 by the 10, which is 1\over 5, and multiply this by the 15 to get 3. Then add this 3 back to the 15 to get 18, and multiply by 10 to get the answer, 180. Al-Hawārī then does it again, switching the roles of the 12 and 15. In general, to multiply a number of the form 10+a ( a is a digit) by another number b, one calculates ({a\over 10}b+b)\cdot 10. Al-Hawārī gives a second example, 13\times 17, that gives fractions. Try it in your head! 3\over 10 of 17 is 51\over 10, or 5{1\over 10}. Add this to the 17 to get 22{1\over 10}. Finally, multiply by 10 to get the product, which is 221.

link to parallel text 109.7 Another trick for mental calculation is called “denomination”. Al-Hawārī’s example is to multiply 6 by 12. First denominate (we would say “divide”) one of the numbers with their sum. Dividing 6 by 18 gives 1\over 3. Then multiply this by the other number, 12, to get 4. Last, multiply this 4 by the sum, 18, giving 72.

Written algebraically, to multiply a by b one performs these operations: {a\over a+b}\cdot b\cdot (a+b). This method can only be useful if the two numbers have a common divisor. Otherwise, the multiplication of a\over a+b by b will require one to find a\cdot b, which is the original problem.

link to parallel text 109.16 A variation on this rule can be written in modern notation as a\cdot b=(a-{a\over a+b}a)(a+b). Al-Hawārī gives Ibn al-Yāsamīn as his source, though this rule is not in that author’s Grafting of Opinions. We do not know its origin.

link to parallel text 109.19 Another denomination method is to divide one of the two numbers by a power of ten, and then multiply the result by the other number. This is then multiplied by the power of ten to get the product. Al-Hawārī first works out the example 24\times 8. He divides the 8 by 10 to get 4\over 5. Then {4\over 5}\times 24=19{1\over 5}. Multiplying this by 10 gives the product 192. The advantage here is that the numbers are kept small until the end when the power of ten is multiplied back.

link to parallel text 110.5 The Arabic word we translate as “power of ten” is ʿaqd. The use of this word in arithmetic originated in finger-reckoning. Historian A. S. Saidan wrote: “In Arabic ʿuqda is the name of the finger-joint, and ʿaqd in this sense should mean ‘to bend the finger-joint’”.[24] He later described the word in more detail:

This placement [of the fingers] is called ʿaqd, plural ʿuqūd. Thus the finger-reckoner understood numbers as formed of places, namely units, tens, hundreds, etc., each place having one or the other of the nine ʿuqūd: one, two,… nine. With this understanding the word ʿuqūd came to mean what we may now call digits. But in usage ʿaqd and place were not always clearly distinguished.[25]

The first occurrence of the word ʿaqd in al-Hawārī’s book is in a rule of Ibn al-Bannāʾ given above at 87.17. There it takes the meaning of “power of ten” or “place”. The other seven instances of the word are in the present chapter covering finger reckoning rules, from 109.19 to 111.3. When reading “power of ten” here one should keep in mind its association with the positioning of the fingers.

The definition at 110.5 is not accurate. The only non-zero rank should be a ten or a hundred or the like, not just the first (awwal). The way the condition is stated, a number like 310 would be a “simple power of ten”.

link to parallel text 110.12 Al-Hawārī then works out 12\times 15, which he solves by working through 15\div 10=1{1\over 2}, 1{1\over 2}\times 12=18, 18\times 10=180. He then works it out again, this time subtracting 5 from the 15 so that the first operation gives a whole number. He takes 15-5=10, then 10\div 10=1, 1\times 12=12, and 12\times 10=120. To compensate for the subtracted 5, he calculates 5\times 12=60, and this is added to the 120 to get 180, which is the product.

link to parallel text 111.1 Next al-Hawārī finds 3\times 15 similarly, but by adding rather than subtracting. Adding 2 to the 3 gives 5, and dividing that by 10 gives 1\over 2. This 1\over 2 may not be a whole number, but it is easier to work with than 3\over 10. Then {1\over 2}\times 15= 7{1\over 2}, and “we raise each [digit] by ten” to get 75. From this we must subtract 2\times 15, or 30, to get 45, which is the answer. With practice, methods like these prove to be quite useful.

link to parallel text 111.7 Recall the “repetition” method at 106.11, which was covered just before the techniques of mental multiplication. That method requires that the number of digits in the multipliers be the same, and that all the digits in each multiplier be equal, as in al-Hawārī’s example 333\times 444. This next method, called “nines”, works when one of the two multipliers consists of all 9s. Al-Hawārī works through the example 444\times 999. We put them on two lines, and above them we put a row of dots equal to the sum of the number of places of the two numbers, in this case six:

First, 4\times 9=36. The 6 replaces the right-hand dot, and the 3 replaces the middle dot of the remaining dots:

Next, the difference 9-4=5 replaces the dots between the 3 and 6:

Last, the remaining dots are replaced with 4s:

Then 444\times 999=443{,}556.

To see why this works, note that multiplying a number of the form aaa (i.e., 100a+10a+a) by 999 is the same as aaa by 1000-1, which is of the form aaa000-aaa. The digits of the answer must then be a, a, a-1, 9-a, 9-a, and 10-a.

link to parallel text 112.5 Next is another method of multiplying by a number expressed with all 9s. This one has no restriction on the other number. Al-Hawārī’s example is 999\times 9{,}354. Add to the 9,354 as many 0s as there are 9s in 999 to get 9,354,000. Then subtract the 9,354 to get 9,344,646, which is the desired product.

link to parallel text 112.16 The method called “squaring” derives from the fact that [{1\over 2}(b+a)]^2-[{1\over 2}(b-a)]^2=ab. The method is much simpler to apply mentally than the modern formula suggests. To multiply 17 by 19, al-Hawārī squares half their sum, 18, to get 324. From this he subtracts a square of half the difference between them, which is 1, to get 323. This is the desired product.

Incidentally, in his Lifting the Veil Ibn al-Bannāʾ appropriates this rule as the foundation for arithmetical proofs for the rules for solving the three composite algebraic equations. The same rules are stated and illustrated in the present work starting at 214.7, but without proofs.[26]

link to parallel text 113.6 Another “squaring” method entails squaring one of the two numbers and multiplying or dividing the result by their ratio. To multiply 25\times 15, al-Hawārī squares the 25 to get 625. Because 25 is the greater of the two numbers, its square is multiplied by the ratio of 15 to 25, or 3\over 5, to get 375, which is the answer. He then works it out by squaring the 15 to get 225. Because 15 is the smaller number, its square is divided by the ratio 3\over 5, again giving 375. Algebraically, a\cdot b=a^2\cdot {b\over a}.

link to parallel text 113.16 In this next method, Ibn al-Bannāʾ makes use of rules that can be written in modern notation as ab=b^2-(b-a)b and ab=a^2+(b-a)a, for a<b. Al-Hawārī multiplies 36 by 14 using both rules. For the first rule, the difference 22 is multiplied by 36 to get 792. This is taken from 1296, a square of 36, leaving the answer 504. For the second rule, the difference 22 is multiplied by 14 to get 308. This is added to 196, a square of 14, to get the same answer, 504.

Al-Hawārī did not pick a good example to illustrate the utility of this trick. In one step in the first rule he has to multiply 22 by 36, which itself is no easier than finding 36 by 14 directly. The rules have an advantage if the difference between a and b is a nice number, like in the example of 16 by 26. The difference is 10, which multiplied by 26 easily gives 260. Subtract this from 676, a square of 26, to get 416. Or multiply the 10 by the 16 to get 160, and add this to 256, a square of 16, to again get 416.

link to parallel text 114.4 The last trick deals with multiplying multiples of powers of 10. To multiply 30 by 140 al-Hawārī first multiplies 3 by 14 to get 42, and to this he adds back the zeros to get the answer, which is 4,200.

link to parallel text 115.1 Students should memorize the multiplications of the whole numbers from 1 to 10.

link to parallel text 117.1 Section I.1.5. Division.

link to parallel text 117.2 Ibn al-Bannāʾ may not have taken into account non-integers in his definition of multiplication at 95.2 above, but he did so for division. He copied two definitions from Ibn al-Yāsamīn, one at 117.2, which calls for the decomposition of the dividend into equal parts, and the other at 117.5, based instead on ratio. Al-Hawārī copies Ibn al-Bannāʾ’s explanations from Lifting the Veil that the first “applies to discrete quantities”, that is, to whole numbers, while the second “concerns continuous quantities”, which in his case are the numbers of the arithmeticians that include fractions and irrationals. This second definition is echoed in Ibn al-Bannāʾ’s definition of a fraction, below at 133.1, but there the numbers in the ratio are assumed to be whole numbers. Neither definition is known from Greek sources (but see above at 95.2 for a discussion of Abū l-Wafāʾ’s attempt to link the first definition of division to Euclid and Nicomachus). At 117.7 Ibn al-Bannāʾ then gives the definition of division that is observed by “most people”, which instead counts how many divisors are in the dividend.

link to parallel text 117.9 The two meanings of division that he then offers stem from the two definitions at 117.2 and 117.7. The first is “the division of a type by another type, like dirhams by men”, and the second is “the division of a type by the same type”. This distinction corresponds to the notion of a change in terms for the two kinds of multiplication, above at 95.3.

link to parallel text 117.16 The examples of the two meanings, here and at 118.1, mirror the examples for multiplication at 95.10 and 95.12. For the first meaning, Ibn al-Bannāʾ divides 15 dirhams equally among five men, and for the second, he divides a piece of wood of fifteen spans into pieces of wood of three spans. The explanations given here correspond to those for multiplication at 95.3 and 95.6: decomposing (the 15 dirhams) and uniting (the 15 spans into groups of three) are the opposites of the combining (5 dirhams of 3 men) and dissecting (5 dirhams into thirds of a dirham) that we saw for multiplication.

link to parallel text 118.14 A distinction was often made in medieval Arabic arithmetic between dividing a greater number by a smaller number and dividing a smaller number by a greater number. Books written in the finger-reckoning tradition typically “divide” (qasama) the greater by the smaller, and “relate” (nasaba) the smaller to the greater.[27] Nasaba is the verb associated with nisba, the word for “ratio”. Ibn al-Bannāʾ and al-Hawārī use qasama similarly, but they follow al-Ḥaṣṣār and Ibn al-Yāsamīn by “denominating” (from sammā) the smaller with the greater. To denominate 3 with 7 means to give 3 the denomination, or name, “sevenths”. Saying the result “three sevenths” is like saying “three cents”, but with a different denomination. The prepositions are different between division and denomination, too. One divides a number by (ʿalā) a smaller number, while one denominates a number with (min) a greater number.

link to parallel text 118.17 Al-Hawārī illustrates the method of dividing a greater number by a smaller number with the example 245\div 12. First, the numbers are put on two lines like this, with the highest power terms lined up:

The 12 goes into the 24 of the 245 two times, so we write a 2 under the 12. Because the 2 and 4 of 245 are exhausted, they are replaced with zeros:

Next, shift the 12 one place to the right:

The 12 does not go into the 5 at all, so a zero is put below and the remainder is 5:

The remaining 5/12 is “two sixths and half a sixth”, which is written as 1~\,2\over 2~\,6. This kind of fraction is explained at 138.1. The final answer is {1~\,2\over 2~\,6}\,20, read right-to-left.

link to parallel text 119.18 The rules Ibn al-Bannāʾ gives from here up to and including 124.8 are from finger-reckoning. The first technique is to partition the dividend and then add the respective quotients. Al-Hawārī’s example is to divide 44 by 11. He breaks up the 44 into and , then he divides 22 by 11 to get 2 for each of the 22s. The quotient is then 2+2=4. This technique might seem more useful with a problem like dividing 399 by 19. Thinking of 399 as 380+19, we divide each number by 19 to get 20+1=21, which is the answer. In general, (a+b)\div c=(a\div c)+(b\div c).

The second technique is to factor the divisor and divide by each of them in turn. To divide 96 by 12, al-Hawārī decomposes the 12 into 2\cdot 6. He finds that 96\div 2=48, and this he divides by the 6 to get 8, which is the answer. In general, a\div (b\cdot c)=(a\div b)\div c.

The third technique is to “reconcile” (wafiqa) the dividend and the divisor, which means to cancel common divisors. To divide 35 by 15, al-Hawārī takes a fifth of the dividend and the divisor to change the problem into 7\div 3, which gives “two and a third”. In general, ab\div ac= b\div c.

link to parallel text 120.20 In “apportionment” (muḥāṣṣa), a certain quantity is deducted proportionally from two or more amounts. In al-Hawārī’s example three people want to give a total of ten dinars to a bankrupt friend, and they do it in such a way that each of them gives an amount proportional to how much each of them has, and they have 4, 5, and 6 dinars, respectively. In modern notation, we want to find three numbers a, b, and c such that a+b+c=10 and {a\over 4}={b\over 5}={c\over 6}.

Al-Hawārī begins by adding up the wealth of the three people: 4+5+6=15. These “surpass them”; in other words, they surpass the three “apportioned parts” that will be given, and which will add up to ten dinars. For the first friend’s gift al-Hawārī multiplies the 10 by the 4 dinars to get 40 dinars, and he divides the result by 15 to get his share, or “apportioned part”, which is 2{2\over 3} dinars. The same procedure gives the other shares. For the second friend this is 5\cdot 10\div 15=3{1\over 3} dinars, and for the third it is 6\cdot 10\div 15=4 dinars. This works because the first must give 4/15 of the total amount of 10 dinars, the second 5/15, and the third 6/15, and the proportions add to 1.

link to parallel text 121.14 Al-Hawārī mentions four variations on this method, each one being a reordering of the operations. For example, where the share of the first friend was calculated above by 10\times 4, then \div 15, the first variation is to calculate 4\div 15, then \times 10. The second is to calculate 10\div 15, then \times 4. In the third, we find 15\div 4, then we divide 10 by the result. In the fourth we find 15\div 10, then we divide 4 by the result. These variations were probably devised for cases in which one or another would be easier. For example, if instead of 4, 15, and 10, we had 8, 26, and 22, it would be easier to apply the first variation, by dividing 8 by 16 to get a half, then multiplying this by the 22 to get 11. The original way has us multiplying 8 by 22 and then dividing by 16, which is more difficult.

link to parallel text 122.2 In case there are fractions, multiply the numbers corresponding to the wealth of the friends by the least common multiple of their fractional parts. And if the fractions have common divisors, divide by those factors. The amount of money given to the bankrupt friend remains the same.

By “they are all different (mutabāyna)” in the passage at 122.10 Ibn al-Bannāʾ means they are relatively prime. Also, we translate ishtirāk, a word whose ordinary meaning is “common”, as “common divisor”.

Al-Hawārī considers the situation in which a total of 12 dinars is given by three friends who have 4{1\over 3}, 5{1\over 4}, and 6{1\over 6} dinars, respectively. He first calculates the least common multiple of the denominators by collecting factors: a 3 from the third, two 2s from the 4, and nothing from the 6 because we already have a 2 and a 3. So the least common multiple is 12. The 4{1\over 3}, 5{1\over 4}, and 6{1\over 6} are then multiplied by 12 to get 52, 63, and 74. Next he checks to see if there is a common factor among these numbers that can be cancelled out. In this case, the numbers are relatively prime. He then finishes the problem, following the rule in the first example.

link to parallel text 123.18 Next, Ibn al-Bannāʾ gives the rule for expressing the result of a denomination as a related fraction (see below at 135.10 and 138.1 for this type). We saw one example of this notation already at 119.7, where “two sixths and half a sixth” was written as 1~\,2\over 2~\,6.

link to parallel text 123.22 The example is to denominate 11 with 15. Decompose the 15 into 5 and 3 and put them under a line like this:

Then divide the 11 by the 3 and put the remainder, 2, above it:

Then divide the quotient 3 from 11\div 3 by the 5. Since 3 is less than 5, 3 is the remainder, and it too is written above:

This is spoken as “three fifths and two thirds of a fifth”, and we might write it in our notation like this: {3\over 5}+{2\over 3\cdot 5}.

This process was followed in situations where the denominating number is composite and greater than ten. It was convention that the denominators descend. One said “three fifths and two thirds of a fifth”, 2~\,3\over 3~\,5, rather than “two thirds and a fifth of a third” 1~\,2\over 5~\,3. Also, it was preferred to start with the largest possible denominator, so we read “half a sixth” more often than “a third of a fourth”.

link to parallel text 124.8 There are three “lesser known” ways to denominate. Al-Hawārī’s example of the first is to denominate 4 with 12. He switches the numbers and divides 12 by 4, then denominates 1 with the result. Here 12\div 4=3, and the denomination gives 1\over 3. This works best in simple cases where there is a common factor. It would not work so well with a problem like 11\div 15, since 15\div 11=1{4\over 11}, and it would be too difficult to denominate 1 by this number. This method uses the property that {a\over b}=1/{b\over a}.

His example for the second way is to denominate 9 with 15. Denominating 1 with the 15 gives “a third of a fifth”, or 1~\,0\over 3~\,5. Multiplying this by 9 gives “three fifths”, or 3\over 5. Here, to find a\div b, one first finds 1\div b, then the result is multiplied by a.

In the third way, al-Hawārī denominates 10 with 16. The idea is to turn denomination into division by multiplying the 10 by some number to make it greater than 16. He chooses 8, to get 10\times 8=80. Dividing this by 16 gives 5, and to compensate for the multiplied 8 he denominates the result with 8 to get 5\over 8. This works best when the new numerator (here 80) is a multiple of the denominator, and the method is convenient only when there is a common factor (in this case 2). In modern notation, the motive becomes somewhat lost: if a<b then {a\over b}={ac\over b}/c.

link to parallel text 124.20 Ibn al-Bannāʾ now turns to the decomposition of numbers into their prime factors. This will be useful later in the manipulations of the denominators of fractions. He gives several rules, some of which warrant explanation because of the way they are worded.

The first case is a number that “does not begin with units”. Recall that the units are 1, 2,…, 9. The 0, signifying nothing, stands for an empty place. So a number that “does not begin with units” has a 0 in the units place, and is divisible by 10, 5, and 2. This is what he means when he says that it “has a tenth and a fifth and a half”.

link to parallel text 125.5 One way of finding factors is to consider the residue after casting out nines or eights or sevens. Ibn al-Bannāʾ first covers various cases for a number whose units digit is even. The result of casting out nines is also the remainder after dividing by 9, so from this we can tell if the number is divisible by 3 or 9. If it is cast out entirely by nines – in other words, if nothing is left – then the number “has a ninth and a sixth and a third”, or, in our language, it is divisible by 9, 6, and 3.

Al-Hawārī gives the example of 36. He does not mention that it is also divisible by 18. There is no need to say it is divisible by 2 since the number is already known to be even. If the residue is 3 or 6 then the number is divisible by 3, and because we already knew it was even, it is also divisible by 6. Al-Hawārī gives the examples 66 and 42.

If the residue is some other number (1, 2, 4, 5, 7, or 8), then cast out eights. If it is cast out entirely then the number is divisible by 8 and 4, and if the residue is 4, then it is divisible by 4. If the residue is some other number, then cast out sevens. Because 7 is prime, the only rule is that if it is cast out entirely then the number is divisible by 7.

Similar rules are then given for odd numbers.

link to parallel text 126.14 Subsection on finding deaf parts.

If all these rules have been applied and the original number is still not decomposed completely, “then look for deaf parts by dividing by them”. (We explain the term “deaf part”, meaning “prime”, below at 134.2.)

link to parallel text 127.9 To find deaf parts, Ibn al-Bannāʾ explains how to draw a table of odd numbers to make the sieve. This is the famous “sieve of Eratosthenes”, after the third century BC Greek mathematician. Ibn al-Bannāʾ’s description ultimately derives from Nicomachus’s Arithmetical Introduction.[28] Al-Hawārī draws the table to find the prime numbers less than 145. The first number in the table is 3, which is prime. Putting a bar over every third number after that gives:

The next number in the table without a bar is the 5, so it is prime. Put a bar likewise above every fifth number after it, starting with 15, 25, etc. Then do the same for the next number without a bar, which is 7, then for 11. There is no need to do this for 13 or any greater prime, since their squares are greater than 145. The table should now look like this:

Numbers with a bar are composite, and the remaining numbers are prime. Al-Hawārī writes that “these deaf parts are counted only by one”, or as we would say, prime numbers are divisible only by 1. One can say, for example, that 5 counts 40 because one can count to 40 by fives. But one cannot count to a prime number by anything but ones.

link to parallel text 128.1 Al-Hawārī writes that he is filling the table to 145, but below that he writes that we know, presumably from the table, that 151 is prime. The Oxford and Medina manuscripts show the table to 145, while the Tunis and Tehran manuscripts show it to 199. The table in the Istanbul manuscript is drawn with 15 columns and 10 rows. The copyist must have been confused. He began by filling in the odd numbers, from 3 to 37, and then continued by writing in only prime numbers from 83 to 157. The rest of the squares are blank, and no marks are drawn above any number.

link to parallel text 129.1 Section I.1.6. Restoration and reduction.

In problems involving proportion it is often necessary to restore or reduce a number to another number. “Restoration” (al-jabr) is what we do when we want to increase a number to a greater number, and “reduction” (al-ḥaṭṭ) is for reducing a number to a smaller number. These operations are applied, for instance, to set the “coefficient” of the highest power in an algebraic equation to 1 (at 217.1). Al-jabr is also the word used in the simplification of subtractions and equations in algebra (in the text at 211.2, 220.5, 223.1, and in our commentary at 223.1). Restoration and reduction of fractions is covered at 154.1.

link to parallel text 131.1 Chapter I.2. Fractions.

Fractions in Arabic arithmetic are numbers that result from partitioning the unit. Ibn al-Bannāʾ, for example, writes in the passage at 137.1 the fraction “fifteen…parts of twenty-four parts of the unit” ( 15\over 24). The unit is partitioned into 24 parts, and the fraction is 15 of those parts. But to define fractions, Ibn al-Bannāʾ feels obliged to respect the Greek notion that the unit is indivisible. Because of this he identifies fractions with ratios of whole numbers, so the fraction just mentioned is considered to be the ratio of 15 to 24. A ratio according to Euclid is “a sort of relation in respect of size between two magnitudes of the same kind”.[29] Ratios are not mathematical objects in themselves, but are only relations between such objects. This way, the unit maintains its integrity. Ibn al-Bannāʾ’s definition only serves to provide Arabic fractions with a semblance of a Greek foundation. It had no impact on actual calculations, and in Arabic arithmetic books, al-Hawārī’s included, fractions were understood to be fractional portions of the unit.[30]

Euclid defines “part” and “parts” at the beginning of Elements, Book VII, the first of his three books on number theory: “A number is a part of a number, the less of the greater, when it measures the greater…” For example, four is “a part” of twelve because twelve can be broken into three equal parts, each one of them four. He continues, “…but parts when it does not measure it”.[31] Eight, for example, is “parts” of fourteen because, if we consider the number two as one part, then eight is four of the seven parts making up fourteen, or, as we might say, eight is four-sevenths of fourteen. Similarly, five is “parts” of twelve where one part is the unit.

The Greek term meaning “part” is translated into Arabic as juzʾ. This is the same word meaning “part” used in the statements of simple fractions in Arabic with (usually prime) denominators greater than ten, like Ibn al-Bannāʾ’s 15\over 24 quoted above. The big difference is that in Greek “a part” is always a positive integer that is part of a greater integer, while in Arabic fractions it is the unit itself that is partitioned.

We describe the different ways Arabic arithmeticians expressed fractions below at 134.2. It is there that we explain Ibn al-Bannāʾ’s phrase “the part and its name”.

link to parallel text 133.3 Al-Hawārī goes further than his teacher in addressing the ontology of fractions. Had he worked with the common notion of fractions as parts of a divisible unit, he could have written that a fraction like “three sixths” is named in terms of sixths, much like “three cats” is named in terms of cats. In both cases the “three” modifies the name. But al-Hawārī follows Ibn al-Bannāʾ in formally identifying fractions with ratios. He argues that the ratio of three to six is not named in terms of either number, nor in terms of the two together.[32]

Al-Hawārī’s distinction between “sensible” and “intelligible” ultimately comes from Aristotle. For Aristotle, mathematical objects are sensible objects. We experience lines, squares, spheres, and numbers through our senses as attributes of the physical things we see and touch. But a ratio, being a relation between two objects, cannot be apprehended through the senses. It is an intelligible object that can only be imagined in the mind.[33] Al-Hawārī’s statement that ratios are intelligible objects is the only philosophical observation he makes in the book. We suspected that he may have copied it from somewhere, so we checked the obvious places: Aristotle, Ibn Sīnā, and Ibn al-Bannāʾ. Aristotle makes no comment on the ontology of ratios in his extant works, nor did we find anything in Ibn al-Bannāʾ’s writings. But we did find this remark of Ibn Sīnā on relations in general: “As for predicating [the quiddity of the relative] with respect to another, this occurs only in the mind”.[34]

We set this aside for a moment and continue with the rest of the passage. Al-Hawārī next paraphrases Ibn al-Bannāʾ’s Lifting the Veil[35] when he explains the word “fraction” by comparing a fractional number with fractured land. Perhaps he was thinking of the incremental increase of fractions with the same denominator, like 1\over 17, 2\over 17, 3\over 17, etc., that model the incremental strata of some rocky landscape. He also compares fractions with geometric magnitudes, probably because lines, surfaces, and bodies can be partitioned into arbitrarily many parts. Al-Hawārī contrasts these continuous magnitudes with “discrete quantities”, echoing the notion in Aristotle and Euclid that numbers do not admit fractions due to the indivisibility of the unit.

We contend that Ibn al-Bannāʾ related nearly the entire passage at 133.3, from “For example…” to “…similar abstractions”, verbally to al-Hawārī. Ibn al-Bannāʾ was familiar with Ibn Sīnā’s work, and al-Hawārī paraphrased rather than quoted Ibn al-Bannāʾ’s comparison of fractions with terrain from Lifting the Veil.

link to parallel text 134.1 Section I.2.1. The names of fractions and numerating them.

link to parallel text 134.2 Fractions in most Arabic books on Indian arithmetic, al-Hawārī’s included, are borrowed from finger-reckoning. This particular system has its origin in the ancient Egyptian practice of expressing fractions as sums of unit fractions, like writing 2\over 5 as the sum of 1\over 3 and 1\over 15. It was later modified, probably by Greek calculators, with approximation techniques based in sexagesimal arithmetic.[36]

At some point before the ninth century CE the restriction to unit fractions was dropped. One could now say fractions like “five sevenths” and “two ninths”. But in Arabic fractions could only be formed from nine names, or “heads” (ruʾūs): “a half”, “a third”, “a fourth”, up to “a tenth”. There were no words in Arabic for “eleventh”, “twelfth”, “thirteenth”, etc., so fractions with denominators greater than ten were expressed when possible by combinations of the heads. For instance, al-Khwārazmī wrote “a fifth and four fifths of a fifth” for 9\over 25 and “a ninth and a tenth” for 19\over 90.[37] The various ways of combining the heads already present in ninth century texts remained common throughout the medieval period, and are explained by al-Hawārī in the present section.

Numbers in medieval Arabic that could be expressed in words were called “expressible” (munṭaq).[38] These included integers and fractions reducible to combinations of the heads. Numbers inexpressible in words, for which only approximations could be found, were often called “deaf” (aṣamm). These included irrational roots, and also fractions irreducible to combinations of the heads, that is, fractions whose denominators contain prime factors greater than ten. The problem of deaf fractions was overcome some time before the early ninth century with the introduction of “parts”. For example, al-Khwārazmī wrote 4\over 13 as “four parts of thirteen parts of a dirham”.[39] With the dirham divided into thirteen equal parts, the fraction is four of those parts. So now, in addition to being able to say “four fifths” and “four sevenths”, one could say “four parts”. Thus, “a part” became the tenth name for the naming of fractions.

This accounts for the origin of the term “deaf parts” for prime numbers, above at 126.14. To express the result of a denomination as a related fraction, explained at 123.18 above, one needs to find the prime factorization of the denominated number. If this number contains a factor not divisible by 2, 3, 5, or 7, then one needs to look for “deaf parts”, that is, to find the greater prime denominators required for the fraction. The example at 151.14 results in “fifty parts of one hundred thirty-seven parts and five ninths of a part of one hundred thirty-seven parts”. The deaf parts here are the 137 parts. Prime numbers do not quite coincide with the denominators of deaf fractions, since the former include 2, 3, 5, and 7, but the association was close enough.

In arithmetic, the naming of fractions with parts typically terminates with “of a unit/one” or “of a dirham” (units were often counted in dirhams, a silver coin, in many calculations). In one problem al-Karajī writes “thirty-five parts of eighty-three parts of a unit” and then, just after, “thirty-four parts of eighty-three parts of a dirham”.[40] Often, as in al-Hawārī, the designation is left off altogether. So his “a part of eleven” is short for “a part of eleven parts of one/a unit/a dirham”. Also, it is not just abstract units that are partitioned in Arabic fractions. In the inheritance problems solved by al-Khwārazmī it is sometimes a share of the estate, as in “twenty-three parts of fifty-nine parts of a share”,[41] and in the context of algebra Abū Kāmil writes “fifteen parts of thirty-nine parts of a thing”, where the “thing” is the name of the first degree unknown.[42]

Grammatically, the ten names function like other Arabic nouns. Saying “three fourths” or “three parts” is like saying “three apples”. The “fourths”, “parts”, and “apples” are the names, or kinds of object counted, while the “three” tells how many there are (the same applies also to “shares” and “things”). Our own words “numerator” and “denominator” reflect this idea. The word “denominator” derives from the Latin dēnōmināre, “to call, to name”, and our word “name” comes from the related Latin word nōmen. Our “numerator” is the “number” of this name.

There was no rule that the language of parts could only be used for deaf fractions. Al-Khwārazmī and Abū Kāmil routinely use it to name fractions like 25\over 36, 14\over 15, 5\over 12, and 4\over 25. And when convenient for the calculations the fraction can even be improper, like al-Khwārazmī’s “twenty-eight parts of thirteen of a dirham”.[43] Al-Hawārī, though, works with combinations of the heads whenever possible, and his numerator is always less than his denominator.

Although schoolbooks today often still explain fractions in terms of parts, mathematicians define them in terms of division. The value of al-Khwārazmī’s “four parts of thirteen parts of a dirham” may be equal to the result of dividing 4 by 13, but the two ways of regarding fractions are not equivalent. With quotients in mind, we allow numerators and denominators to be irrational, like 1\over \sqrt{2} and \sqrt{2}\over 2. But these numbers cannot be fractions in medieval Arabic. The first makes no sense because one cannot partition the unit (or anything else, for that matter) into an irrational number of parts. And the second does not work because the number of parts making up a fraction cannot be irrational. To have \sqrt{2} parts is just as meaningless as having \sqrt{2} loaves of bread. One can perform the corresponding divisions, however, to get the acceptable “root of a half” ( \sqrt{1\over 2}). We explain this below at 188.1.

link to parallel text 135.1 When the denominator is a composite number greater than ten it was common to express it with “two or more names”. Al-Hawārī’s first example is “two eighths and a seventh of an eighth”, which in notation looks like this: 1~\,2\over 7~\,8. We can write this as a modern fraction by working it out to get 15\over 56. Although it may sound overly complicated to us, saying “two eighths and a seventh of an eighth” gives a good idea of the magnitude of the fraction. We know that “two eighths” is a fourth, and we are adding to that the small amount “a seventh of an eighth”. The fraction is expressly presented as being just a little over 2\over 8, while with our 15\over 56 this is not immediately clear. One of us even found himself thinking in terms of these fractions while measuring wood with a ruler scaled in inches. The fractional part was clearly three fourths and a fourth of a fourth, and it would have been much less transparent as well as superfluous to convert 1~\,3\over 4~\,4 to 13\over 16. We can all grasp the magnitudes of the fractions 1\over 2, {1\over 3},\ldots ,{1\over 10}, and combinations of these are often easier to understand than fractions with large composite denominators.

An example in which the related fraction is not easier to grasp is Ibn al-Bannāʾ’s artificial example 2~\,4~\,5\over 3~\,5~\,6 (at 135.10 below), which is equal to 89\over 90. And just as our fractions do not admit of a unique representation ( {1\over 2}={2\over 4}={3\over 6} etc.), related fractions can be written in many ways. One example is al-Hawārī’s 3~\,4\over 6~\,9 in the example at 171.6, which is equal to 1\over 2.

Al-Hawārī uses the language of parts when the denominator is not prime in only two instances, at 155.2 and 156.7. There the denominators are 87 and 93.[44]

link to parallel text 135.8 To expand on Ibn al-Bannāʾs remark, al-Hawārī copies Lifting the Veil from 135.10 to 137.10 to explain how to find the numerators of three kinds of fractions: related, distinct, and portioned. These types are covered again immediately after by al-Hawārī himself, along with other types. Finding the numerator is necessary to perform every kind of operation: addition, subtraction (including casting out, above at 93.9), multiplication, division/denomination, conversion, and finding square roots.

link to parallel text 135.10 Ibn al-Bannāʾ works with the example 2~\,4~\,5\over 3~\,5~\,6 to show how to find the numerator and denominator of a related fraction. These fractions are expressed with more than one name, or denominator, so that each part is related to, or is a fraction of, what precedes it. In this example, the 5 in the top is multiplied by the second denominator, also a 5, to get 25, and this is added to the 4 to get 29. So the first two terms of the fraction are equivalent to 29\over 5\cdot 6, or 29\over 30. Then this 29 is multiplied by the 3 to get 87, which is added to the 2 above, resulting in the numerator 89. The fraction is the same as our 89\over 90.

link to parallel text 136.8 Sometimes two fractions are gathered with the conjunction “and” (wa), like the example “five sixths and four fifths”. In notation, one is written next to the other like this: {4\over 5}\,{5\over 6}. This is called a distinct fraction, and its value is the sum of the individual fractions. To find the numerator, Ibn al-Bannāʾ multiplies the 5\over 6 by the denominator 5 of the other fraction to get 25\over 30 (though he does not name this fraction). He then multiplies the 4\over 5 by the 6 to get 24\over 30. Together they are 49\over 30, or forty-nine “parts of thirty” (or “fifths of sixths” or “sixths of fifths”), so the numerator is 49.

link to parallel text 137.1 A portioned fraction is one like “three fourths of five sixths”, written as 5~\bullet ~3\over 6 ~\bullet ~4. This may be equivalent to the product of the fractions, but it was conceived and stated as a fraction of a fraction. In modern notation the example becomes 15\over 24.

link to parallel text 137.11 Al-Hawārī now returns to commenting on the Condensed Book. Ibn al-Bannāʾ names five different kinds of fraction, and al-Hawārī shows how to find the numerator of each.

link to parallel text 137.13 Simple fractions. The numerator of a simple fraction, like “a seventh”, written 1\over 7, is the number above the line. From the example at 139.2, we know that for Ibn al-Bannāʾ simple fractions are not restricted to those with a numerator of 1. So a fraction like “four sixths” ( 4\over 6) is also simple, and its numerator is the 4 above the line.

Combined fractions. This may not have been considered as a separate type since neither al-Hawārī nor Ibn al-Bannāʾ describe it. A combined fraction is a fraction whose representation has all 0s in the numerator except on the extreme left, like Ibn al-Yāsamīn’s 21\,\,0~\,0~\,0\over \,2~\,6~13~7. In modern notation, the fraction is 21\over 2\cdot 6\cdot 13\cdot 7, or 21\over 1092, and the numerator is 21. (Ibn al-Yāsamīn preferred to write his fractions in reverse order, so he shows it like this: \,0~\,\,0~\,0~21\over 7~13~6~\,2.[45])

link to parallel text 138.1 Related fractions. A related fraction is the common type with “two or more names”, which we have seen several times already. Al-Hawārī’s example is 2~\,3~\,4~\,5\over 3~\,5~\,7~\,8. Ibn al-Bannāʾ gave two rules, the first being the same as the one from Lifting the Veil cited above at 135.10. The fraction can be written in modern form as {596\over 3\cdot 5\cdot 7\cdot 8}={596\over 840}. The second rule is explained as clearly as the first.

link to parallel text 139.1 Distinct fractions. Distinct fractions are those that are added, or gathered together. They have already been described above at 136.8. Al-Hawārī finds the numerator for “five sevenths and half a seventh and four sixths”. In notation this is {4\over 6}~{1~\,5\over 2~\,7}, which we would write as {11\over 14}+{4\over 6}. Combining them is equivalent to our method of cross multiplication, so the numerator is 11\cdot 6+4\cdot 14= 122.

link to parallel text 139.10 Portioned fractions. The numerator of fractions of fractions is the product of the individual numerators. These were described above at 137.1.

link to parallel text 140.1 Excluded fractions of disconnected type. These are called “excluded” because they are expressed as a fraction removed or excluded from a greater fraction. For an explanation of the terms “less”, “diminished”, and “excluded”, see our remarks at 219.4 below. They are “disconnected” because the second fraction is not taken of the first fraction, but is independent of it. The example here is “six eighths less a ninth”, written as {1\over 9}~\scalebox {-1}[1]{\ell }~{6\over 8}. This fraction is drawn in the Medina manuscript as Too late. (See our commentary above at 86.1 for an explanation of the sign for “less”.) In modern notation we write it as {6\over 8}-{1\over 9}. By “is not taken from what precedes it”, al-Hawārī means that this is not {6\over 8}-{1\over 9}\left({6\over 8}\right). The procedure is like that for distinct fractions, but involves taking the difference rather than the sum of the products. Here 6\times 9=54, and 1\times 8=8. The numerator is 54-8=46.

link to parallel text 140.14 Excluded fractions of connected type. Al-Hawārī converts the example “a half less its third” to the disconnected version “a half less a sixth”. He then finds the numerator of “six sevenths and half a seventh less its third”. By “its third” he means that you take away a third of the six sevenths and half a seventh. In notation it is written just like an excluded fraction of disconnected type, {1\over 3}~\scalebox {-1}[1]{\ell }~{1~\,6\over 2~\,7}, though the meaning is different. In modern notation we would write 1~\,6\over 2~\,7 as 13\over 14, and the whole fraction is equivalent to our {13\over 14}-{1\over 3}\left({13\over 14}\right). Take the numerator of the greater part, which is 1~\,6\over 2~\,7, to get 13, and multiply it by the denominator of the smaller part, which is 3, to get 39. Then multiply the two numerators: 1\times 13=13. The numerator is then 39-13=26, and the fraction is equal to 26\over 42.

The word “connected” for excluded fractions takes the same meaning as the word “related” for related fractions, in that the fraction that follows is taken as a portion of the preceding fraction. So it is natural that in cases where the diminished amount is itself a related fraction the notation was sometimes written with a single bar. For this example, if the “less” were instead an “and”, it would have been “six sevenths and half a seventh and a third of a half of a seventh”, or 1~\,1~\,6\over 3~\,2~\,7. When the last fraction is taken away instead of added, as in “six sevenths and half a seventh less its third”, it might be written as 1~\scalebox {-1}[1]{\ell }~1~\,6\over 3~~~\,2~\,7 instead of the ambiguous {1\over 3}~\scalebox {-1}[1]{\ell }~{1~\,6\over 2~\,7}. Four of the five manuscripts we consulted show the latter, but the Tehran manuscript shows the fraction with the bar extending all the way across: Too late (this manuscript shows the Eastern forms of the numerals).

It was not a problem to have one notation with two possible meanings. While working through a problem, one knows whether the fraction is connected or disconnected. Notation was used by the individual to perform calculations, not to communicate the work to others or to preserve a record for future consultation. A rhetorical version served those purposes, and there the ambiguity is erased by the language.

link to parallel text 141.7 Al-Hawārī then gives some “additional remarks” of Ibn al-Bannāʾ’s. Recall from 86.1 above that an expression like “ten less eight less seven less five less two” is 10-(8-(7-(5-2))). Here Ibn al-Bannāʾ explains that when the word “and” (wa) appears before each “less”, then the numbers are all taken away from the first term. Al-Hawārī’s example is “five and a third less its fourth and less its seventh and less its fifth”, and is written in notation as {1\over 5}~\scalebox {-1}[1]{\ell }~{1\over 7}~\scalebox {-1}[1]{\ell }~{1\over 4}~\scalebox {-1}[1]{\ell }~{1\over 3}\,5. We might write it as 5\,{1\over 3}-{1\over 4}\left(5\,{1\over 3}\right)-{1\over 7}\left(5\,{1\over 3}\right)-{1\over 5}\left(5\,{1\over 3}\right). The figure can be reduced to {1\over 5}\,{1\over 7}\,{1\over 4}\,\scalebox {-1}[1]{\ell }\,{1\over 3}\,5, with a single “less”.

The “particle of exclusion” is most often the word “less” (illā), as it is here. But sometimes in Arabic mathematics other words, meaning “except” or “other than” (ghayr or siwa), are used. The Arabic word for “detached” (munfaṣil) is related to the word we later translate as “apotome” (munfaṣila) in the chapter on roots. See our remarks below at 173.10.

link to parallel text 142.6 Here al-Hawārī gives a disconnected example. If we said “five and a third less a fourth of one and less a seventh of one and less a fifth of one” it would be equivalent to the modern 5{1\over 3}-{1\over 4}-{1\over 7}-{1\over 5}, or 5{1\over 3}-{83\over 140}={1991\over 420}. He does not show the notation for this.

link to parallel text 142.10 Again quoting Ibn al-Bannāʾ, al-Hawārī explains the meaning of the repeated “less” and cites the same rule given at 86.9 for whole numbers. He does this for the connected type (like “…less its fifth”) and the disconnected type (like “…less a fifth of one” or simply “…less a fifth”). He does not mention that the work starts from the last term.

link to parallel text 143.1 The “three of its fourths” must become disconnected from the “five sixths”. So the fraction is rephrased as an excluded fraction of disconnected type.

link to parallel text 143.5 Al-Hawārī finds the numerator for the mixed fraction {3~5\over 4~6}\,5, or 5\,{23\over 24} in modern notation. He first multiplies the whole number 5 by the two denominators to get 120. He then adds this to the numerator of the fraction, which is 23, to get 143. We can write the number as the fraction 143\over 24.

link to parallel text 143.13 If the whole number comes after the fraction (that is, it is placed on the left), then the number is the fraction of the whole number. Al-Hawārī’s example is 10\,{6\over 8}\,{4\over 7}, which is “four sevenths and six eighths of ten” which we might write as \left({4\over 7}+{6\over 8}\right)\cdot 10. The numerator of the fractions is 74, since together they are 74\over 56. Multiply this 74 by the 10 to get the numerator 740.

link to parallel text 144.4 If the whole number is written between two fractions, then it might be attached to the first fraction or to the second. If it is attached to the first, or “to what precedes it”, then one proceeds as in al-Hawārī’s example {3\over 6}\,5\,{4\over 9}, which is “four ninths of five, and three sixths”. That is, take 4\over 9 of 5, then add 3\over 6. Al-Hawārī multiplies the 4 by the 5 to get 20, the numerator of the first part. We can now look at it as the addition of 3\over 6 to 20\over 9. He effectively cross multiplies, adding 20\times 6 to 3\times 9 to get the numerator 147.

link to parallel text 145.1 The whole number might be attached to the second fraction, or “what is after it”. Al-Hawārī’s example is “two thirds of seven and four sevenths”, or {4\over 7}\,7\,{2\over 3}. In modern notation this is equivalent to {2\over 3}\cdot \left(7\,{4\over 7}\right). First we find the numerator of the mixed fraction 7\,{4\over 7} to get 53. The problem is now {2\over 3}\cdot {53\over 7}. Then the 53 is multiplied by the 2 to get 106, which is the numerator.

As we explained above at 140.14, the ambiguity of the notation is not problematic for the person working through the calculations.

link to parallel text 146.3 Al-Hawārī often does not cancel common factors in his fractions. The most glaring examples could have been simplified easily, like “four sixths” at 139.2, “six eighths” at 140.8, and “three sixths” at 144.10. But here Ibn al-Bannāʾ states that one must decompose the numerator and denominator into their prime factors and cancel the common factors, and al-Hawārī explains it for the case of portioned fractions.

link to parallel text 147.1 Section I.2.2. Adding and subtracting fractions.

link to parallel text 147.2 Now we turn to operating on fractions, beginning with addition. To add two fractions Ibn al-Bannāʾ gives the rule that one multiplies the numerator of each fraction by the denominator of the other, and then the sum of the results is divided by the product of the denominators. Al-Hawārī’s example is to add {6\over 8}\,{4\over 5}\,3 to 1~3~\,\,4\,\over 2~8~10. He gives instructions to put the addend over the augend. The operations are clearer if we write the first fraction as 182\over 40 and the second as 71\over 160. He multiplies 182 by 160 to get 29,120. Next he multiplies 71 by 40 to get 2,840. Adding 29,120 to 2,840 gives 31,960, which is the numerator of the sum. The denominator is the product of the denominators of the two numbers: 40 \times 160=6{,}400. So the sum is 31,960\over 6,400, which he prefers to express as “four and nine tenths and seven eighths of a tenth and half an eighth of a tenth”, and to write as {1~7~\,\,9\,\over 2~8~10}4.

Al-Hawārī then remarks “And its answer is given by (bi-) five”. This might be read as an approximation, since the answer, as we would write it, is 4{159\over 160}. But phrases like this follow the sample calculations for the other operations on fractions (subtraction, multiplication, division, denomination), and the “answer” in these cases is often not close to the number given. At 149.8, he writes “and its answer is given by one” for the number that as a decimal is approximately .3246. At 150.2, he writes “And its answer is given by two” for the number 1{5\over 6}, and at 151.4 he has “And its answer is four” for 3{1\over 63}. All we can say is that the stated “answer” is the smallest whole number greater than the calculated value. Then in two cases, at 147.15 and 151.14, he writes “And its answer is given by removal/subtraction (ṭarḥ)”. In these two questions one of the numbers is diminished. We have not been able to decipher the meanings of these phrases, which in any case are given after the exact answer to the question is found.

link to parallel text 147.15 Al-Hawārī’s next example is to subtract {1\over 3}~\scalebox {-1}[1]{\ell }~2\,{7\over 10} from {5\,\bullet \, 3\over 6\,\bullet \, 4}\,4. In our notation this is {111\over 24}-{32\over 30}. Al-Hawārī rhetorically works through (111\cdot 30)-(32\cdot 24)=2,562. This is the numerator of the difference. The denominators of the two fractions are already given as 6, 4, 3, and 10. Al-Hawārī silently cancels a 6, and instead of working with “thirds of fourths of tenths” he switches to the more customary “halves of sixths of tenths”. The result is stated as “three and five tenths and three sixths of a tenth and half a sixth of a tenth”, or {1~3~\,\,5\,\over 2~6~10}3.

link to parallel text 149.1 Section I.2.3. Multiplying fractions.

link to parallel text 149.2 Ibn al-Bannāʾ characterizes, or defines, the multiplication of fractions as “the portioning of one of the two fractions by the amount of the other”. Said a little differently, one takes the portion of one number according to the fraction of the other, which is of course also the meaning of portioned fractions described above at 137.1. So to multiply three fourths by two thirds, for example, one takes two thirds of the three fourths to get one half. As noted in the text, this definition is different from the definition of multiplication for whole numbers given at 95.2.

Both definitions work for the multiplication of a whole number by a fraction. The first is to find “the whole number by the amount of the fraction”. To multiply 10 by 3\over 5, for example, take 3\over 5 of 10 to get 6. By the second definition, one duplicates the fraction as many times as there are units in the whole number. Duplicating 3\over 5 ten times again gives 6.

link to parallel text 149.6 The rule for multiplying fractions is just what we know it should be: multiply the numerators and divide the result by the product of the denominators. Al-Hawārī sets up his examples the same way he did for adding and subtracting fractions, with one number placed above the other. To multiply {1\over 3}\,{3\over 4} by {1~\,4~\,3\over 5~\,6~\,9} he multiplies the numerators 13 and 111 to get 1443, which he divides by the denominators 3, 4, 5, 6, and 9. This gives “four ninths and a fourth of a fifth of a sixth of a ninth”, or 1~\,0~\,0~\,4\over 4~\,5~\,6~\,9. He would have arrived at this form by canceling the common 3 to get 481\over 4\cdot 5\cdot 6\cdot 9, and then by following the rule for denomination given above at 123.18.

link to parallel text 150.2 In the next example the numbers are set up similarly. The notational versions are written one above the other, the numerators are found, and the multiplication follows. Here the numerators of {1\over 8}\,4\,{1\over 3} and 10\,{2\,\bullet \, 1\over 3\,\bullet \, 5} are 33 and 20, respectively, so their product is 660. The denominators are given as 8, 3, 3, and 5. Everything but a leftover 2 and a 3 cancel, so the answer is “one and five sixths”.

link to parallel text 151.1 Section I.2.4. Division and denomination.

link to parallel text 151.2 This rule is also what we would expect. To divide or denominate a\over b by or with c\over d, divide/denominate ad by/with bc. Al-Hawārī’s example for division is to “divide six and a third by four fifths of seven eighths of three”:

The numerator of the top fraction is 19, and this is multiplied by the denominator of the bottom number, which is 40, to get 760. The other numerator and denominator are 84 and 3, and their product is 252. The answer is what we get from dividing 760 by 252, which is “three and a seventh of a ninth”, or {1~\,0\over 7~\,9}\,3. Here the denominators 7 and 9 do not come from the denominators in the given numbers. To get them, note that 252 goes into 760 three times with a remainder of 4. The fraction 4/252 is the same as 1/63, and 63 is 7\times 9. Thus the remainder is “a seventh of a ninth”.

link to parallel text 151.14 The example for denomination is to “denominate three and a fourth less two ninths of it with six and two eighths and three fifths”. The calculations are easier to follow if we rewrite the problem as {91\over 36}\div {274\over 40}. After cancelling eights, the denominator of the quotient is 1233, which is 9\cdot 137.

link to parallel text 152.9 There is no need to do all this work if the denominators of the two fractions are the same. In this case, just divide the numerator of the top number by the numerator of the bottom number. Al-Hawārī again gives examples for both division and denomination, and his calculations are clear.

link to parallel text 153.7 If the numerators are equal, then for division one divides the denominator of the divisor by the denominator of the dividend, and similarly for the case of denomination. The example for division is to “divide five by five sixths”. Al-Hawārī notes that the denominator of a whole number is 1, so he divides 6 by 1 to get 6. Similarly, denominating 5\over 6 with 5 gives “a sixth”.

link to parallel text 154.1 Section I.2.5. Restoration and reduction.

This chapter is the version for fractions of Section I.1.6 (at 129.1 above). Ibn al-Bannāʾ breaks up his explanations of restoration into six “problems”: (1) restoring a fraction to a fraction, (2) restoring a fraction to a whole number and a fraction, (3) restoring a fraction to a whole number, (4) restoring a whole number to a whole number and a fraction, (5) restoring a whole number and a fraction to a whole number, and (6) restoring a whole number and a fraction to a whole number and a fraction. He omits the two cases of (a) a whole number to a fraction, and (b) a whole number and a fraction to a fraction, since in both cases the first number is necessarily greater than the second, so the problem belongs to reduction. There are also six types of reduction, this time omitting the two that necessarily belong to restoration.

The rule is the same as that given above in Section I.1.6. Reviewing one example each for restoration and reduction should suffice. For the third type of restoration, al-Hawārī restores “two thirds of five sevenths so that it gives ten”. The answer is found by dividing 10 by 5~\bullet ~2\over 7~\bullet ~3, which gives 21. This means that to restore 5~\bullet ~2\over 7~\bullet ~3 to 10, one multiplies it by 21. The example for the first type of reduction is to “reduce seven tenths so that it becomes a third”. Denominate 1\over 3 with 7\over 10 to get 10\over 21, which is “three sevenths and a third of a seventh”, or 1~\,3\over 3~\,7. So, to reduce 7\over 10 to 1\over 3 one multiplies the 7\over 10 by 1~\,3\over 3~\,7.

link to parallel text 157.1 Section I.2.5. Converting.

Sometimes one wants to change a fraction from one name to another, like from thirds to fifths. Two thirds, for example, is the same as three and a third fifths. We might awkwardly write this as 3{1\over 3}\over 5, but it would have been expressed by al-Hawārī as “three fifths and a third of a fifth”, which in notation is 1~\,3\over 3~\,5. Al-Hawārī begins the section on converting (taṣrīf) by quoting from Lifting the Veil.

link to parallel text 157.2 Ibn al-Bannāʾ speaks about two kinds of conversion. The first type “concerns only the name”. Ibn al-Bannāʾ’s example is to convert “five sixths and three fourths” ( {3\over 4}~{5\over 6}) to tenths. “Naming the fraction in tenths” results in “one and five tenths and five sixths of a tenth”, or {5~\,\,\,5\,\over 6~\,10}\,1. Al-Hawārī will illustrate Ibn al-Bannāʾ’s rule below at 158.11.

link to parallel text 157.9 The second type “concerns how many of that name, taken as units, are in the whole [fraction]”. The example asks how many tenths are in {3\over 4}~{5\over 6}. Ibn al-Bannāʾ relates this to the problem of converting whole numbers in the section on multiplication above at 95.6. The answer is found by multiplying the fraction by 10, so it “does not require division by the denominator of the converted number” (158.1).

link to parallel text 158.7 Al-Hawārī then returns to give the rule in the Condensed Book. His example is to convert six eighths and four tenths to ninths. The numerator and denominator of the fraction “to be converted” are 92 and 80. The procedure is to multiply the 92 by the 9, giving 828, and then this is divided by the 80. We would write the quotient as 10{28\over 80}, but in Arabic it is {4~\,\,\,3\,\over 8~\,10}\,10. Al-Hawārī writes “four eighths” instead of “a half” because he wants to keep the same denominators as the original fractions. Dividing this by 9 is easy: 10\div 9 is {1\over 9}1, and to divide 4~\,\,\,3\,\over 8~\,10 by 9 one just puts a 0\over 9 on the right: 4~\,\,\,3\,~\,0\over 8~\,10~\,9. Gathering them together gives the answer: {4~\,\,\,3\,~\,1\over 8~\,10~\,9}\,1. For this problem, al-Hawārī did not bother to rearrange the denominators in descending order.

link to parallel text 161.1 Chapter I.3. Roots.

link to parallel text 163.1 Section I.3.1. Taking a root of a whole number and a root of a fraction.

link to parallel text 163.2 The term for “square root” in Arabic was simply “root” (jadhr or jidhr). A root of a number might be rational, like \sqrt{25}=5 and \sqrt{4\over 9}={2\over 3}, or it might be irrational, like \sqrt{3} and \sqrt{{17\over 25}\,1}. In Ibn al-Bannāʾ and al-Hawārī, as in many Arabic authors, the word for “rational” is munṭaq, which literally means “expressible”. One can say \sqrt{4\over 9} as “two thirds”, while one cannot say the result of taking a square root of 3 other than as “a root of three”. Our authors write ghayr munṭaq (“inexpressible”) to mean “irrational”. Since irrational numbers in medieval mathematics are all surds, we translate this term by “surd” (recognition of transcendental numbers came only in the nineteenth century). Some other Arabic texts write aṣamm (“deaf”) for “irrational”, but our authors use this word to mean “prime” (see our comments at 66.7 and 134.2 above). The corresponding words in other languages have the same association with speech. In Greek the word for “rational” was rhêtos, meaning “something that may be said or spoken”, while “irrational” was either alogos (“deprived of speech”) or arrhêtos (“unsayable”). In medieval Latin the word surdus, meaning “deaf”, was used for “irrational”, and Italian texts show the related word sordo. The Latin surdus is the origin of our word “surd”.

Roots are expressed in medieval Arabic in a slightly different way than they are in English today. Where we speak of “the root of ten”, they said “a root of ten”. In modern arithmetic numbers are unique. There is only one 3, one \sqrt{10}, etc. But just as in Greek arithmetic,[46] in medieval arithmetic numbers admitted multiplicity. One can have twelve roots of ten, for example. So when an Arabic arithmetician takes the square root of a number, the result is expressed with the implied indefinite article. Read “a root of ten” to mean a single \sqrt{10}. This may seem to be just a linguistic curiosity, but it becomes relevant in the duplication of roots and for understanding their preference for single roots as opposed to multiple roots, discussed below at 179.1. Al-Hawārī will, however, sometimes write “the root” when he is pointing to a specific quantity in a calculation. This is done with the same intention as when he writes, for example, “we add the four in the hundreds rank…” (at 75.11). Other words, like “difference”, “square”, and “ratio”, will lack the definite article, too. Sometimes we insert a “the” where there is none in the Arabic to make the reading easier. We did this in the passage at 174.1, for example, where neither “the” in our translation “the difference between the squares” is in the original (with the exception of “the result”, the other instances of “the” in that passage are present in the Arabic).

link to parallel text 163.4 Ibn al-Bannāʾ differentiates between irrational roots that can be expressed with the word “root” once, and those that require the word more than once. A number in the first category, like \sqrt{3{9\over 10}}, is said to be “rational in square” because its square is rational, while “a root of a root of ten” is an example of a “medial” root. In our notation this number is \sqrt{\sqrt{10}}, which we usually write as \root 4 \of {10}. Rational numbers are also rational in square, so when we want to speak of an irrational number that is rational in square we say it is “rational in square only”.

The phrase “rational in square” is more literally translated as “expressible in power (fī l-quwwa)”. The phrase fī l-quwwa is a translation of the Greek term dynámei, which ordinarily means “in power” or “in value”, but was applied in mathematics to mean the size of the square on a line or of a number.

The word “medial” (muwassaṭ) is translated from the Greek mésos. Reinterpreting Euclid’s geometric definition of mésos in Elements Proposition X.21 in arithmetical terms, medial numbers are those that take the form \sqrt{\sqrt{p}}, where p is a non-square rational. These “roots of roots” are called “medial” because they are the mean proportion between 1 and a number rational in square only. For example, the medial \sqrt{\sqrt{10}} satisfies 1:\sqrt{\sqrt{10}}::\sqrt{\sqrt{10}}:\sqrt{10}. Ibn al-Bannāʾ’s characterization, if taken to the letter, diverges from the Greek. Other numbers, like \sqrt{\sqrt{\sqrt{3}}} and \sqrt{2+\sqrt{8}} are also expressed with the word “root” more than once, though it is unlikely that he considered them to be medial, too.

link to parallel text 163.11 Vowels are generally not indicated when writing Arabic, so the word for “root” shows only the three consonants j-dh-r. Some pronounce this as jidhr and others jadhr. Ibn al-Bannāʾ preferred the latter.

link to parallel text 163.14 Ranks, as well as numbers, are said to either have a root or not have a root. Every other rank, starting with the units, has a root (the units, hundreds, ten thousands, etc.), while the remaining ranks have no root (the tens, thousands, hundred thousands, etc.). As Ibn al-Bannāʾ explains, a rank has a root “if there is a number in it that has a root”. For example, the number 400 is in the hundreds rank, and it has a root, but none of the numbers 1,000, 2,000, up to 9,000 have a root.

link to parallel text 164.3 – 165.16 Here al-Hawārī copies Ibn al-Bannāʾ’s rules for determining when a whole number might have a (rational) square root. For example, if the units digit (the first digit) is a 5, and the tens digit is not a 2, then the number does not have a root.

Squaring numbers of the form 9n+r for r=1, 2,\ …, 8 and then casting out nines shows that the remainder, when it is not cast out entirely, must be a 1, 4, or 7. Determining the remainders of the squares after casting out eights and sevens works similarly.

link to parallel text 166.1 The examples that al-Hawārī and Ibn al-Bannāʾ give for extracting roots are too small to show fully how the rule works, and no figures are shown in the book. Fortunately, the same method is explained by Ibn al-Yāsamīn with greater numbers and with some figures. We give one of his examples here.[47]

To find a root of 876,096 we begin by writing the number on a line, and we mark the first place with an “r” for “root”, the second place with an “n” for “no root”, alternating “r”s and “n”s to the last place of the number:

We start with the last “r” on the left. The number below it is 87, and the greatest square that can be subtracted from it is 81. So we put its root, 9, under it, and we subtract 81 from 87 and replace the 87 with the remainder:

The 9 below is then doubled and shifted one place to the right:

Now we look for the greatest digit n such that when we multiply it by the number of the form “18 n” (i.e., 180+n) it will cancel as much of the 660 above it as possible. This digit is 3. We put the 3 below the 0, and then start the multiplication. Instead of multiplying the 3 by the 183 at once, Ibn al-Yāsamīn, Ibn al-Bannāʾ, and al-Hawārī perform the multiplication one digit at a time. First we multiply the 3 by the 1 to get 3, and this is subtracted from the 6 above:

Next, we multiply the 3 by the 8 to get 24, and this is subtracted from the 36 above:

Then the 3 is multiplied by itself, and the 9 is subtracted from the 120 above:

Now the 3 is doubled, and the 186 is shifted one place to the right:

Next, we want the greatest digit n such that when it is multiplied by “186 n” it cancels as much of the 11196 above as possible. This digit is a 6. We put it under the 6, and we multiply it by 1866, again one digit at a time. Multiplying it by 1 gives 6, and we take 6 away from the 11 above it:

Then we multiply the 6 by the 8, and we take 48 away from the 51 above it:

The 6 is then multiplied by the 6 next to it to get 36, and this is subtracted from the 39 above:

And finally, the 6 is multiplied by itself to get 36, and this cancels the 36 above. The answer is found by adding the 6 to half of the 1860 that follows it, giving 936. So \sqrt{\hbox{876,096}}=936.

link to parallel text 166.9 To find a fractional approximation to a root of a non-square number, Ibn al-Bannāʾ gives different rules depending on whether or not the remainder is greater than the integer part of the root. The rule for remainders less than or equal to the root is the well-known approximation n+{r\over 2n}, where n is the integer part of the root and r is the remainder. The rule for greater remainders is the same, but based at the next root up. In modern notation, the value obtained by this rule is n+{r+1\over 2n+2}, and we can calculate that it is equal to (n+1)+{r^{\prime }\over 2(n+1)} where the remainder r^{\prime } is now negative, counting back from (n+1)^2 instead of forward from n^2.

Ibn al-Bannāʾ gives proofs/derivations for this method in Lifting the Veil. These are not copied by al-Hawārī, so we translate part of one proof here to show the reasoning behind it. This part covers the case where the approximation to the root is smaller than the true root. Ibn al-Bannāʾ’s arguments are arithmetical, and to make the reading easier we give modern algebraic equivalents in brackets. In it, he approximates the fraction f that is added to the known root n of a square n^2 so that together they equal the unknown root m of a greater square m^2. The “surplus” m^2-n^2 is our remainder r.

And its cause is that a root of the number is divided into two parts, a root of the smaller and a fraction [ m=n+f]. Multiplying that by itself is like multiplying each one by itself and one of them by double the other [ (n+f)^2= n^2+f^2+f\cdot 2n]. So the surplus between[48] the number and the square of the smaller is equal to a square of the fraction and a product of the fraction by double a root of the smaller [ m^2-n^2=f^2+f\cdot 2n]. You are allowed to drop a square of the fraction, and you make the surplus equal to a product of the fraction by double a root of the smaller [ f^2 will be small, so we can suppose that m^2-n^2=f\cdot 2n]. So divide the surplus by double a root of the smaller, resulting in the fraction by approximation [ f={m^2-n^2\over 2n}].[49]

He continues: “And it is clear that the resulting fraction is greater than the true fraction, so the approximation is always in excess of the required root of the number”.[50] This is true for both rules, whether the remainder exceeds the root or not. Reinterpreting the rules in terms of modern functions, the approximations for the first rule lie on the line tangent to f(x)=\sqrt{x} at x=n^2, where n is the integer part of the root. The approximations from the second rule lie on the line tangent to f(x)=\sqrt{x} at x=(n+1)^2. We know that the approximations will always be greater than the actual values because the graph of f(x)=\sqrt{x} is concave down. Ibn al-Bannāʾ’s reasoning is simpler.

Examples are calculated at 167.8, 167,14, and 168.1. Because n can be any approximation to the root, not just a whole number, Ibn al-Bannāʾ understood that the rule can be iterated. This is done beginning at 168.8.

link to parallel text 166.13 Al-Hawārī’s first example of calculating a square root is to find \sqrt{625}. He sets it up and begins the process according to Ibn al-Bannāʾ’s instructions:

The next digit to be found is the units digit. Because al-Hawārī presumes that 625 is a perfect square, he knows that it must be a 5: no other unit would produce the 5 in 625.

link to parallel text 167.3 He then multiplies the 5 by 4 to get 20, which leaves a 2 when confronted with the 22 above it. Then the 5 is squared, which cancels the 25 left above:

By “the five and the doubled two after it” he means the 5 in the units place on the lower line, and half of the 4 (really 40) to its left. The root is then 25.

link to parallel text 167.8 Al-Hawārī’s next example is not a perfect square. He finds an approximation to a root of 20 using the rule from 166.9:

Since the number above is not exhausted and is not greater than the root, he adds to the root, which is the 4 below, the remainder above divided by twice the root, or 4\div (2\cdot 4), to get 4{1\over 2}. A square of 4{1\over 2} is 20{1\over 4}, which is close to 20.

link to parallel text 167.14 The same rule is applied to approximate a root of 54, since the remainder, 5, is less than the root, 7.

link to parallel text 168.1 Al-Hawārī’s next example is to find a root of 92. On the dust-board, it would have progressed like this:

This time, the remainder 11 is greater than the root 9: so he adds 1 to the 11 to get 12, and he adds 2 to double the root to get 20. Then the fractional part is 12/20, or 3/5. The approximation is then {3\over 5}\,9. As al-Hawārī notes, its square is {4~\,0\over 5~\,5}\,92. We would write it as 92{4\over 25} or 92.16.

link to parallel text 168.8 In the approximation method just described, one finds the integer part of the root first, and the approximation is calculated as a fraction to be added on. This approximation is greater than the desired root, so Ibn al-Bannāʾ’s rule for obtaining an even closer approximation is calculated by subtracting off another fraction.

link to parallel text 168.10 In the example of approximating \sqrt{92}, the “close smaller square” is a square of the integer part of the first approximation 9{3\over 5}, or 9, which is 81, and which is less than 92. The “close greater square” is a square of 9{3\over 5}, or 92{4\over 25}, which is greater than 92.

link to parallel text 168.16 To obtain a better approximation with this greater square, he denominates the 4\over 25 with double the root 9{3\over 5} to get 1\over 120, or “half a sixth of a tenth”. He subtracts this from the root 9{3\over 5} to get {1~\,5~\,\,\,5\,\over 2~\,6~\,10}\,9, or, in our terms, 9{71\over 120}. This approximation is more accurate: its square as a decimal is 92.0000694…

link to parallel text 169.1 Another way to get a better approximation is to multiply the original number by some large square number, take its root, and then divide by a root of that large square. The example is to approximate a root of 12 by first multiplying the 12 by 16, taking its root, and then dividing the result by 4. The approximation reached this way is {1~\,3\over 4~\,7}\,3, whose square as a decimal begins 12.00127… If we were to approximate \sqrt{12} by the rule at 166.9 we would get {1\over 2}3, whose square is 12.25.

link to parallel text 169.8 Ibn al-Bannāʾ extends the technique for finding roots of whole numbers by allowing one to put down numbers with fractional parts, and he gives two examples in Lifting the Veil that are copied by al-Hawārī at 169.10 and 169.17. The statement that “the remainder will be smaller than the remainder with whole numbers” is best illustrated in the example at 169.17. The remainder after the first step is 3\over 4, which is smaller than the 3 that would have been the remainder had he worked with whole numbers. When working with fractions, there is no smallest possible (positive) remainder because one can always find a closer fraction to the root.

link to parallel text 169.10 In Lifting the Veil, Ibn al-Bannāʾ shows how to use the standard rule from 166.1 with fractions to find \sqrt{729} and \sqrt{625}. Al-Hawārī copies them in reverse order, doing \sqrt{625} first. Instead of choosing the greatest whole number whose square does not exceed the number above, now a fractional part can be added to that number. The work proceeds like this:

Ibn al-Bannāʾ finds the answer by taking half of the 50 in the last figure.

link to parallel text 169.17 The next example is to find a root of 729. The work goes like this:

From the second to the third figure he calculates that \frac{3}{4} of 100 is 75, so he adds this to the 29 to get 104. In the last step 2\cdot 50+2^2=104, so the number on top is exhausted. The answer is then half of the 50 together with the 2, which is 27.

link to parallel text 170.6 The example of calculating \sqrt{100} by adding half the zeros back to \sqrt{1} is clear.

link to parallel text 170.10 Al-Hawārī gives examples of taking roots of fractions. There are two methods to do this. The first is that “you multiply the numerator by the denominator and you divide a root of the result by the denominator”. This method is preferred when the denominator is not a perfect square, because you avoid dividing by an ugly fractional approximation of a root. The second way is to “divide a root of the numerator by a root of the denominator”. This way is easier if the denominator is a perfect square. In modern notation, in the first method the square root of the fraction a\over b is found by calculating \sqrt{ab}\div b, and in the second by \sqrt{a}\div \sqrt{b}.

link to parallel text 170.14 Ibn al-Bannāʾ classifies four kinds of fraction, depending on whether the numerator and/or the denominator is a square.

As a single fraction, the example at 170.16 is 25\over 36, and at 171.1 it is 49\over 4. The fraction at 171.6 is 27\over 54, which is the same as 1\over 2. The approximation for its root is equal to 1451\over 2052, which is correct to four decimal places. Applying the standard method to 1\over 2 gives an approximation of 3\over 4, which is much less accurate.

The fraction at 171.15 is 175\over 16. For the first method, al-Hawārī multiplies the denominator by the numerator to get 2800, and he takes a root of the product. The result, {1~\,48\over 2~\,53}\,52, is then divided by 16. For the second method, he calculates a root of 175 as {3\over 13}13, and he divides this by 4 to get {4\over 13}3. He is wrong when he writes “This method is closer than the first [method]”. The first method is correct to five places, while the second is correct only to three.

The fraction at 172.12 is 9\over 14.

link to parallel text 173.4 Binomials and apotomes.

The only mention of binomials and apotomes in Ibn al-Bannāʾ’s Condensed Book is his rule for finding their roots at 173.4. He does not explain what binomials and apotomes are, and he omits their classification into the six types and how to find examples of them. All that should precede the calculations. He may have inserted this rule as an afterthought, since the heading for the current section, given above at 163.1, does not cover roots of irrational numbers: “on taking a root of a whole number and a root of a fraction”.

link to parallel text 173.10 Ibn al-Bannāʾ provided the missing material in his Lifting the Veil, and al-Hawārī copied it from there into his book. This includes the definitions of binomials and apotomes (173.12-18), the classification of the six types (174.1-13), and the finding of the six binomials and six apotomes (174.14-175.10). Al-Hawārī then finds a root of a sample binomial and its apotome by Ibn al-Bannāʾ’s rule (175.11-21).

Then, at 176.1, al-Hawarī gives a variation on Ibn al-Bannāʾ’s rule and applies it to find roots of examples of each of the six types of binomial and apotome (176.6-177.15). After calculating each root, al-Hawārī adds a brief description of it. These descriptions are translated from Euclid’s Elements, and they are found with this same particular wording in a number of Arabic geometry and arithmetic books written before al-Hawārī’s time. In the translation, we put the descriptions in quotation marks to indicate the borrowing. We do not know al-Hawārī’s immediate source for these descriptions, but they may originate from al-Ḥajjāj’s translation of the Elements.[51] We single them out in part because the meanings of the terms “medial” and “bimedial” become altered in the arithmetical setting of al-Hawārī’s book. In Euclid’s geometrical setting medial/bimedial areas are intended, meaning that it is their sides (square roots) that are medial. Al-Hawārī’s \sqrt{20} (at 177.5), for example, corresponds to a medial area because its “side”, \sqrt{\sqrt{20}}, is medial. Reading the Arabic literally, it seems that the number \sqrt{20} itself is being called medial.

We should give two related definitions before proceeding. Two (positive) numbers are said to be commensurable if their ratio is rational, like \sqrt{12} and \sqrt{27}, or any two rational numbers. Two numbers are commensurable in square if their squares are commensurable. Examples include \sqrt{\sqrt{12}} and \sqrt{\sqrt{27}}, 5 and \sqrt{13}, and any two commensurable numbers. Our authors write “numerical ratio” (nisba ʿadadiya, at 174.17) or “in ratio” (min nisba, at 180.8) to mean “commensurable”, and a word whose ordinary meaning is “different” (mutabāyna, at 180.11 and 182.5) for “incommensurable”. There may not have been standard terms for these meanings.

The nomenclature and theory of binomials and apotomes that was part of Arabic arithmetic derive from a numerical reading of Book X of Euclid’s Elements.[52] Our word “binomial”, like the original Greek (ek) duo onomatōn, means “two names”. In Arabic, the term for “binomial” is dhū l-ismīn, meaning “two unified names”. A binomial is a number that can only be expressed in the form “ x and y”, where the incommensurable numbers x and y are either rational or rational in square.[53] Examples are “eight and a root of sixty” and “a root of five and a root of three”, which we would write as 8+\sqrt{60} and \sqrt{5}+\sqrt{3}, respectively. Our 8+\sqrt{60} contains the arithmetical operation of addition, but as we explain at 219.1 below, the premodern “eight and a root of sixty” consists instead of two numbers gathered together.

Ibn al-Bannāʾ’s verb for forming a binomial is waṣala, “to join” two numbers together (pp. 174-5). Sometimes the related noun muttaṣil, “joining”, takes the place of the usual term for “binomial”. At 188.19, Ibn al-Bannāʾ spells it out as “[a] joining of two names”, and in three instances just after that we find it as merely “joining”. On that page we translate these variants as “binomial” to make the reading clearer, while at 177.9 and 177.15 the word is better rendered as “joining”.

Al-Hawārī does not show notation for roots, but other authors do. In his 1370 commentary on Ibn al-Bannāʾ’s Condensed Book, Ibn Qunfudh writes these two binomials in notation as Too late and Too late.[54] Reversing the order of the terms, we can transcribe them as “ 8~\sqrt{60}” and “ \sqrt{5}~\sqrt{3}”, respectively. The letter jīm (Too late) without the dot, for jadhr (“root”), is placed above the 60, 5, and 3 to indicate square root. We have already seen this way of writing gathered numbers – one after the other – several times for distinct fractions, first at 136.8.

Our word “apotome” comes from the Greek word apotomē, meaning “something cut off”, and the Arabic translation is munfaṣil, meaning an amount from which something has been “detached”. Examples are “eight less a root of sixty” and “a root of five less a root of three”. We write these in modern notation as 8-\sqrt{60} and \sqrt{5}-\sqrt{3}, but again, our operation of subtraction distorts the premodern idea of a diminished quantity. An apotome was regarded as a single quantity from which something has been removed. The “eight less a root of sixty”, for example, should be thought of as a deficient eight, and not as the subtraction of something from eight. We explain this more thoroughly below at 219.4. Ibn Qunfudh shows these two examples in notation with the word illā (“less”) between the numbers: Too late and Too late.[55] We transcribe them as 8~\ell ~\sqrt{60} and \sqrt{5}~\ell ~\sqrt{3}. This notation is the same as we have seen for excluded fractions of disconnected type, first at 140.8, and with an image from the Medina manuscript in our commentary at 140.1. Further, Ibn al-Bannāʾ and al-Hawārī use the term munfaṣilat with the meaning of “detached” for the excluded parts of these fractions in the passages at 141.7, 142.6, and 142.10.

Binomials and apotomes are not restricted to the arithmetic of fractions and roots. In algebra, too, the same concepts and notation are behind polynomials. At 183.15, al-Hawārī makes a direct comparison between the quadratic irrationals of this section and algebraic expressions: “The principle behind multiplying appended and deleted terms will be covered in [the chapter on] algebra”.

Euclid’s theory of binomials and apotomes is expressed in a geometric context, so he necessarily takes into account the dimensions of his magnitudes. For example, instead of the multiplication of two numbers, he writes of the formation of a rectangle from two sides. He defines a binomial as a line, and lines by their nature cannot have “square roots”. So, Euclid defines its square root as the side of a square equal in area to a rectangle having the binomial as one of its sides and a rational line (his version of a unit) for the other side. This theory was transferred to the setting of numbers in Arabic arithmetic. Numbers are homogeneous and thus dimensionless, so some of Euclid’s distinctions become superfluous.

Because we will be interspersing the notation for binomials and apotomes with calculations in modern notation, we will express the binomials and apotomes with “ +” and “ -” from now until the end of the section. To transcribe them in notation that better reflects the notation in the texts, omit the “ +” and replace the “ -” with an “ \ell”. We write them this way, right-to-left, in the conspectus in Appendix A.

link to parallel text 174.1 In Lifting the Veil, Ibn al-Bannāʾ explains the classification of the six types of binomial and apotome, and al-Hawārī gives examples of each. There are three basic types. Writing the greater term first, in order they are p\pm \sqrt{q}, \sqrt{p}\pm q, and \sqrt{p}\pm \sqrt{q} where both p and q are rational, and where one of these values appears under a root it is not a perfect square. Each of the basic types is divided into two subtypes stemming from a characteristic of its root. Writing the general binomial in modern notation as \sqrt{P}+\sqrt{Q}, in which P and Q are both rational, P>Q, and at most one is a square, al-Hawārī’s rule for calculating \sqrt{\sqrt{P}+\sqrt{Q}} at 176.1 can be written as

A root of the associated apotome, \sqrt{\sqrt{P}-\sqrt{Q}}, is

In one of the two subtypes \sqrt{P} is commensurable with \sqrt{P-Q}, so the two terms can be combined into a single root. This holds for the first three of the six types of binomial/apotome. In the other subtype the two are incommensurable, and this holds for the last three of the six types.

Ibn al-Bannāʾ’s classification follows that of Euclid, who gives constructions for each of the six types of binomial in Propositions X.48-53, and for the six types of apotome in Propositions X.85-90.

Just as the digits in a number come in ranks, so do numbers expressed with roots. Two numbers are of the same rank if they are the same “distance” from being rational. Thus \sqrt{5} and \sqrt{21} are of the same rank, as are \sqrt{\sqrt{5}} and \sqrt{\sqrt{21}} and the pair \sqrt{\sqrt{\sqrt{5}}} and \sqrt{\sqrt{\sqrt{21}}}.

link to parallel text 174.14 Euclid then gives constructions for roots of the six binomials in Propositions X.54-59, and for roots of the six apotomes in Propositions X.91-96. Ibn al-Bannāʾ gives numerical rules in place of Euclid’s constructions. We rewrite them below in modern notation. Al-Hawārī does not give examples here, so we provide our own. In these descriptions it is assumed that a, b, and c are positive rational numbers, and numbers under roots are positive non-squares.

A number is a first binomial if it can be written in the form a+\sqrt{a^2-b^2}, and a first apotome if it can be written as a-\sqrt{a^2-b^2}. If we let a=5 and b=2, then the binomial is 5+\sqrt{21} and the corresponding apotome is 5-\sqrt{21}.

A number is a fourth binomial if it can be written in the form a+\sqrt{a^2-b}, where b is not a perfect square. If a=5 and b=8, then the fourth binomial is 5+\sqrt{17} and the associated apotome is 5-\sqrt{17}.

A number is a second binomial if it can be written in the form \sqrt{a^2(a^2-b^2)}+(a^2-b^2). A second apotome can be written in the form \sqrt{a^2(a^2-b^2)}-(a^2-b^2). For example, if we let a=3 and b=2, the binomial is \sqrt{45}+5 and the associated apotome is \sqrt{45}-5. Medieval mathematicians would say “a root of forty-five” instead of “three roots of five” because the latter is a collection of three numbers. They preferred the single number \sqrt{45}. See below at 179.1 for more on this. Ibn al-Bannāʾ made an error in defining this type by writing “a root of their difference” instead of simply “their difference”.

Ibn al-Bannāʾ also made an error in defining the fifth binomial. In our notation his description translates into \sqrt{a^2+b^2}+a. In fact, it should be of the form \sqrt{a^2+b}+a, where \sqrt{a^2+b} is incommensurable with \sqrt{b}. If we make a=5 and b=3, the fifth binomial is \sqrt{28}+5 and the apotome is \sqrt{28}-5. Al-Hawārī’s example of the fifth binomial at 177.5 below is correct.

A number is a third binomial if it can be written in the form \sqrt{ca^2}+\sqrt{c(a^2-b^2)}. If we let a=5, b=2, and c=3, then the third binomial is \sqrt{75}+\sqrt{63} and the associated apotome is \sqrt{75}-\sqrt{63}.

A number is a sixth binomial if it can be written in the form \sqrt{a^2+b}+\sqrt{b}, and if the two roots are incommensurable. If a=6 and b=7, then the sixth binomial is \sqrt{43}+\sqrt{7} and the associated apotome is \sqrt{43}-\sqrt{7}.

link to parallel text 175.11 Al-Hawārī finds roots of the first binomial 8+\sqrt{60} and its associated apotome 8-\sqrt{60} (from a=8 and b=2) by the method given by Ibn al-Bannāʾ at 173.4. We can rewrite this as the modern formula:

For the binomial, he subtracts 1\over 4 of a square of the smaller ( 1\over 4 of \sqrt{60}^2 is 15) from 1\over 4 of a square of the greater ( 1\over 4 of 8^2 is 16), leaving 1. Its root is 1. This 1 is added to half of the greater term ( 1\over 2 of 8 is 4) to get 5, and it is also subtracted from the 4 to get 3. Then \sqrt{5}+\sqrt{3} is a root of 8+\sqrt{60}. He then finds a root of the associated apotome 8-\sqrt{60} by the same procedure to get \sqrt{5}-\sqrt{3}.

176.1 – 177.15 Al-Hawārī gives the variation on the rule for finding roots of binomials and apotomes that we described above at 174.1, and he applies it to his examples for each of the six types.

link to parallel text 176.1 To find a root of the first binomial 8+\sqrt{55} (with a=8 and b=3), take the difference of their squares 8^2-(\sqrt{55})^2 to get 9, and take its root to get 3. Now figure {1\over 2}(8+3) and {1\over 2}(8-3), and take their roots. A root of the binomial is then \sqrt{5{1\over 2}}+\sqrt{2{1\over 2}}, and a root of the apotome 8-\sqrt{55} is \sqrt{5{1\over 2}}-\sqrt{2{1\over 2}}. Euclid (Propositions X.54, 91) and the Arabic translation call the root of the binomial “a binomial” and the root of the apotome “an apotome”, which is indeed what they are. Al-Hawārī expands these phrases to “one of the binomials” and “one of the six apotomes”.

link to parallel text 176.10 Next, al-Hawārī finds a root of \sqrt{112}+7, which is an example of the second binomial ( a=4, b=3). Following the same procedure, (\sqrt{112})^2-7^2=63. Then {1\over 2}(\sqrt{112}+\sqrt{63}) and {1\over 2}(\sqrt{112}-\sqrt{63}) are \sqrt{85{\frac{3}{4}}} and \sqrt{1{\frac{3}{4}}}. (For these steps, al-Hawārī calculated \sqrt{112}+\sqrt{63}=\sqrt{343} and \sqrt{112}-\sqrt{63}=\sqrt{7}. Instructions on how to add and subtract roots are given below, starting at 179.1.) So a root of the binomial \sqrt{112}+7 is \sqrt{\sqrt{85{\frac{3}{4}}}}+\sqrt{\sqrt{1{\frac{3}{4}}}}.

Euclid (Proposition X.55) and the Arabic translation call a root of the second binomial “the first bimedial”. Euclid defines this term in Proposition X.37: “If two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational; and let it be called a first bimedial straight line”. Reinterpreting this in our arithmetical context, the root is called a bimedial because it is composed of the two medial numbers \sqrt{\sqrt{85{\frac{3}{4}}}} and \sqrt{\sqrt{1{\frac{3}{4}}}}. They are commensurable in square only because the ratio of their squares, \sqrt{85{\frac{3}{4}}}/\sqrt{1{\frac{3}{4}}}, is 7, but their ratio \sqrt{\sqrt{85{\frac{3}{4}}}}/\sqrt{\sqrt{1{\frac{3}{4}}}} is \sqrt{7}, which is irrational. The two medial numbers contain a rational rectangle because their product, \sqrt{\sqrt{85{\frac{3}{4}}}}\cdot \sqrt{\sqrt{1{\frac{3}{4}}}}=3{1\over 2}, is rational.

A root of the second apotome \sqrt{112}-7 is \sqrt{\sqrt{85{\frac{3}{4}}}}-\sqrt{\sqrt{1{\frac{3}{4}}}}. Euclid (Proposition X.92) and the Arabic translation call this “a first apotome of a medial [straight line]”.[56] Because the root was considered to be a diminished \sqrt{\sqrt{85{\frac{3}{4}}}}, it does not consist of “two names”. Thus it is medial and not bimedial.

link to parallel text 176.15

Al-Hawārī’s example of the third binomial is \sqrt{32}+\sqrt{14} ( a=4, b=3, and c=2). Its root is \sqrt{\sqrt{24{1\over 2}}}+\sqrt{\sqrt{1\over 2}}. This root is called a “second bimedial” in Euclid (Proposition X.56) and “the binomial of the second bimedial” in the Arabic translation. The second bimedial is defined in Elements, Proposition X.38: “If two medial straight lines commensurable in square only and containing a medial rectangle be added together, the whole is irrational; and let it be called a second bimedial straight line”.[57] The bimedial numbers \sqrt{\sqrt{24{1\over 2}}} and \sqrt{\sqrt{1\over 2}} are commensurable in square only because the ratio of their squares \sqrt{24{1\over 2}}/\sqrt{1\over 2} is 7, while their ratio is the irrational number \sqrt{7}. Their product \sqrt{\sqrt{24{1\over 2}}}\cdot \sqrt{\sqrt{1\over 2}} is \sqrt{3{1\over 2}}, which is irrational. A “medial area” (i.e., a “medial rectangle”) “is the area which is equal to the square on a medial straight line”.[58] In arithmetical terms, the \sqrt{3{1\over 2}} is a medial area because its square root \sqrt{\sqrt{3{1\over 2}}} is medial. In arithmetic there is no use for such a designation, because, unlike geometric magnitudes, numbers do not possess dimension. A “medial area” translated to arithmetic becomes simply a number rational in square only, or, in other terms, a number of the form \sqrt{p} where p is a non-square rational.

A root of the corresponding apotome is \sqrt{\sqrt{24{1\over 2}}}-\sqrt{\sqrt{1\over 2}}. Euclid (Proposition X.93) and the Arabic translation call this “a second apotome of a medial [straight line]”.

link to parallel text 176.20 Next, al-Hawārī finds a root of 7+\sqrt{30}, which is an example of the fourth binomial ( a=7, b=19). Following the same procedure he gets \sqrt{3{1\over 2}+\sqrt{4{\frac{3}{4}}}}+\sqrt{3{1\over 2}-\sqrt{4{\frac{3}{4}}}}. Again, following Euclid (Propositions X.57, X.94) the Arabic translation calls this “the major” and the corresponding apotome “the minor”. A major is defined in Elements Proposition X.39: “If two straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole straight line is irrational: and let it be called major”.[59] In al-Hawārī’s example, \sqrt{3{1\over 2}+\sqrt{4{\frac{3}{4}}}} is incommensurable in square with \sqrt{3{1\over 2}-\sqrt{4{\frac{3}{4}}}} because the ratio of their squares, (3{1\over 2}+\sqrt{4{\frac{3}{4}}})/(3{1\over 2}-\sqrt{4{\frac{3}{4}}}), is 2{4\over 15}+\sqrt{4{31\over 225}}, which is irrational. The sum of their squares is 7, which is rational. The rectangle contained by them, or their product, is \sqrt{7{1\over 2}}, which is a medial area when interpreted in terms of geometry. The minor is defined similarly for the apotome in Proposition X.76.

link to parallel text 177.5 The example for the fifth binomial is to find a root of 3+\sqrt{20} ( a=3, b=11). Its root by the same procedure is \sqrt{\sqrt{5}+\sqrt{2{\frac{3}{4}}}}+\sqrt{\sqrt{5}-\sqrt{2{\frac{3}{4}}}}. Euclid (Proposition X.58) called this “[the line whose] power is a rational and a medial [area]”.[60] Here the rational area is the 3, and the medial area is the \sqrt{20}. The Arabic translation is quite literal: “[the number whose] power is a rational and a medial”, with the result that term “medial” would be misinterpreted in the arithmetical setting. The number \sqrt{20} is rational in square, not medial like its square root.

Euclid (Proposition X.95) calls the corresponding apotome \sqrt{20}-3 “[a straight line] which produces with a rational [area] a medial whole”.[61] In our numbers, the root \sqrt{\sqrt{5}+\sqrt{2{\frac{3}{4}}}}-\sqrt{\sqrt{5}-\sqrt{2{\frac{3}{4}}}} produces by squaring it the quantity \sqrt{20}-3 which, if one adds back the rational area 3, gives the medial area \sqrt{20}. The Arabic translation describes it as “the joining with a rational to become a whole medial”.

The Greek and Arabic words behind “power” in the descriptions “[the line/number whose] power is a rational and a medial” warrant some explanation. The Greek term is dynaménē, deriving from dynamis, a word meaning “power” or “value”, and in a mathematical setting, “square”. In geometry, a dynamis is a characteristic of a line. One did not make reference to a dynamis directly by naming its opposite vertexes, but rather to a side in respect of dynamis. As Jens Høyrup proposed, a dynamis is “a square identified with its side” or “a line seen under the aspect of a square” (Høyrup 1990, 210). The word often occurs in the dative form, dynámei (“in power”, or “in square”), as we described above at 163.4 for the phrase “rational in square”. Thus, our added words “[the line/number whose]” in the description of the fifth binomial are implied in Euclid’s description.

The forms of the word dynamis in Euclid were translated into Arabic by corresponding forms of quwwa, a word which also means “power”. This word, too, is a characteristic of a side or a number. Gustav Junge and William Thomson, in an appendix to their edition and translation of the medieval Arabic translation of Pappus of Alexandria’s Commentary on Book X of Euclid’s Elements, wrote this of quwwa:

The Dictionary of Technical Terms (Calcutta, A. Sprenger, Vol. II, p. 1230, top.) defines it as “Murabbaʿu-l-Khaṭṭi”, i.e., “the square of the line”, “the square which can be constructed upon the line”, and goes on to say that the mathematicians treat the square of a line as a power of the line, as if it were potential in that line as a special attribute. Al-Ṭūsī (Book X, Introd., p. 225, l. 9.) says: — “The line is a length actually (reading “bi-l-fiʿli” for “bi-l-ʿalqi”)[62] and a square (murabbaʿun) potentially (bi-l-quwwati) i.e., it is possible for a square to be described upon it. (Pappus 1930, 181)

So here, too, an Arabic reader familiar with Euclid would have understood the description as meaning “[the number whose] power is a rational and a medial”. But for a young arithmetic student studying al-Hawārī’s book, we guess that the meaning of this description, and of the others, too, would have been obscure, and would have been read as being mere labels.

As a last remark on this passage, al-Hawārī writes the fraction “three fourths” in his calculations of the fourth and sixth binomials and apotomes, but here he writes it in the colloquial way reminiscent of the unit fractions from finger-reckoning as “a half and a fourth”.

link to parallel text 177.11 The example for the sixth binomial is \sqrt{10}+\sqrt{11} ( a=1, b=10), and its root is \sqrt{{1\over 2}+\sqrt{2{\frac{3}{4}}}}+\sqrt{\sqrt{2{\frac{3}{4}}}-{1\over 2}}. It is called by Euclid (Proposition X.59) “[the line whose] power [consists of] two medial [areas]”,[63] where in this case the medial areas are the \sqrt{11} and the \sqrt{10}. The Arabic translation is again literal: “[the number whose] power is a bimedial”. And as before, the word “bimedial” takes a different meaning in an arithmetical context.

Euclid (Proposition X.96) calls the root of the corresponding apotome, here \sqrt{{1\over 2}+\sqrt{2{\frac{3}{4}}}}-\sqrt{\sqrt{2{\frac{3}{4}}}-{1\over 2}}, “[a straight line] which produces with a medial [area] a medial whole”.[64] The root produces by squaring it the apotome \sqrt{11}-\sqrt{10}. This is a deficient, or incomplete, \sqrt{11}, so to make it whole we add to it the medial area \sqrt{10} that it lacks to produce the whole medial area \sqrt{11}. The Arabic translation has “the joining with a medial to become a whole medial”.

link to parallel text 179.1 Section I.3.2. Adding and subtracting roots.

Before plunging into the rules and examples for operating on roots, we must clarify some distinctions made in medieval arithmetic that are lost in modern notation. Consider the following:

     A.

Duplicate a root of five three times. (explained at 186.1)

     B.

Three roots of five.

     C.

Multiply three by a root of five. (explained at 183.1)

     D.

A root of forty-five.

While we might translate each of these into 3\sqrt{5}, all four were conceived of differently in their medieval forms. Both (A) and (C) are operations to be performed, while (B) and (D) are quantities. To “duplicate a root of five three times” in (A) means to collect together three copies of \sqrt{5}. This wording is well-suited for integer multiples, but it cannot work for irrationals. One cannot duplicate a number \sqrt{23} times, for instance. For fractional multiples, al-Hawārī’s wording is different. For them, he “partitions” the root (at 186.8).

The immediate and intermediate result of (A), duplicating a root of five three times, is (B), three roots of five. This quantity was not regarded as a single number like our 3\sqrt{5}. It is a collection of three numbers, all of them \sqrt{5}. Just as the binomial “eight and a root of fifty-five” is the pair of numbers 8 and \sqrt{55} gathered together, “three roots of five” is the gathering of \sqrt{5}, \sqrt{5}, and \sqrt{5}. Likewise, below at 179.20 we will see that “half a root of twenty” was regarded as being “less than one root”, while “two roots of five” are “more than one root”. The “half” and the “two” tell us how many roots there are. They are not the coefficients or scalar multiples that we work with in {1\over 2}\sqrt{20} and 2\sqrt{5}.

Multiples of a root and fractions of a root are routinely converted to whole, single roots before operating on them. To convert “three roots of five” to a single number the 3 is multiplied by the \sqrt{5}. To do this the 3 is squared, and the result is multiplied by 5 to get \sqrt{45}. In other words, one performs (C), and the result is (D). The multiplication in (C) is a different operation from (A). The most evident difference is that (C) allows any number as a multiplier. One can multiply \sqrt{7} by \sqrt{5}, for instance, to get “a root of thirty-five”. The result of the multiplication must be a single root. It cannot be anything like our \sqrt{7}\sqrt{5} since this symbolic amount, like our 3\sqrt{5}, has no equivalent in medieval arithmetic.

link to parallel text 179.2 We noted above at 176.10 that \sqrt{112}+\sqrt{63}=\sqrt{343}. This is because, as we would work it out, \sqrt{112}+\sqrt{63}=\sqrt{16\cdot 7}+\sqrt{9\cdot 7}=4\sqrt{7}+3\sqrt{7}=7\sqrt{7}=\sqrt{343}. Ibn al-Bannāʾ gives the rule that two roots can be added or subtracted only if their product is a perfect square. In this example 112\cdot 63 =7{,}056, which is a square of 84. On the other hand, \sqrt{5} cannot be added to \sqrt{12} to form a single root because 5\cdot 12=60 is not a square number.

link to parallel text 179.7 Once it is known that two roots can be added, Ibn al-Bannāʾ gives a rule for finding their sum. Al-Hawārī first applies it to the example of adding \sqrt{3} to \sqrt{27}. Here 3\cdot 27=81 is a square, so the roots can be added. Two roots of 81 are 18, and this is added to 3+27 to get . Then \sqrt{3}+\sqrt{27}=\sqrt{48}. In modern notation, \sqrt{a}+\sqrt{b}=\sqrt{a+b+2\sqrt{ab}}.

This rule is found in many arithmetic and algebra books, extending back to the Book on Algebra of Abū Kāmil (late ninth century). It is a practical rule for converting two roots into a single root. It is not covered in Euclid’s Elements, and it is not connected with any theory of quadratic irrationals.

link to parallel text 179.11 Al-Hawārī gives a second method of finding the sum. Using the same example he calculates {\sqrt{27}\over \sqrt{3}}=3 (division of roots is explained below, at 187.2), and he adds 1 to get 4. This 4 is then multiplied by the \sqrt{3} to get \sqrt{48}. The same procedure works if we reverse the roles of 3 and 27. In modern notation, \sqrt{a}+\sqrt{b}=\left(\sqrt{a\over b}+ 1\right)\cdot \sqrt{b}.

link to parallel text 179.16 Another example: add \sqrt{2} to \sqrt{8} to get \sqrt{18}.

link to parallel text 179.20 Al-Hawārī borrows his problem “add half a root of twenty to two roots of five” from Ibn al-Bannāʾ’s algebra book, where it is solved the same way.[65] In his explanation al-Hawārī notes that “half a root of twenty is less than one root”, so he transforms it to one root, which is \sqrt{5}. (The verb behind “transform” here, and elsewhere in our translation, is radda, which ordinarily means “to return”, or “to send back”.) At this point, al-Hawārī could add this to the two roots of five to get three roots of five, which would lead to \sqrt{45}. But his goal is to illustrate the technique of forming single roots from fractions and multiples of roots, so he continues: “And two roots of five are more than one root, so we transform them to one root”, which is \sqrt{20}. Then the problem is the same as adding \sqrt{5} to \sqrt{20}. This is worked out by the rule at 179.7 to get \sqrt{45}.

link to parallel text 180.6 Al-Hawārī, or more likely Ibn al-Bannāʾ, speaks here about adding roots of different ranks, like adding a root to a root of a root (i.e., to a fourth root). The square root must be made into a root of a root before adding. But there is no nice rule to tell if two fourth roots can be added to get a single fourth root. Al-Hawārī seems to think that the rule for square roots works here too, for he writes below, in the passage at 180.15, that “their surface is likewise not a square, so we add them with the coordinating conjunction”, where this conjunction is the word “and”.

link to parallel text 180.10 If the roots cannot be added, then one puts the word “and” between them. The example is to add \sqrt{3} to \sqrt{15}. Since 3\cdot 15=45 is not a perfect square, we cannot write them as a single root. Their sum is thus expressed as the binomial “a root of three and a root of fifteen”. The “and” (wa) does not mean “plus”. See the discussion at 219.1 below. Also, “their surface” is their product. See our comments at 66.17 above.

link to parallel text 180.15 Here al-Hawārī asks for the sum of 1\over 2 of \sqrt{\sqrt{80}} and 1~\bullet ~1\over 4~\bullet ~3 of \sqrt{684}, where the 1~\bullet ~1\over 4~\bullet ~3 is “a third of a fourth”. First, the 1\over 2 of \sqrt{\sqrt{80}} is “a root of a root of five” ( \sqrt{\sqrt{5}}), and the 1~\bullet ~1\over 4~\bullet ~3 of \sqrt{684} is “a root of four and three fourths” ( \sqrt{{\frac{3}{4}}4}). To add them we make them the same rank, so we write \sqrt{{\frac{3}{4}}4} as \sqrt{\sqrt{{1~4\over 2~\,8}\,22}} by squaring the {\frac{3}{4}}4. As we just mentioned, the rule al-Hawārī seems to apply to determine if the two can be added does not work. The sum of the roots is expressed as “a root of a root of five and a root of a root of twenty-two and four eighths and half an eighth” ( \sqrt{\sqrt{5}}~\sqrt{\sqrt{{1~4\over 2~\,8}\,22}}).

link to parallel text 181.4 The first rule for subtracting roots is similar to the first rule for adding roots. Al-Hawārī subtracts \sqrt{8} from \sqrt{32} by first multiplying 8\cdot 32=256, taking its root to get 16, and doubling it to get 32. This is subtracted from the sum of the two numbers 8+32=40, leaving 8. The difference of the roots is then \sqrt{8}. In modern notation, \sqrt{a}-\sqrt{b}=\sqrt{a+b-2\sqrt{ab}}.

link to parallel text 181.10 The second rule for subtracting roots is like the second rule for adding them. Al-Hawārī’s example is to subtract \sqrt{12} from \sqrt{27}. He divides \sqrt{27} by \sqrt{12} to get {1\over 2}1 (division of roots is explained below, at 187.2). Subtracting 1 leaves 1\over 2. Multiplying this by \sqrt{12}, the divisor, gives \sqrt{3}, which is the answer. In modern notation, \sqrt{a}-\sqrt{b}=\left(\sqrt{a\over b}-1\right)\cdot \sqrt{b}.

link to parallel text 182.1 Ibn al-Bannāʾ again mentions that if the roots are less than one root or more than one root, they must be returned to one root, like in the addition problem above at 179.20. And if their ranks are different they must be made the same, as above at 180.15.

link to parallel text 182.4 Here al-Hawārī gives examples in which the roots cannot be subtracted. For example, subtracting \sqrt{8} from \sqrt{10} can only be expressed as “a root of ten less a root of eight”. The “less” (illā) does not mean “minus”. See below at 219.4.

link to parallel text 183.1 Section I.3.2. Multiplying roots.

link to parallel text 183.4 Although we have not found any Arabic book that characterizes the multiplication of irrational roots in a way like Ibn al-Bannāʾ did for the multiplication of whole numbers (95.2) and fractions (149.2), the basic rule was well known. In the latter ninth century Abū Kāmil stated the rule and gave two proofs of it, while other authors, like al-Baghdādī, merely state the rule and give examples.[66] Al-Hawārī’s first example is to multiply \sqrt{8} by \sqrt{9}. The answer is a root of the product of the numbers, \sqrt{72}.

link to parallel text 183.7 Next, al-Hawārī multiplies \sqrt{\sqrt{5}} by \sqrt{\sqrt{7}}. Since the roots are of the same rank (both are roots of roots), again he multiplies the numbers and applies the roots. The answer is \sqrt{\sqrt{35}}. The next example, at 183.11, is \sqrt{\sqrt{\sqrt{3}}}\cdot \sqrt{\sqrt{\sqrt{8}}}=\sqrt{\sqrt{\sqrt{24}}}.

link to parallel text 183.15 Here al-Hawārī multiplies 3 by the apotome \sqrt{7}~\ell ~2. He distributes just the way we would, first multiplying the 3 by \sqrt{7} to get \sqrt{63}, then the 3 by 2 to get 6. The answer is \sqrt{63}~\ell ~6.

link to parallel text 183.20 Ibn al-Bannāʾ gives the rule for multiplying a rational number by a root that al-Hawārī has already applied above in the calculations at 179.20, 180.15, 181.12, and 183.15. The first example is to multiply 3 by \sqrt{7}, which he had just done as part of the previous problem. Al-Hawārī calculates 3^2=9, and 9\cdot 7=63, so the answer is \sqrt{63}. He then continues through 185.12 with examples of multiplying numbers by multiples and fractions of roots. After that, starting at 186.1, he will duplicate and partition roots.

link to parallel text 184.4 Next, to find 2\cdot \sqrt{\sqrt{3}}, he finds 2^2=4, 4^2=16, and 16\cdot 3=48, so the answer is \sqrt{\sqrt{48}}.

link to parallel text 184.11 To multiply 2 by two roots of 7 is like multiplying the 2 by a pair of \sqrt{7}s. The two \sqrt{7}s must be turned into a single root first, and together they are \sqrt{28}. Now the 2 can be squared and multiplied by 28 to get the answer, which is \sqrt{112}.

link to parallel text 184.18 Similarly, to multiply 5 by three roots of a root of 2 is like multiplying 5 by three \sqrt{\sqrt{2}}s. The three roots together are \sqrt{\sqrt{162}}, and multiplying the 162 by a square of a square of 5, which is 625, gives the answer \sqrt{\sqrt{101{,}250}}.

link to parallel text 185.6 Al-Hawārī multiplies 2\over 3 by half of \sqrt{20}. The latter is the same as \sqrt{5}, so the 2\over 3 is squared to give 4\over 9, and this is multiplied by the 5, resulting in \sqrt{{2\over 9}2}.

link to parallel text 185.12 To multiply \sqrt{5} by half of \sqrt{\sqrt{40}} one first turns the half into a whole by squaring the 1\over 2 to get 1\over 4, and then squaring this to get 1\over 16, and then multiplying it by the 40 to get \sqrt{\sqrt{{1\over 2}2}}. Also, the \sqrt{5} must be converted into a root of a root, and this is \sqrt{\sqrt{25}}. Multiplying {1\over 2}2 by 25 gives {1\over 2}62, so the answer is \sqrt{\sqrt{{1\over 2}62}}.

link to parallel text 186.1 To duplicate a root means to collect together copies of it. In the second of two examples, to duplicate \sqrt{7} five times means to gather five \sqrt{7}s into a single root. This is done by multiplying 5 by \sqrt{7}. The answer is \sqrt{175}.

link to parallel text 186.8 The same works for partitioning a root, which means to take a fraction of a root. Half a root of ten is found by multiplying \sqrt{10} by 1\over 2. Likewise, a third of four eighths of \sqrt{\sqrt{60}} is found by multiplying the fraction by the root.

link to parallel text 187.1 Section I.3.2. Dividing and denominating roots.

link to parallel text 187.2 Dividing or denominating a root by a root follows the same general rules as multiplying roots. Al-Hawārī begins by dividing \sqrt{20} by \sqrt{3}. First he finds 20\div 3=6{2\over 3}, then he takes its root to get \sqrt{{2\over 3}6}. The next example works the same way. Al-Hawārī wrote at 187.7 and at 187.14 “divide” where he should have written “denominate”.

link to parallel text 187.10 To divide \sqrt{\sqrt{6}} by \sqrt{\sqrt{2}}, divide the 6 by 2 to get 3, then apply the roots to get \sqrt{\sqrt{3}}. The next example, dividing \sqrt{\sqrt{18}} by \sqrt{\sqrt{32}}, works the same way.

link to parallel text 188.1 One should only perform a division with single roots, so multiples and fractions of a root must first be converted to single roots. Also, the ranks of the roots of the dividend and divisor must be the same. We mentioned above at 134.2 that there are no medieval counterparts to fractions with irrational numbers, like our 1\over \sqrt{2} and \sqrt{2}\over 2, but one can perform the divisions. To divide 1 by \sqrt{2} the ranks should be made equal, so 1 is converted to \sqrt{1}. Then 1 is divided by 2 to get 1\over 2, and finally a root is applied to get \sqrt{1\over 2}. Here, 1\over 2 is valid as a fraction, and a root can be taken of any number. To divide \sqrt{2} by 2 one follows the same rule by first rewriting the 2 as \sqrt{4}, to again get \sqrt{1\over 2}.

link to parallel text 188.6 To divide \sqrt{\sqrt{14}} by \sqrt{2}, the \sqrt{2} has to be written as \sqrt{\sqrt{4}}. Then 14\div 4=3{1\over 2}, so the answer is \sqrt{\sqrt{{1\over 2}3}}.

link to parallel text 188.11 Dividing two \sqrt{15}s by 2 should be trivial – one simply gets a single \sqrt{15}. But al-Hawārī illustrates the general rule by converting the two roots into a single root and then dividing the result by 2. The two roots together are \sqrt{60}, and the 2 as a root is \sqrt{4}. Then 60\div 4 = 15, so the answer is \sqrt{15}. The next example, dividing half of \sqrt{24} by \sqrt{2}, works similarly.

link to parallel text 188.18 To divide a quantity by a binomial or an apotome one multiplies both amounts by the conjugate to eliminate the square root. Al-Hawārī’s first example is to divide 12 by the binomial 5~\sqrt{3}. He multiplies the 12 by the apotome 5~\ell ~\sqrt{3} to get 60~\ell ~\sqrt{432}. Then he multiplies 5~\sqrt{3} by 5~\ell ~\sqrt{3} to get 22. Dividing the 60~\ell ~\sqrt{432} by 22 results in {8\over 11}\,2~\ell ~\sqrt{9~~9\over 11~\,11}. We would write this as 2{8\over 11}-\sqrt{108\over 121}, or 2{8\over 11}-{6\over 11}\sqrt{3}, or even 30-6\sqrt{3}\over 11. Restating the latter two rhetorically would result in improperly expressed amounts. These only became possible with Descartes.[67]

Dividing 10 by 3~\ell ~\sqrt{7} is done the same way. The numerator and denominator are multiplied by the binomial 3~\sqrt{7}, which leads to the answer 15~\sqrt{175}.

link to parallel text 191.1 Part II. On the rules by which one arrives at knowledge of the required unknown from the posited known.

Ibn al-Bannāʾ covers three problem-solving methods in Part II: the rule of three, double false position, and algebra. Ibn al-Bannāʾ and other Arabic authors call the first of these “the four proportional numbers”. It is often applied directly to solve a problem, and it is also the core of the method we call single false position. In that method, a convenient value is chosen for the answer (the false posited value, or position), and the true value is found via proportion by a calculation on the posited value and its error. Ibn al-Bannāʾ does not cover single false position in his Condensed Book.

Double false position was called the method of “the two errors” (al-khaṭaʾān). Like other arithmeticians in the Maghreb, Ibn al-Bannāʾ calls it the method of “scales” (kiffāt) because of the shape of the diagram drawn to work out the problems. In this method, two false values are posited for the answer, and the true value is found by a calculation involving the values and their respective errors. This method relies on proportion, so it is covered along with the rule of three in Chapter II.1.

Algebra is covered in Chapter II.2. The Arabic name for algebra is al-jabr wa-l-muqābala, meaning “restoration and confrontation”, or sometimes al-jabr for short. This is the origin of our word “algebra”. In an algebraic solution, an unknown is named in terms of designated names of the powers of the unknown, and an equation is set up in terms of these names and solved. In the other methods the calculations are worked out on known numbers, and without equations.

Ibn al-Bannāʾ and al-Hawārī explain the rules for each of the three methods, but they give sample problems only for double false position. We give examples translated from other books of the four proportional numbers, single false position, and algebra in Appendix B.

link to parallel text 193.1 Chapter II.1. Working out [problems] with proportion.

The Arabic word for mathematical “ratio” is nisba. This word is also used in non-mathematical settings to mean any kind of relation between two objects, and even kinship between people. Nisba is also the Arabic word for “proportion”. The two meanings can be distinguished by the context.

link to parallel text 195.2 Ibn al-Bannāʾ lists the following kinds of proportion in Lifting the Veil:[68]

Geometric. This is the usual proportion involving four numbers a, b, c, and d in which the ratio a:b is equal to the ratio c:d, or, in notation, a:b::c:d. Numerically, it is equivalent to {a\over b}={c\over d}.

Arithmetic. “The difference between the first and the second is equal to the difference between the third and the fourth”.[69] The four numbers a, b, c, and d are in arithmetic proportion if b-a=d-c.

Harmonic. Ibn al-Bannāʾ writes “For three numbers, the ratio of the extremes is as the ratio of the two differences between the middle and each one of the extremes. This ratio is composed of the geometric and the arithmetic, since, on the one hand, it comes from the difference between the middle and each one of the extremes, resembling the arithmetic, and, on the other, from proportionality, resembling the geometric”.[70] Three (descending) numbers a, b, and c are in harmonic proportion if a:c::a-b:b-c. Ibn al-Bannāʾ gives his source for the definition as Nicomachus’s Arithmetical Introduction.[71]

Ex-aequali. Given two sets of numbers a_1, a_2,\ldots a_n and b_1, b_2,\ldots b_n, if a_i:a_{i+1}::b_i:b_{i+1} for each i, then a_1:a_n::b_1:b_n. Although this reads like a proposition, it is presented as Definition V.17 in Euclid’s Elements. Nicomachus does not cover it.

Harmony. This one comes from Nicomachus, who describes it as “the most perfect proportion, that which is three-dimensional and embraces them all”.[72] Both Nicomachus and Ibn al-Bannāʾ remark that this proportion is important in music and astronomy. Briefly, four numbers are “in harmony” if they are of the form a, 2ab\over a+b, a+b\over 2, and b. See Nicomachus’s Introduction, Chapter II.29 for the different properties it satisfies.

As Nicomachus notes, geometric proportion “is the only one in the strict sense of the word to be called a proportion, because its terms are seen to be in the same ratio”.[73]

In his list at 195.2 al-Hawārī omits geometric proportion, and he switches the order of the last two. Apparently he omitted geometric proportion by mistake. Not only does he dispense with all four proportions from his list in this passage, and then follow Ibn al-Bannāʾ by discussing geometric proportion, but the omission causes a misreading. Ibn al-Bannāʾ writes “[the] origin of these last three proportions is the first, which is in the four numbers mentioned in the book”.[74] By leaving off geometric proportion al-Hawārī’s paraphrase “all three derive from it and it does not derive from them” implies that the last three derive from arithmetic proportion instead of geometric proportion.

link to parallel text 195.7 The four proportional numbers.

Ibn al-Bannāʾ focuses on geometric proportion for the rest of this chapter. Four numbers, a, b, c, and d, are proportional if a:b::c:d, or, equivalently, if ad=bc.

link to parallel text 195.9 Whenever three of the numbers are known, the fourth can be found. Continuing with modern notation, suppose we want to find x in the proportion x:a::b:c or a:x::c:b.

link to parallel text 195.14 In x:a::b:c, the x and b are of the same kind because they have the same positions in their respective ratios. Likewise, a and c are of the same kind. The number b is isolated because it is not of the same kind as either of the other two known numbers. This is multiplied by a, the number in ratio with x. The product is then divided by the third number c to give the value of x.

link to parallel text 195.16 Al-Hawārī gives examples of this rule for the proportion 3:6::4:8, treating each number in turn as the unknown and finding it from the other three by the rule.

link to parallel text 196.16 A proportion can be converted to a number of other proportions. The four types mentioned by Ibn al-Bannāʾ are taken in order from Definitions V.12 through V.15 in Euclid’s Elements. If a:b::c:d, then the following proportions also hold:

Switching (badala): a:c::b:d.

In Elements Def. V.12 Heath translates this as “alternate ratio”.

Reversing (khalafa): b:a::d:c.

In Elements Def. V.13 Heath translates this as “inverse ratio”.

Combining (rakiba):

        (a+b):a::(c+d):c or (a+b):b::(c+d):d.

In Elements Def. V.14 Heath translates this as “composition of a ratio”.

Separating (faṣala):

        |a-b|:a::|c-d|:c or |a-b|:b::|c-d|:d.

In Elements Def. V.15 Heath translates this as “separation of a ratio”.

One can also combine two or more of these operations. For example, combining after switching gives (a+c):a::(b+d):b. Euclid gives one more in Definition V.16: the “conversion” of the proportion gives a:(a-b)::c:(c-d), presuming a>b and c>d. But this can be obtained by reversing after separating, which may be why Ibn al-Bannāʾ did not include it.

Al-Hawārī may have chosen to copy this passage from Lifting the Veil because, as he mentions below at 198.4, one can derive the rule for double false position from a known proportion by switching, separating, and then switching again.

link to parallel text 197.4 Al-Hawārī gives variations on Ibn al-Bannāʾ’s rule at 195.9 that switch the order of operations. The jurist mentioned at the end of this passage is otherwise unknown.

link to parallel text 198.2 The method of scales (double false position)

The method of “scales”, or double false position, “comes from the art of geometry” because of its use in geometric proportion. The method is in fact purely numerical. In another part of Lifting the Veil, Ibn al-Bannāʾ explains the term “geometric proportion”: “It is said to be geometric proportion because it is more specific to the quantities that are considered in geometry”.[75]

Double false position works for problems in which the operations on the unknown described in the enunciation give a number proportional to the unknown. Sometimes a problem can be adjusted to satisfy this condition (our authors do not do this), and other times, as we shall see starting at 201.14, the method itself can be adjusted.

link to parallel text 198.4 Al-Hawārī mentions that the proportion just stated by Ibn al-Bannāʾ is found by switching, separating, and then switching a known proportion. We explain this in terms of the specific numbers in the problem below at 199.1, so read that first and then come back here. The “first scale”, or the first posited value, is 15. The calculated “portion” is the result of performing the operations on the 15, and this is 6{1\over 4}. The “assigned number” is the 10, and the value of the unknown is found at the end to be 24.

We begin with the ratio stated by Ibn al-Bannāʾ, “the ratio of the error of each scale to the difference between the scale and the unknown number is as the ratio of the assigned number to the unknown”:

Switching gives

Separating gives

Switching again gives

This is now “the ratio of the portion to its scale is as the ratio of the assigned number to the unknown”, which al-Hawārī took from Lifting the Veil.

From here, Ibn al-Bannāʾ continues in Lifting the Veil to derive the rule for solving problems by double false position, though al-Hawārī does not copy it into his book. Like all his derivations/proofs, Ibn al-Bannāʾ expresses the steps in arithmetical terms, without the use of algebra. In other words, he does not name any of the quantities and then manipulate the subsequent algebraic expressions. For example, he says “so the two results from multiplying each error by the excess of the other scale are equal”,[76] where by contrast we would name the unknown as, say, x, the two posited values or scales as x_1, x_2, and their respective errors as e_1, e_2, and then write e_1\cdot (x_2-x)=e_2\cdot (x_1-x). The names of the powers of the unknown in premodern algebra were generally not used outside the context of algebraic problem-solving, and Ibn al-Bannāʾ’s derivation is not an exception. We leave it to the reader to take our last equation and solve for x to find a modern rendering of the procedure.

link to parallel text 198.7 Double false position is worked out with a diagram, shown below. It has the shape of a balance, or “scales”.

link to parallel text 199.1 Al-Hawārī’s first problem is: “A quantity: taking away its third and its fourth leaves ten. How much is the quantity?” See our commentary below at 211.8 for the different meanings of the word māl, translated here as “quantity”, in Arabic arithmetic.

The first step is to draw the scales and put the assigned number above the dome:

This method calls for two values to be posited. Al-Hawārī makes the first value 15. This is placed in the right scale like this:

The operations are then performed on the posited value: subtracting 5 (a third of 15) and 3{\frac{3}{4}} (a fourth of 15) from the 15 leaves 6{1\over 4}. This is placed next to the 15, inside the balance:

We confront this 6{1\over 4} with the desired 10, and we find that it falls short by an error of 3{\frac{3}{4}}. Because it falls short, it is written below the scale:

Al-Hawārī then chooses 12 for the second value. He follows the same process, this time putting the numbers in the left scale. The 12 is written in, then the calculations are made: subtracting 4 (a third of 12) and 3 (a fourth of 12) from the 12 leaves 5. Confronting this 5 with the desired 10, we find that it falls short by 5. The second error, 5, is likewise written below the scale:

Now we are ready to calculate the value of the unknown. First, each posited value is multiplied by the other error, and their difference is taken. Multiplying the 15 by 5 gives 75, and 3{\frac{3}{4}} by 12 gives 45. Their difference is 30. This is divided by the difference between the two errors, which is 5-3{\frac{3}{4}}=1{1\over 4}. The result is 24, which is the answer.

link to parallel text 200.1 The next example is: “A quantity: we take the sum of its third and its fifth, and we add to it half of the remainder, so it comes to twenty-three”. To explain what is meant by “the remainder”, we resort to modern algebraic notation. If we name the quantity x, then its third and its fifth are {1\over 3}x+{1\over 5}x. The remainder is what we get after subtracting this from the quantity, so let r=x-({1\over 3}x+{1\over 5}x). We could then express the enunciation as the equation {1\over 3}x+{1\over 5}x+{1\over 2}r=23, but of course this is not how our author proceeds with his solution.

Al-Hawārī chooses 40 for the first value. Adding its third and its fifth gives 13{1\over 3}+8=21{1\over 3}. The remainder from 40 is then 18{2\over 3}. Adding to the 21{1\over 3} half the remainder, or 9{1\over 3}, gives 30{2\over 3}. This 30{2\over 3} is not the 23 we want. Since the calculated number this time exceeds the desired amount, we put the error 7{2\over 3} above the scale.

Next, al-Hawārī tries 45. Taking its third and its fifth and adding half the remainder gives 34{1\over 2}, so the error is 11{1\over 2}. Since the 34{1\over 2} is greater than the desired 23, the error is again put above the scale:

The same procedure is followed as in the last example. He calculates 7{2\over 3}\cdot 45=345 and 40\cdot 11{1\over 2}=460. Subtracting the smaller from the greater leaves 115. This is divided by the difference between the errors, which is 11{1\over 2}-7{2\over 3}=3{5\over 6}, to get 30, which is the unknown quantity.

link to parallel text 201.1 In the third example, al-Hawārī illustrates the case in which one value gives a result that falls short of the desired number and the other gives a result that exceeds it. The problem is to find a quantity such that if “we add its tenth to the difference between its fourth and three of its fifths \ldotsit comes to nine”. We resort again to modern algebra to clarify what is being asked. If we call the quantity x and perform the operations, we get the equation ({3\over 5}x-{1\over 4}x)+{1\over 10}x=9. Again, there is no algebra in the enunciation, and algebra plays no role in the solution.

The first posited value is 12, and the result of the operations is 5{2\over 5}. This falls short of the desired 9 by 3{3\over 5}. The second posited value is 25. The result of the operations is 11{1\over 4}, which exceeds the desired 9 by 2{1\over 4}. The two errors are placed below and above the scale like this:

Again we perform the two multiplications, but this time we add the results rather than subtract them, since one error is above the scale and the other is below. So 3{3\over 5}\cdot 25=90 and 12\cdot 2{1\over 4}=27. Their sum is 117, and this is divided by the sum of the errors, which is 2{1\over 4}+3{3\over 5}={1~\,4\over 4~\,5}\,5, which for us would be 5{17\over 20}. The result of the division is 20, which is the unknown quantity.

Al-Hawārī has now covered three cases: (a) both calculated portions fall short of the desired number, (b) both exceed it, and (c) one falls short and the other exceeds. We can, if we want, write a single formula to cover all three cases because our arithmetic allows for negative numbers.

link to parallel text 201.14 Ibn al-Bannāʾ gives a second method of working with scales. The diagram and values are drawn as described above, but the calculations are now asymmetric with respect to the scales, so that it works even when the same value is posited for both scales (as in the example at 203.1). With only one error, it functions more like a complicated version of single false position. The asymmetry also opens it up to problems in which only one posited value and its error are given, but not the number above the dome! The example for this is given at 203.11.

“This second method is used only when there is a proportional relation”, as al-Hawārī writes at 202.3. He means that in cases where two or more values are placed in each scale, they must be chosen proportionally according to the conditions of the problem. This method, in fact, will also work where there is only one value placed in each scale, like the three examples given above.

link to parallel text 202.3 The first example is “ten: you divide it into two parts so that a third of one of them is a fourth of the other”. The enunciation of this problem copies word-for-word the first problem in Ibn al-Bannāʾ’s algebra book.[77] There, of course, it is solved by algebra, with the naming of one of the parts as “a thing”, which plays the role of our x. It is easier for us to understand what is being asked in the enunciation by converting it to modern algebra: If we name the parts a and b, then a+b=10 and {1\over 3}a={1\over 4}b. This second condition is equivalent to saying that the ratio of one part to the other is as the ratio of 3 to 4.

No naming occurs in double false position. The first step in al-Hawārī’s solution is to put the 10 above the dome, and then to pick two numbers so that a third of one is a fourth of the other. He picks 3 and 4. They are placed in the right scale, with the 3 above the 4. Their sum is 7, and this is placed next to them. Confronting 7 with 10, we see that it falls short by 3, so a 3 is put below the scale. Next, 6 and 8 are posited for the second two numbers. Their sum is 14, which exceeds the 10 by 4. The completed diagram looks like this:

The rule for finding the answer will seem kind of bizarre. Al-Hawārī multiplies the 6 on the left by the error 3 on the right to get 18. Then he multiplies the 14 on the left by the 3 on the right, above the 4, to get 42. Because the first error is below the scale (whether the second error is above or below does not matter), he adds them to get 60. Dividing this by the 14 gives 4{2\over 7}, which is the smaller number, and the greater is 10-4{2\over 7}=5{5\over 7}. We leave it as an exercise for the reader to show with modern algebra, or by geometry if you want, that this method works.

link to parallel text 203.1 Because of the asymmetry of the calculations, one can work with the same value twice rather than pick a different one. In this next example, al-Hawārī solves “ten: you divide it into two parts. You divide the greater by the smaller, so the result is four”. This time the ratio of the parts is 4 instead of \frac{3}{4}. He picks as his first values the numbers 3 and 12. Their sum is 15, which exceeds the 10 by 5. So a 5 is put above the scale. With the same numbers on both sides, the diagram looks like this:

Now the 3 on the left is multiplied by the error 5 on the right to get 15, and the 15 on the left multiplied by the 3 on the right make 45. We take their difference because the error is above the scale. So 45-15=30. We divide this by the 15 to get 2, which is the smaller part, and the greater part is 8.

link to parallel text 203.11 Ibn al-Bannāʾ writes that this second method will also work when only one of the posited values and its error are given in the problem. He gives the example “a quantity: we subtract its third and its fourth from a third of sixty and its fourth, leaving fourteen”. If we call the unknown x, this would yield the equation ({1\over 3}60+{1\over 4}60)-({1\over 3}x+{1\over 4}x)=14. This derives from the problem “A quantity: you added its third and its fourth, so it gave some number ”. We can think of 14 as the error for the posited value 60. Pick another number for the second value, like 72, add its third and its fourth to get 42, and fill out the diagram:

Since the error 14 exceeds, we calculate 60\cdot 42 - 72\cdot 14=1512, and we divide this by the 42 to get 36, which is the answer. The number above the dome would have been 21, but we were not asked to find that.

link to parallel text 204.3 Ibn al-Bannāʾ dictated three alternatives to the basic rule of double false position illustrated in the examples above at 199.1, 200.1, and 201.1. These explanations would have been better placed just after those examples. Al-Hawārī does not give examples of these alternatives, so we give here the calculations from his previous examples. The diagrams are the same.

link to parallel text 204.6 For the example at 199.1, the difference between (the numbers in) the scales is 15-12=3. Multiplying this by one of the errors, 5, gives 15. Dividing this by the difference between the errors, or 1{1\over 4}, gives 12. Adding to this the 12 in the scale associated with the error 5 gives 24, which is the answer.

For the example at 200.1, we calculate 45-40=5, then 5\cdot 11{1\over 2}=57{1\over 2}. This is divided by 11{1\over 2}-7{2\over 3}=3{5\over 6} to get 15. Subtracting this from the 45 gives 30, which is the answer.

For the example at 201.1, we calculate 25-12=13, then 13\cdot 2{1\over 4}=29{1\over 4}. Dividing this by 2{1\over 4}+3{3\over 5}=5{17\over 20} gives . Subtracting this from the 25 gives 20, which is the answer.

link to parallel text 204.12 The calculations for these same three examples by the second alternative are as follows:

For 199.1: 15-12=3; 5+3{\frac{3}{4}}=8{\frac{3}{4}}. Multiplying them, 3\cdot 8{\frac{3}{4}}=26{1\over 4}; 26{1\over 4}\div (5-3{\frac{3}{4}})=26{1\over 4}\div 1{1\over 4}=21. Keep this in mind. One option is: 21+3=24; its half is 12, and adding the 12 in the scale gives the answer 24. The other option is: 21-3=18; its half is 9, and 9+15=24.

For 200.1: 45-40=5; 11{1\over 2}+7{2\over 3}=19{1\over 6}. Multiplying them, 5\cdot 19{1\over 6}=95{5\over 6}; 95{5\over 6}\div (11{1\over 2}-7{2\over 3})=95{5\over 6}\div 3{5\over 6}=25. Keep this in mind. One option is: 25+5=30, its half is 15, and 45-15=30 is the answer. The other option is: 25-5=20, its half is 10, and 40-10=30.

For 201.1: 25-12=13; 3{3\over 5}-2{1\over 4}=1{7\over 20}. Multiplying them, 13\cdot 1{7\over 20}=17{11\over 20}, 17{11\over 20}\div (3{3\over 5}+2{1\over 4})=17{11\over 20}\div 5{17\over 20}=3. Keep this in mind. One option is: 3+13=16, its half is 8, and 8+12=20, which is the answer. The other option is: 13-3=10, its half is 5, and 25-5=20.

link to parallel text 205.1 The calculations for these examples by the third alternative are as follows:

For 199.1: 15-12=3; 3\cdot 10=30; 30\div (5-3{\frac{3}{4}})=24, which is the answer.

For 200.1: 45-40=5; 5\cdot 23=115; 115\div (11{1\over 2}-7{2\over 3})=30, which is the answer.

For 201.1: 25-12=13; 13\cdot 9=117; 117\div (3{3\over 5}+2{1\over 4})=20, which is the answer.

link to parallel text 205.5 If an indeterminate problem does not restrict any number to a particular class such as squares or cubes, then to be non-trivial it must give conditions on at least three unknown numbers. In a solution by scales one of these numbers can reside above the dome, so there will be at least two values in each scale. Unlike problems with a proportional relation, two of these values can be chosen independent of each other. Thus the first method must be used.

This first indeterminate problem is of a common type in Arabic and later Latin and Hebrew arithmetic, and it is found in abstract form in Diophantus.[78] In this type two or more men want to buy a horse, with conditions on how much money each one has. Problems of this type are usually solved by algebra, like problem [5] in Appendix B.

Because the problem is indeterminate, any number can be chosen for the first man, i.e., the amount of money the first man has. Ibn al-Bannāʾ chooses 4 dirhams. He then first posits 2 dirhams for the second man. From these two numbers the price of the horse and the money of the third are found, 5 and 9 respectively. The 4, 2, and 9 are placed in the first scale, and the 5 is put above the dome. In the second scale the first man is given 4 again, and the value for the second man is posited as 6 dirhams. Then the price of the horse is calculated from the last condition as 7, so the third man must have 3 dirhams. The 4, 6, and 3 are placed in the second scale, and the 7 is placed above the dome.

The price of the horse is then calculated a second time from the last condition. For the right scale, the horse costs 9+{1\over 4}\cdot 4=10, which exceeds the calculated 5 by 5, so a 5 is put above the scale. In the left scale, the same condition gives the price of the horse as 4 dirhams, which is 3 less than the calculated 7, so the error of 3 is placed below the scale. Here is the diagram:

Now the money of each man can be calculated using the rule from the first method. The first man has 4 dirhams, as we decided in advance. The second man’s money is found by the calculation 5\cdot 6 + 3\cdot 2=36, divided by 5+3=8, giving 4{1\over 2} dirhams. The third man’s money is found the same way: 5\cdot 3+3\cdot 9=42, and dividing by the same 8 gives 5{1\over 3} dirhams. By the first condition, the price of the horse is 4+{1\over 2}(4{1\over 2})=6{1\over 4} dirhams.

Ibn al-Bannāʾ does not prove in Lifting the Veil that this variation with two numbers above the dome works. He could have derived it through a manipulation of proportions as he did for the standard version of the method, or he could have followed Qusṭā ibn Lūqā in proving it by geometry. We give an explanation in terms of modern elementary algebra. Regardless of how many men want to buy the horse, once the money of the first is chosen and the money of the second is posited, all other values are calculated. If we call the money of the second y, then the problem is equivalent to finding y where the price of the horse can be written two ways, each of them of the form ay+b. In our problem, these are {1\over 2}y+4 and {3\over 2}y-{1\over 2}. When Ibn al-Bannāʾ calculates the values for the horse to put above the dome he gets the same values we get by plugging his numbers into {1\over 2}y+4. When he recalculates the values, he gets the same results we get by plugging them into {3\over 2}y-{1\over 2}.

In the kinds of problems that are usually solved by double false position, the (posited) unknown is proportional to the desired value (the number placed above the dome), or, in modern algebra, \alpha y=\beta for constants \alpha and \beta. In the present problem, the condition is not stated as one of proportionality. It effectively equates the results of two calculations, which we can write as ay+b=a^{\prime }y+b^{\prime }. But this equation can be rearranged as (a-a^{\prime })y=b^{\prime }-b, and applying double false position to this proportional relation, with the single number b^{\prime }-b above the dome, gives the same errors for posited values of y, and thus the same answer, as Ibn al-Bannāʾ’s way.

link to parallel text 206.5 Since all solutions are proportional to this one, al-Hawārī notes that if we do not want fractions, we can multiply all the numbers by 4. This way, the first man has 16 dirhams, the second has 18 dirhams, the third has 21 dirhams, and the horse costs 25 dirhams.

link to parallel text 207.1 We can also assign an arbitrary value for the price of the horse, and pick a value for what the first man has, and work through the calculations. Then we pick a different value for the horse, but the same value for the first man, and work through the calculations again. The same rule then gives what the second and third man have.

link to parallel text 207.6 This next problem asks for the number of each kind of bird given the total cost and the price of each kind. Problems of this type were posed and solved in China and India before the rise of Islam, and they later appear in medieval Europe, including Fibonacci’s Liber Abaci.[79] In our problem, 40 birds are purchased for 40 dirhams. There are three kinds of bird: starlings, which cost 1/8 of a dirham each; chickens, which cost 2 dirhams each; and geese, which cost 3 dirhams each. How many of each kind of bird was purchased?

link to parallel text 207.9 Ibn al-Bannāʾ states two conditions for this problem. The first is that the number of each kind of bird must be a whole number. Then, since the total price is whole, and starlings are the only bird that cost a fraction of a dirham, their number must be a multiple of 8. The second condition is that the product of the price of the cheapest bird by the number of birds should be less than the total cost, and the product of the price of the most expensive bird by the number of birds should be greater than the total cost.

Working with these conditions, Ibn al-Bannāʾ finds that the number of starlings cannot be 8 or 16, but that 24 is a possibility. Then, two values are posited for the number of chickens, 8 and 14. The number of geese must be the difference from 40, or 8 and 2, respectively. The total cost is calculated, 37 and 43 dirhams, respectively, making the errors 3 and 3. The diagram is set up like this:

Next to the number of birds, al-Hawārī has put their cost. Only the Oxford manuscript shows the 43 and 37, and they were put below the scales because there was no room inside.

The number of geese is calculated as {2\cdot 3+8\cdot 3\over {3+3}}=5, and the number of chickens is {14\cdot 3+8\cdot 3\over 3+3}=11. One can use modern algebra to show why this works, too. It is simpler than the case of problems with two numbers above the dome.

For an explanation as to why our authors write “the starlings”, “the chickens”, and “the geese” for what we would write as “the number of starlings”, “the number of chickens”, and “the number of geese”, see the last three paragraphs of our commentary at 95.3 above.

link to parallel text 208.12 Last, the possibility that the number of starlings is 32 is considered. This is shown to be impossible.

By “problems involving multiplication” Ibn al-Bannāʾ probably means problems in which unknowns are multiplied by unknowns. In such problems, the calculated portions will not have a proportional relation to the posited values, so the method of scales will not work. Problems [3] and [4] in Appendix B, solved by algebra, are of this type. Problems involving divisions of unknowns cannot be solved by scales, either. Problem [6] in Appendix B is of this type, and it too is solved by algebra.

link to parallel text 209.1 Chapter II.2. Algebra.

Arabic algebra is the third and last method of numerical problem solving presented by Ibn al-Bannāʾ. As with single false position, al-Hawārī does not give any example of a problem worked out by the method. Instead, he only covers the kinds of calculations that crop up in the course of working through algebraic solutions. These include the solutions to the six simplified equations, operations on polynomials, and the simplification of equations. To make up for the lack of problems, we present some translations from other books in Appendix B.

link to parallel text 211.1 Section II.2.1. The meaning of algebra (al-jabr wa-l-muqābala).

The name given to algebra in medieval Arabic is al-jabr wa-l-muqābala, literally “restoration and confrontation”. Sometimes the name was shortened to just al-jabr, and via transliterations in medieval Latin and Italian it led to the English word “algebra”. The words al-jabr and al-muqābala occur individually in the solutions to problems by algebra, specifically for particular steps in the simplification of equations, and it is these meanings that are briefly described in the next line, along with the term “equalization” (muʿādala). By “its types” Ibn al-Bannāʾ means the six types of simplified equation, described at 213.1.

link to parallel text 211.2 Ibn al-Bannāʾ gives brief explanations of restoration (al-jabr), confrontation (al-muqābala), and equalization (muʿādala). Al-Hawārī will clarify these terms later, via examples in the section on addition and subtraction beginning at 219.1. It would not do much good for him to expound on them here anyway, before the basic elements of algebra are covered.

The rules for algebra, as explained by Ibn al-Bannāʾ and illustrated by al-Hawārī, are best understood in the context of the general structure of algebraic solutions. Three basic stages are followed in the solution of a problem by algebra in medieval Arabic:

Stage 1.

An unknown number is named in terms of the names of the powers of the unknown, usually as a “thing” (like our “ x”). Then the conditions of the problem are applied to the name to produce an equation;

Stage 2.

The equation is simplified to one of six types (given below) via restoration and/or confrontation;

Stage 3.

The simplified equation is solved using the specified rule for that type.

link to parallel text 211.8 The name given to the first degree unknown is a “thing” (shayʾ), though it is sometimes called a “root” (jidhr, or, as Ibn al-Bannāʾ preferred, jadhr (see 163.11)). This is like our x, though whether rhetorically or in notation, it was interpreted differently from its modern counterpart. The second degree unknown, which corresponds to our x^2, is called a māl. This word ordinarily takes the meaning of “sum of money”, “possession”, or “wealth”. Because there is no good English translation of the word, and because it was used in a technical sense in Arabic algebra unconnected with its everyday meanings, we leave it untranslated. And, because of the different plural forms for Arabic nouns, we write the plural of māl with the English suffix: māls.

Units, things, and māls were considered to be three different types or species of number, just like units, tens, hundreds, etc. are different species (see our description above at 68.18). This idea goes back at least to al-Khwārazmī in the early ninth century. He introduced the powers of the unknown by writing: “I found that the numbers which are needed in algebraic calculation are of three types, which are roots and māls and simple number…”[80]

The word māl also commonly takes another meaning in arithmetic, where it can be an amount of money, or, in most cases, a generic “quantity”. The word māl is used to mean a “quantity” in two problems translated in Appendix B: Ibn al-Bannāʾ solves problem [2] by single false position, and al-Ḥaṣṣār solves problem [7] by three different methods. The word also means “quantity” in four of the problems al-Hawārī solves by double false position, above at 199.1, 200.1, 201.1, and 203.11. For instance, leaving māl untranslated, the problem athyperlinkc199.1199.1 is “a māl: taking away its third and its fourth leaves ten. How much is the māl?” Such problems were also frequently solved by algebra. In the enunciation māl is a common noun meaning “quantity” or “amount”, while in an algebraic solution it means instead the proper name given to the second power of the unknown “thing”. Which meaning is intended is clear from the context.

The word “root” likewise takes different meanings in arithmetic and algebra. In arithmetic it means “square root”, while in algebra it is also the name sometimes given to the first degree unknown. Again, the meaning is made clear by the context.[81]

Units in Arabic arithmetic are often counted in dirhams, a denomination of silver coin. Sometimes the word refers to money, but often the problem is stated in more abstract terms, and it is clear, then, that “dirham” is a substitute for a generic, arithmetical “unit”. This is most evident in problems that are stated in terms that cannot possibly involve money, such as problems in al-Khwārazmī in which the dirham stands for one man, or in geometry problems in Abū Kāmil in which the dirham stands for a unit length.[82] In the passage at 158.1, “dirham” appears to refer to the arithmetical unit. In all other instances before the chapter on algebra it refers to the coin, and in the chapter on algebra it is the unit.

We shall encounter higher powers below, especially in Section 4 starting at 225.1. The cube of a thing is called a “cube” (kaʿb), the fourth power is a māl māl, the fifth power is usually a māl cube, and higher powers are some combination of “māl” and “cube”. These species can be written any way one likes, as long as the powers add up. For example, at 226.3 al-Hawārī writes that the eighth degree term can be called a māl māl māl māl, a cube māl cube, or a cube cube māl.

link to parallel text 211.13 The most common ways of equating numbers and magnitudes in Arabic mathematics were with words deriving from sawiya (“to be equivalent, be equal”), mithl (“equivalent”), the prefix ka- (“as, like”), and the implied verb “to be”.[83] Algebraic equations, however, were stated in Arabic with the unusual verb ʿadala, meaning “equal” or “well-balanced”. Another feature that distinguishes equations from other kinds of equating is that the two sides of an equation are stated with the names of the algebraic powers. These two features are evident, for example, in the equation “a māl and five things equal (taʿdil, conjugated from ʿadala) ten dirhams and two māls less a thing” at 224.1, which in modern notation would be x^2+5x=10+2x^2-x.

We explain below in our comments in the section on addition and subtraction, beginning at 219.1, that modern notation is not compatible with medieval algebraic expressions. For this reason, it is better to write the expressions and equations of Arabic algebra with a transliteration of the notation that was current in the western part of the Islamic world in the time of al-Hawārī, even if he does not show it in his book. This notation is an extension of Indian notation to cover algebraic expressions, operations, and equations.

This last equation is a good example for introducing the Arabic notation. Instead of x^2+5x=10+2x^2-x, we write it as {m\atop 1}~{t\atop 5}=\hbox{10}~{m\atop 2}~\ell ~{t\atop 1}. In the first term, the m\atop 1, the “m” stands for māls and the 1 indicates that there is one of them. The “t” in t\atop 5 stands for “things”, and the 5 below is how many there are. (Even when the first degree term is called “roots”, the letter in the notation is an abbreviation for “things”.) Simple numbers, here 10, do not need a designation, so there is no letter above them. There is no sign for addition on the left side or between the first two terms on the right because the terms are gathered together. This is the same kind of grouping we saw for distinct fractions at 136.8, where “five sixths and four fifths” is shown as {4\over 5}\,{5\over 6}, and for binomials in our commentary at 173.10, with examples from Ibn Qunfudh like “ \sqrt{60}~8”. Last, the \ell is the same sign for “less” that we saw for excluded fractions and apotomes, and which we explain and show above at 86.1, 140.1, and 173.10. It indicates that the t\atop 1 is lacking or missing from the \hbox{10}~{m\atop 2}.

The Arabic algebraic notation originated in the western part of the Islamic world (i.e., west of Egypt) no later than the end of the twelfth century.[84] Several textbooks with chapters on algebra show it, including those written by Ibn al-Yāsamīn (d. 1204), al-Mawāḥidī (14th c.), Ibn Qunfudh (1370), al-Qalaṣādī (d. 1486), and Ibn Ghāzī (1483), but it is not uncommon to find purely rhetorical presentations of algebra like al-Hawārī’s, either. Like the Indian notation for numbers that al-Hawārī and others show, the algebraic notation only appears occasionally in books as figures to show what should be written on the dust-board. The notation was not part of the running text.

Here are a couple of sample equations from a manuscript of a book that shows the notation, Ibn Ghāzī’s Aim of the Students. At one point he writes “five māls and four things and three in number equal two māls and three things and six in number, and its figure is Too late”.[85] Reading the notation right to left, the first term shows a mīm, the first letter in the word māl, above a 5, for the “five māls”. The next term shows a shīn, the first letter in shayʾ (“thing”) above a 4, and this is followed by a lone 3. The sign for “equals” is the black line shaped like a backwards “L”. It is a lām, the last letter in taʿdil (“equals”). The left side of the equation is written similarly. Our transliteration, read left to right, is {m\atop 5}~{t\atop 4}~\hbox{3}={m\atop 2}~{t\atop 3}~\hbox{6}, which corresponds to the modern 5x^2+4x+3=2x^2+3x+6.

An example from the same manuscript with “less” is: “ten less four things equals sixteen less six things, so the figure for this is Too late”.[86] We write it as \hbox{10}~\ell ~{t\atop 4}=\hbox{16}~\ell ~{t\atop 6}, and in modern notation it is 10-4x=16-6x. Like we have seen for arithmetic, in the manuscripts either the whole word illā is written, or the initial alif is dropped so it looks like an upside-down \ell \,. Note that here the Arabic “0” is written as a dot, so the “10” looks like “1 \cdot”. Because it takes a little while to become accustomed to the Arabic notation, we will continue for a while to give the version in modern notation as well. See below at 229.8 for a brief description of how this notation was interpreted.

Today, we have just one form of simplified quadratic equation, ax^2+bx+c=0, where our a, b, and c can be positive, negative, or zero. The solution to this equation, x={-b\pm \sqrt{b^2-4ac}\over 2a}, is a formula expressed in terms of the coefficients. The corresponding solutions in medieval algebra also use the “coefficients”, but because only positive numbers were acknowledged they had six types of simplified equation: three of them “simple” (mufrad), with a single term on each side, and the other three “composite” (murakkaba), where one side has two terms. These are listed below, with al-Hawārī’s examples under each generic form:

Simple equations
Type 1: some māls equal some roots
      “three māls equal seven things” ( {m\atop 3}={t\atop 7}; 3x^2=7x).
Type 2: some māls equal some numbers
      “five māls equal twenty” ( {m\atop 5}=\hbox{20}; 5x^2=20).
Type 3: some roots equal some numbers
      “three roots equal twelve” ( {t\atop 3}=\hbox{12}; 3x=12).
Composite equations
Type 4: some māls and some roots equal some numbers
      “a māl and ten roots equal twenty-four” ( {m\atop 1}~{t\atop 10}=\hbox{24};
      x^2+10x=24).
Type 5: some māls and some numbers equal some roots
      “a māl and four equal five roots” ( {m\atop 1}~\hbox{4}={t\atop 5}; x^2+4=5x).
Type 6: some māls equal some roots and some numbers
      “a māl equals four roots and five” ( {m\atop 1}={t\atop 4}~\hbox{5}; x^2=4x+5).

There is no notation in al-Hawārī’s chapter on algebra, not even for numbers. Recall that in the first part of the book, on arithmetical calculation, he used the Indian figures only to illustrate work that was to be performed on the dust board or other surface. Like the previous chapter covering proportion and double false position, the calculations in the chapter on algebra are given entirely in words because their execution in notation has already been explained in the first part of the book.

link to parallel text 213.1 Section II.2.2. Working out the six types (of equation).

link to parallel text 213.2 The third stage in the solution to a problem by algebra is to solve the simplified equation. Each of the six types of equation is solved by its own rule. The solutions to the three simple types require a single division to get the māl or thing/root.

link to parallel text 213.7 The example for type 1 is “three māls equal fifteen things”, or {m\atop 3}={t\atop 15}. In modern notation, this is 3x^2=15x.

Students who had not studied algebra before would not be familiar with the wording of algebraic equations. So like al-Khwārazmī and many others, al-Hawārī explains each of his equations by translating it into the enunciation of an arithmetic problem, taking advantage of the two meanings of the words “māl” and “root”. In the equation “three māls equal fifteen things”, the word “māl” takes its algebraic meaning as the name of the second power of the unknown. Al-Hawārī reformulates it as an arithmetic problem: “the meaning of this problem is: what quantity (māl), if we take its (square) root fifteen times, gives a sum equal to three times the quantity?” Here the word māl takes the arithmetical meaning of “quantity” or “amount”, and “root” means square root. Note how the equation and the enunciation are worded differently, too. The two sides of the equation are collections of the names of the powers that are declared equal to each other, while the enunciation expresses an operation (“we take its root \ldots”) with a specified outcome (“gives a sum \ldots”). Also, the equation is stated with the algebraic “equal” (taʿdil, conjugated from ʿadala) while the arithmetical version uses an “equal” (musāwiya) common to arithmetic.[87]

Here al-Hawārī calls the equation a “problem” (masʾala), the same word used for other kinds of arithmetic problems or questions. This is also the word used for equations by al-Khwārazmī, Abū Kāmil, and other early algebraists. The specific Arabic word for an algebraic “equation” (muʿādala) first appears in the latter ninth century in Qusṭā ibn Lūqā’s Arabic translation of Diophantus’s Arithmetica. It is common to find both “problem” and “equation” in later Arabic books and chapters on algebra, including Ibn al-Bannāʾ’s and al-Hawārī’s books. Other examples of “problem” are in the passages at 213.13, 214.1, 217.1, 217.10, etc., and “equation” is found in the passages at 217.1, 227.14, 227.17, and 228.1.[88]

link to parallel text 213.13 Al-Hawārī’s equation “two māls equal eighteen”, {m\atop 2}=\hbox{18}, corresponds to our 2x^2=18. Al-Hawārī again translates the equation into an arithmetical enunciation: “What māl, if we add itself to it, becomes equal to eighteen?” Like in the first example, the equation “two māls equal eighteen” does not invoke any operation, since it merely asserts the equality of a couple of “māls” with 18. By contrast, the arithmetical version entails an operation: one adds a māl to itself, with an outcome equal to eighteen.

link to parallel text 214.1 Al-Hawārī’s equation “five things equal twenty”, {t\atop 5}=\hbox{20}, corresponds to our 5x=20. In the reformulation, the “thing” becomes a root of the māl: “What māl, if we take five of its roots, gives a result equal to the twenty?” And again, the arithmetical version is framed as an operation and its outcome.

link to parallel text 214.7 The rules for the composite equations are naturally more complicated, and one can easily show that they are equivalent to our quadratic formula. Many authors, like al-Khwārazmī, Abū Kāmil, and al-Khayyām, give proofs to these rules using geometric diagrams in their algebra books. Others justify the rules with purely arithmetical arguments, like Ibn al-Yāsamīn, Ibn al-Bannāʾ (in his Algebra and Lifting the Veil), and al-Fārisī.[89] But it is also common for an author not to bother with proofs at all, like ʿAlī al-Sulamī, Ibn al-Bannāʾ (in his Condensed Book), al-Hawārī, and Ibn Badr (ca. thirteenth century). In these books, only the arithmetical rules for the solutions are given.[90]

link to parallel text 214.9 Al-Hawārī’s type 4 example “a māl and two things equal fifteen”, {m\atop 1}~{t\atop 2}=\hbox{15}, corresponds to our x^2+2x=15. He again explains the equation in terms of an arithmetical enunciation framed with operations, in which “māl” and “root” revert to their meanings in arithmetic.

Some authors occasionally write out the whole phrase “half of the number of things”, but it is usually written simply as “half of the things”, like it is here. In the modern term 2x, the 2 and the x are both numbers which are joined through scalar multiplication. But the medieval “two things” is simply a two of a particular type (“things”). In the present case “the things” is 2, and its half is 1. Then 1^2=1, 1+15=16, \sqrt{16}=4, and 4-1=3, which is the unknown thing/root. The unknown māl is then 9. This type of equation will have just one positive solution.

We can rewrite the solutions to each equation type as a modern formula. In this case, Ibn al-Bannāʾ’s solution to the general equation x^2+bx=c can be transformed into x=\sqrt{({1\over 2}b)^2+c}-{1\over 2}b. But keep in mind that like the rules for double false position, these rules are procedures consisting of a sequence of operations, and not static, written formulas. The rule is spoken, in time, with one step following another, while our formula is apprehended visually, and not necessarily left-to-right (or right-to-left) like rhetorical text (see our comments above at 198.4).

link to parallel text 214.14 In most problems the desired unknown is the “thing”, so the rule given above yields the answer to the problem. But, often one is searching instead for the māl. For instance, if we were asked to find a number given some condition involving its square root, we might name the number x^2 and create an equation with x^2 and its root x. This is why some algebraists give rules for finding the māl directly, even if in practice these rules were not used much. Rules for finding the māl date back to Abū Kāmil’s late ninth century Book of Algebra, and Ibn al-Bannāʾ included them in his own Book on the Fundamentals and Preliminaries in Algebra. Al-Hawārī certainly learned these rules from Ibn al-Bannāʾ, and he explains the calculations well for this equation.

link to parallel text 215.3 The fifth type of equation might have two, one, or no positive solutions. Al-Hawārī’s example with two solutions is “a māl and eight equal six things” ( {m\atop 1}~\hbox{8}={t\atop 6}; x^2+8=6x).

The rule to solve this equation is similar to the rule for the type 4 equation. Take half of the 6 to get 3, square it to get 9, subtract the 8 to get 1, and take its root to get 1. One of the solutions is obtained by adding the 3 to the 1 to get 4, and the other is obtained by subtracting the 1 from the 3 to get 2.

link to parallel text 215.12 Ibn al-Bannāʾ gives the condition under which this equation will have exactly one solution: a square of half the number of roots/things equals the number. Al-Hawārī’s example is “a māl and nine equal six things” ( {m\atop 1}~\hbox{9}={t\atop 6}; x^2+9=6x). Half of the roots is 3, and its square 9 equals the 9 in the equation. The value of the thing is half the roots, or 3, and the māl is the number, 9.

link to parallel text 215.14 Following the first rule at 215.3 for this kind of equation causes one to operate on “nothing” a number of times. This is reminiscent of the operations on zero we saw earlier in the book. It is not necessary to regard nothing or zero as being a number for these operations to be carried out. See our comments on “operating on zero” at 74.17 above.

link to parallel text 216.1 Next, al-Hawārī gives the rule for finding the māl directly for equations with two solutions; in other words, when “the number is smaller than a square of half of the number of roots” (he had not given this condition earlier). He works this out for the first type 5 example, “a māl and eight equal six things”. A square of the number of roots is 36, and its half is 18. Subtracting the 8 leaves 10. Keep this in mind. Then square the 8 to get 64, and subtract it from a square of the remembered 10, which is 100, leaving 36. Its square root is 6. One solution is that the māl is 10+6=16, and the other is 10-6=4.

link to parallel text 216.11 Type 6 equations always have one positive solution. Al-Hawārī’s example is “a māl equals two of its roots and three” ( {m\atop 1}={t\atop 2}~\hbox{3}; x^2=2x+3). Al-Hawārī’s explanation in terms of arithmetic is awkward, since it is the same as the equation, only with the arithmetical term yusāwī (“[is] equal”) replacing the algebraic yaʿdil (“equals”): “what māl is equal to two of its roots and three?” This might be because the two terms in this equation are stated after the single term, and al-Hawārī did not see that he could simply switch them and write “what māl, if we add three to two if its roots, gives a result equal to the māl?” This confusion may also be why he wrote in the equation “two of its roots”, which is the arithmetical formulation, instead of the usual “two roots”.

In the solution, he takes half of the 2 and squares it to get 1. Adding this to the 3 gives 4. Its root is 2, and adding this to half of the 2 gives 3, “which is the unknown thing”.

link to parallel text 216.17 The rule for finding the māl directly is clear.

link to parallel text 217.1 The rules that Ibn al-Bannāʾ gave for the three composite equations presume that the number of māls is 1. Now he explains that if their number is greater or less than 1, we should set the number to 1 by restoring or reducing it, and we must also adjust the other terms of the equation in the same manner. Restoration and reduction have already been described outside the context of algebra for whole numbers in Section I.1.6 at 129.1, and for fractions in Section I.2.5 at 154.1. Once the number of māls becomes 1, then the rules given above can be applied to find the solution(s).

The word we translate as “term” (ism) in this chapter is also the word for “name”. The species “thing”, “māl”, and “cube” are often called “names” (first at 211.8, in the translation), and the same word is used to mean a term of an equation, like “three things”, “a māl”, and “half a cube”. So “thing” as a name for the first degree unknown is an ism, just as “six things” in the equation below is an ism. In the present passage Ibn al-Bannāʾ uses another word, alqāb, also meaning “name”, for a term in an equation. In common use ism is “given name”, while alqāb is “family name”.

link to parallel text 217.10 Al-Hawārī’s first example is “two māls and six things equal thirty-six” ( {m\atop 2}~{t\atop 6}=\hbox{36}; 2x^2+6x=36). Here the two māls must be reduced to a single māl, so he multiplies it by a half. The other terms must be reduced by a half to compensate, so the equation becomes “a māl and three things equal eighteen” ( {m\atop 1}~{t\atop 3}=\hbox{18}; x^2+3x=18).

link to parallel text 217.15 Another way to do this is to divide the two by two, and divide the other terms by two as well. Dividing by two and multiplying by a half may be mathematically equivalent, but they are conceptually different. One is division by a whole number, and the other is multiplication by a fraction.

link to parallel text 218.1 The other example is “half a māl and two things equal six” ( {m\atop {1\over 2}}~{t\atop 2}=\hbox{6}; {1\over 2}x^2+2x=6). Here the half of a māl must be restored to one māl, so he multiplies it and the other terms by two to get “a māl and four things equal twelve” ( {m\atop 1}~{t\atop 4}=\hbox{12}; x^2+4x=12). The other way is to divide each term by a half.

link to parallel text 219.1 Section II.2.3. Addition and subtraction.

In this section, al-Hawārī explains how to add and subtract algebraic expressions. After giving some easy examples mainly taken from Lifting the Veil (219.5-220.4), he pays special attention to examples illustrating Ibn al-Bannāʾ’s rule for simplifying a problem of subtraction in which one or both of the parts is a diminished algebraic amount. These amounts are of the form “A less B”, and can be thought of as algebraic apotomes (220.5-222.4). Al-Hawārī follows this with a subsection on simplifying equations (223.1-224.8), since the techniques there are the same. In both subtractions and equations diminished amounts are “restored” (with al-jabr) and like terms in the two parts are settled by “confronting” them (with al-muqābala). We describe these terms along with “equalization” (muʿādala) below in our comments at 220.5.

And it is here, in the present section, that rewriting the Arabic calculations in modern notation breaks down. The cause of the problem goes back to the way arithmetical operations, and by extension algebraic operations as well, are worded in premodern mathematics. We have been able to overlook the matter so far because it really only becomes an issue for addition and subtraction of quantities that are already composed of more than one term, like the quantities “six eighths less a ninth”, “two and a root of eight”, or “ten less a thing”. This wording is in turn linked with a difference in how multi-term quantities were conceived, and only when the medieval conceptions are addressed will the Arabic “restoration” of diminished amounts and other oddities of medieval practice make sense.

We begin our comments by examining the anatomy of Arabic arithmetical operations, keeping in mind that algebra is a part of arithmetic. In the process, we address how Arabic authors conceived of multi-term amounts, and we expose the inadequacy of modern notation. Later, in our remarks on al-Hawarī’s examples, we continue to work with the transliteration of the Arabic algebraic notation, which is naturally well-suited to represent the calculations.

Al-Hawārī writes at 87.19 “We add the five to the three, giving eight”. Today we write this as 3+5=8, which is spoken as “three plus five equals eight”. Mathematically, the medieval and modern operations are the same, but they are presented differently. The Arabic is phrased with a verb for the operation (in this example “add”, conjugated from jamaʿa) and a word announcing the result (here “gives”, from bi-). The operation is performed in time, with a specified outcome. By contrast, the verb in the modern version is “equals”, and the word “plus” appears in its quasi-prepositional sense to indicate the operation. There was no word in Arabic that means “plus”. The key difference between the Arabic and modern ways of expressing the operation is that the former is presented as a process that is worked through, while the latter is a static equivalence.

This difference becomes problematic for operations involving multi-term amounts, like al-Hawārī’s “add a root of three to a root of fifteen…to get a root of three and a root of fifteen” (at 180.10). Translated into modern notation, the operation does not seem to accomplish anything: \sqrt{3}+\sqrt{15}=\sqrt{3}+\sqrt{15}. Like in the first example, the operation in the Arabic is expressed with the verb “add” (the first “+” in notation). Because the two numbers cannot be written as a single root, the result is a binomial expressed as “a root of three and a root of fifteen” using the common conjunction “and” (wa, the second “+”). This wa is not another word for the arithmetical operation of addition. The result “a root of three and a root of fifteen” is merely a collection of two objects, like saying “an apple and a banana”. We see this kind of aggregation also in distinct fractions, explained above at 136.8 and 139.1. The fraction “five sixths and four-fifths” from 136.8, written {4\over 5}\,{5\over 6}, is a single number expressed as two numbers gathered together. This kind of joining of numbers into an aggregation is even present in the way numbers with different ranks are expressed, like 24, which is spoken as “four and twenty”. Where in Arabic a distinction is made between the operation of addition and the expression of the result with “and”, in our notation the plus sign “ +” is used for both.

Another example, this one involving three multi-term amounts, is Ibn al-Hāʾim’s “add a root of eight and a root of twenty to a root of five and a root of two…so their sum is a root of eighteen and a root of forty-five”.[91] Translated into modern notation this would be

The original Arabic calls for only one operation (the second “+”), but we see a total of four additions in notation. And where the right side should be the outcome of that operation, in our notation the two sides are symmetric, so it is equivalent to

Ibn al-Hāʾim’s operation calls for the addition of two binomials that results in another binomial. We could adjust our notation to distinguish between aggregations and additions, and use an arrow rather than the “=” to show the direction of the operation, like this:

But this unhappy hybrid is no easier to read than our transliteration of the medieval notation. So we will use the latter, which in this case should look something like this:

Here, the horizontal line serves to separate the steps in the working out of the problem. This notation for separating steps is rarely shown in manuscripts. Like we have seen for al-Hawārī’s calculations with Indian numerals, Arabic textbooks describe the operations rhetorically, and those that show the notation generally give only snippets to illustrate what should be written on the dust-board. Of all the medieval Arabic books that show the notation, we know of only one instance, from Ibn Ghāzī’s Aim of the Students, that shows operations and results without any intervening text. Ibn Ghāzī separates each of his steps with horizontal lines, whether there is an operation or not. So there may have been no special sign that was used to announce the result of an operation. Like our medieval sources, we only give snippets of the notation surrounded by text. And we will use the horizontal line only the one time, above, to indicate the outcome of an operation.

Operations themselves were designated in notation by prepositions. Al-Ḥaṣṣār’s problem “Add three fourths to four fifths” is shown in one late twelfth century manuscript as Too late.[92] Between the fractions is the preposition ilā, “to”, the same word appearing in the rhetorical statement of the operation. The Arabic prepositions for addition, subtraction, multiplication, and division are all different, so they suffice to indicate which operation is intended. Al-Ḥaṣṣār performs this calculation rhetorically to get “one and two fifths and three fourths of a fifth”, which is shown in notation as Too late.

Another example is al-Mawāḥidī’s problem “Divide ten by three less a root of three”, shown in notation as Too late.[93] Here the “by” (ʿalā) and “less” are in black, and the numbers are in red. Above the last 3 is the dotless letter jīm (Too late) indicating a square root. We transcribe this as “ 10~\hbox{by}~3~\ell ~\sqrt{3}”. The result is given rhetorically as “five in number and a root of eight and a third”, and is shown in notation as Too late, or “ 5~\sqrt{8{1\over 3}}”. Like with distinct fractions (at 136.8 and 139.1), the aggregation shows no sign for “and”.[94]

In algebra, adding two terms of the same species is done by adding their “numbers”, or what we call their coefficients. For example, adding three things to five things gives eight things (in modern notation, 5x+3x=8x). But, if the species are different, one can only put the conjunction “and” (wa) between them, as we saw with binomials in the chapter on roots. In his abridgment of his own commentary on Ibn al-Yāsamīn’s poem on algebra, Sibṭ al-Māridīnī (15th c.) gives the example “if you add two dirhams to three things, the answer is two dirhams and three things”.[95] In notation, adding 2 to t\atop 3 gives \hbox{2}~{t\atop 3}. The result of the addition is a collection of five items of two different species or types, and it is worded just like we would say “two apples and three bananas”.

Because a term like “three things” was conceived as a collection of three objects, much like “three bananas”, irrational coefficients were not allowed. One can have three bananas or two and a half bananas, but it makes no sense to speak of \sqrt{12} bananas. Arabic algebraists avoided irrational “numbers” of terms by taking a root of the entire term. For example, in solving one problem Ibn al-Bannāʾ writes “you multiply a thing by a root of ten, to get a root of ten māls”. Expressing this in modern notation, he multiplies x by \sqrt{10} to get \sqrt{10x^2}. This way the “number” (the coefficient 10) remains rational, while the root of the entire term is valid as an irrational number (the value of the “thing” in this problem is found to be 10~\sqrt{2{1\over 2}}~\ell ~\sqrt{2{1\over 2}~\sqrt{1000}}).[96] Here is an example of how such a root appears in notation in an Arabic equation, from Ibn Qunfudh’s commentary: Too late. In our transliterated version this is the equation {t\atop 3}~\sqrt{m\atop 2}=18~\sqrt{72}, which corresponds to 3x+\sqrt{2x^2}=18+\sqrt{72} in modern notation.[97]

link to parallel text 219.2 Al-Hawārī’s example of adding appended expressions is to add together a māl, six things, and ten dirhams to get “a māl and six things and ten dirhams”. This result is a collection or aggregation of seventeen items of three different species, and in notation we write it as {m\atop 1}~{t\atop 6}~\hbox{10}. Contrast al-Hawārī’s polynomial with its modern equivalent, x^2+6x+10, where the arithmetical operation of addition appears twice along with scalar multiplication and exponentiation.

Al-Hawārī does not give an example where the addends themselves are already expressed with two or more appended names (i.e., where the terms are connected with “and”), so we give the explanation from al-Karajī’s al-Fakhrī. He writes: “The way to do that is that you add each species to the same species. For example, add five things and four māls to three things and three māls. You add five things to three things to get eight things, and four māls to three māls to get seven māls. The sum is then eight things and seven māls.”[98]

link to parallel text 219.4 Where the word “and” (wa) is used to express the sum of two different species, the word “less” (illā) is used to express their difference. In Arabic grammar this word indicates that the noun that follows is absent from the preceding noun, like in the phrase “all the children except (illā) Amy”. Here “all the children” is called mustathnā minhu, literally “excluded from it”, or better, “diminished”, and “Amy” is called mustathnā, “excluded” or “absent”.

The same idea of something missing from something else applies also to mathematical expressions with illā. Al-Qalaṣādī gives this example in arithmetic: “subtract a root of five from a root of eight. You say the remainder is a root of eight less (illā) a root of five”.[99] Again, in modern notation the operation seems to accomplish nothing: \sqrt{8}-\sqrt{5}=\sqrt{8}-\sqrt{5}. Instead, the result of the subtraction, “a root of eight less a root of five” ( \sqrt{8}~\ell ~\sqrt{5}), is an apotome, and apotomes by their nature contain no operation. This quantity is a diminished \sqrt{8}, from which \sqrt{5} has been removed. In English, we have no word that functions quite like the Arabic illā. It is kind of like saying “an apple less a bite” or “three cents short of a dollar”. We have already seen this use of “less” in the case of excluded fractions, like the example “six eighths less a ninth” at 140.8. The notational version of that example from one manuscript is shown above in our comments for 140.1. See also al-Mawāḥidī’s notation for “three less a root of three” just above.

In mathematics, as in grammar, the two values A and B in expressions of the form “A less B” are called mustathnā minhu (diminished) and mustathnā (excluded), respectively. Al-Hawārī uses these terms in his explanation of excluded fractions beginning at 140.1, in the context of quadratic irrational apotomes at 183.15, and in algebra on pages 219-222 and 230. Sometimes the diminished term is called “appended” (zāʾid) and the excluded term “deleted” (nāqiṣ), as in the passages at 87.11, 183.15, 189.1, 211.2, etc. We translate these two words as “exceeds” and “falls short”, respectively, in the solutions to problems by double false position (first at 198.8).

So where we simplify an equation like x^2-3x=2+x by adding the subtracted 3x to both sides, at 223.2 al-Hawārī restores (with al-jabr) the diminished māl in the left side of the equation {m\atop 1}~\ell ~{t\atop 3}=\hbox{2}~{t\atop 1} to make it complete (restoration is explained more fully below at 220.5) and then adds the t\atop 3 to the other side. Like with “plus”, there is no medieval Arabic word with the meaning of the modern “minus”. For this reason, neither word occurs in our translation.

link to parallel text 219.5 On the previous line Ibn al-Bannāʾ had written “and the different excluded [species] cannot be subtracted”. When an excluded species (a term following the word “less”) has no counterpart from which it can be subtracted, that is, when it is “different” from the others, the exclusion must remain. Al-Hawārī’s example is to “add a māl less a thing to ten dirhams”, or, in notation, to add {m\atop 1}~\ell ~{t\atop 1} to 10. There are no other “things” in the problem from which we can remove the t\atop 1, so the result of the addition is “a māl and ten dirhams less a thing”, or {m\atop 1}~\hbox{10}~\ell ~{t\atop 1}. The answer is eleven objects of two types (one māl and ten dirhams) that are diminished by a thing. Writing it entirely in modern notation makes the operation appear as if all we do is rearrange terms: 10+(x^2-x)=x^2+10-x.

link to parallel text 219.7 When adding expressions with deleted terms, subtract when possible the deleted term from an appended term of the same power. The first example is to add {m\atop 2}~\ell ~{m\atop 1} to 10. This is not a well-formed problem, since the first term, the “two māls less a māl”, is itself not fully resolved. It is of course the same as “a māl”, and the answer to the addition is {m\atop 1}~\hbox{10}.

In the second example {m\atop 1}~\ell ~{t\atop 2} is added to t\atop 10, resulting in {m\atop 1}~{t\atop 8}. The third example is similar: add {m\atop 1}~\ell ~{t\atop 5} to t\atop 10 to get {m\atop 1}~{t\atop 5}. In modern notation this corresponds to adding x^2-5x to 10x to get x^2+5x.

link to parallel text 220.1 Here al-Hawārī explains the particle “less” to designate the result of a subtraction of different species. The result of subtracting a thing from a māl is “a māl less a thing”, or {m\atop 1}~\ell ~{t\atop 1}. The result in the second example is {m\atop 2}~{t\atop 3}~\ell ~\hbox{10}.

link to parallel text 220.5 Al-Hawārī has postponed until this section his explanations of restoration (al-jabr), confrontation (al-muqābala), and equalization (muʿādala) that were briefly described by Ibn al-Bannāʾ above at 211.2. It is here, after all, that he applies them in problems of subtraction, and in the following subsection on simplifying equations.

The subtraction of appended terms in algebraic expressions (those connected with the word “and”) is easy: one simply subtracts like terms. Subtraction is less straightforward when it involves an expression of the form “A less B”, especially when this occurs in the subtrahend. Ibn al-Bannāʾ and al-Hawārī cover these here. Like other Arabic algebraists, al-Hawārī transforms the problem of subtraction into a simpler problem of subtraction by restoring diminished amounts (of the form “A less B”), and confronting like terms in the two sides. These are also the same steps one takes in simplifying equations, which is covered in the next subsection.

As Ibn al-Bannāʾ remarked above at 211.2, the restoration of a diminished term (the “A” in “A less B”) is the same restoration he explained of smaller numbers to greater numbers described at 129.1 and 154.1, and it is also the same restoration we saw of fractions of a māl at 217.1. To pick two examples, both “half a māl” ( m\atop {1\over 2}) and “a māl less three things” ( {m\atop 1}~\ell ~{t\atop 3}) were considered to be incomplete māls. To restore the first of these one multiplies by two, and to restore the second one adds back the missing “three things”.[100]

In his problem (T1), Abū Kāmil explains restoration in the context of simplifying the equation “fifteen things less a māl and a half equal a māl” ( {t\atop 15}~\ell ~{m\atop {1\over 2}1}={m\atop 1}): “And restore the fifteen [things] by the māl and a half so that it is equivalent to fifteen things. Then add the māl and a half to the māl to get: two māls and half a māl equal fifteen things”[101] ( {m\atop {1\over 2}2}={t\atop 15}). Two distinct steps are performed. First, the “fifteen things less a māl and a half”, being a deficient “fifteen things”, is restored to a full “fifteen things”. Then, to balance the equation, “a māl and a half” is added to the other side. The restoration is the first of these two steps.

In the early eleventh century al-Karajī applied the same wording in his explanation of how to simplify a subtraction problem with a diminished subtrahend by restoration in his al-Fakhrī, just before working it out for a few examples: “The way [to do it] is you restore the subtrahend by what is excluded from it, and you add its same to the minuend”.[102]

By the twelfth century many authors had begun to also call the second step, the addition of the same amount to the other side, a “restoration”. This new way of wording it as two restorations was applied to the simplification of both subtractions and equations. Ibn al-Yāsamīn simplifies the problem of subtracting ten less a thing from a thing ( \hbox{10}~\ell ~{t\atop 1} from t\atop 1): “you restore the ten by the deleted thing to get ten. Then restore the thing, which is the greater part, by the same [amount] you restored the ten, to get two things. So it is as if someone had said, subtract ten from two things”.[103] Note that the restorations do not give the answer to the subtraction, but result in an equivalent subtraction problem from which the remainder will be found.

The modern version of Ibn al-Yāsamīn’s problem is x-(10-x), which we work out by distributing the minus sign to get x-10+x, and which reduces to 2x-10. Although medieval algebraists knew that “when subtracting the deleted from another deleted, it is appended” (from above at 87.11), they usually handled the problem of subtracting an apotome through a restoration.

Like Ibn al-Yāsamīn, Ibn al-Bannāʾ also calls both steps restorations. His explanation in Lifting the Veil of his remark from 211.2 applies to equations:

Restoring the deleted to the appended always entails the two sides. This is because whenever you restore the appended by what is excluded from it on one side, once your restoration is completed, that side has become greater than the other side. Thus it is different from what is equated to it, so you restore it by the excess.[104]

Confrontation is performed whenever there are like terms in the two sides. The verb qabila (“to confront”) is rarely used in the course of simplifying an equation or subtraction, however. The tenth-century CE lexicographer Muḥammad ibn Aḥmad al-Khwārazmī explains the step:

And for an example of confrontation, suppose one encounters in a problem: a māl and two roots equal five roots [ {m\atop 1}~{t\atop 2}={t\atop 5}]. You cast away the two roots which are with the māl, and you cast away the same from what it is equal to, to get: a māl equals three roots, and that is nine.[105]

The idea is that the like terms are confronted, or put face-to-face, and their difference is placed on the side of the greater. Like restoration, confrontation is performed in two steps. The smaller amount is deleted from one side, and the smaller is subtracted from the larger on the other side.

Ibn al-Bannāʾ explains confrontation similarly, though more tersely, in Lifting the Veil.[106] What is new in Ibn al-Bannāʾ is the word meaning “equalization” (muʿādala). He writes:

Restoring the deleted to the appended is equalization with restoration, and subtracting the appended from the appended or the deleted from the deleted of things of the same species is equalization with confrontation.[107]

In both cases the “equalization” is the necessary compensation in the other part so that the new subtraction or equation yields the same answer as the original. “Restoring the deleted to the appended” is the adding of the deleted amount to the other side, and “subtracting the appended from the appended or the deleted from the deleted” is apparently what one does to compensate for deleting the smaller of the two like terms (the meaning is not presented clearly for confrontation). In confrontation, one typically subtracts “the appended from the appended”. Al-Hawārī gives an example of subtracting “the deleted from the deleted” in the second solution to the problem at 223.7.

Restoration was also applied to the subtraction of apotomes. Al-Baghdādī explains that one must “restore the exception by what it is deleted from on that side, and add the same to the other side. Then subtract the subtrahend from the minuend”, just before giving the example of subtracting “ten less a root of eighty from a root of eighty less five”.[108]

link to parallel text 220.9 In the first of four examples of subtraction, the minuend is diminished. Here al-Hawārī subtracts “two and a thing” ( \hbox{2}~{t\atop 1}) from “a māl less three things” ( {m\atop 1}~\ell ~{t\atop 3}). Adding the t\atop 3 to the minuend restores the m\atop 1, and adding the same to the subtrahend equalizes the operation. Thus it is called “equalization with restoration”. The problem then becomes to “subtract two and four things from a māl”, or \hbox{2}~{t\atop 4} from m\atop 1. The answer is {m\atop 1}~\ell ~{t\atop 4}~\ell ~\hbox{2}.

link to parallel text 221.1 In the second example, the diminished term is in the subtrahend: subtract \hbox{52}~\ell ~{t\atop 5} from {c\atop 2}~\hbox{30} (we write “c” for “cube”). Restoring the 52 and adding t\atop 5 to the other part changes the question into one of subtracting 52 from {c\atop 2}~{t\atop 5}~\hbox{30}. So far, this is equalization with restoration. Now both parts have simple numbers, 30 and 52, and these should be confronted. Al-Hawārī subtracts the smaller of the two from both sides, making the problem one of subtracting 22 from {c\atop 2}~{t\atop 5}.

link to parallel text 221.8 The step just described is called “confrontation with equalization” (he reverses Ibn al-Bannāʾ’s words) since the 30 is dropped from one part, and is subtracted from 52 in the other. The answer is then {c\atop 2}~{t\atop 5}~\ell ~\hbox{22}. In the modern version of this problem we would not need to transform the subtraction of 52 from 2x^3+5x+30 into the problem of the subtraction of 22 from 2x^3+5x, since for us the latter “problem” 2x^3+5x-22 is also its solution.

link to parallel text 221.13 In this next example, both terms are diminished. To subtract \hbox{12}~\ell ~{t\atop 4} from {m\atop 3}~\ell ~{t\atop 2} al-Hawārī adds t\atop 2 and t\atop 4, totaling t\atop 6, to both sides to turn the problem into one of subtracting \hbox{12}~{t\atop 2} from {m\atop 3}~{t\atop 4}. We can see that he could have just added t\atop 4 to both sides, but our author is following the rule to the letter: add to both sides all amounts that are lacking from diminished terms. From here, confronting the t\atop 4 with the t\atop 2 yields the problem of subtracting 12 from {m\atop 3}~{t\atop 2}, giving {m\atop 3}~{t\atop 2}~\ell ~\hbox{12}.

link to parallel text 222.1 The last example is worked out like the previous one. The problem is to subtract {c\atop 1}~\ell ~{m\atop 2} from \hbox{30}~\ell ~{t\atop 4}. Restoring with equalizing means adding {t\atop 4}~{m\atop 2} to both sides to get the equivalent problem of subtracting {c\atop 1}~{t\atop 4} from {m\atop 2}~\hbox{30}. There is no confrontation to be done here, since all terms are of different species. The answer is then \hbox{30}~{m\atop 2}~\ell ~{c\atop 1}~\ell ~{t\atop 4}.

link to parallel text 223.1 Subsection concerning examples with the two sides of an equation.

Now al-Hawārī shows how to apply restoration (al-jabr) and confrontation (al-muqābala) to simplify equations in stage 2 to one of the six types. The phrase al-jabr wa-l-muqābala, taken from these steps in stage 2, was the name given to the art of algebra. As mentioned above, some algebraists would shorten the name to just al-jabr, and via transliterations in medieval Latin and Italian it became the origin of our word “algebra”.

link to parallel text 223.2 The first example is {m\atop 1}~\ell ~{t\atop 3}=\hbox{2}~{t\atop 1}. Restoring the m\atop 1 amounts to turning the {m\atop 1}~\ell ~{t\atop 3} into m\atop 1. To equalize, t\atop 3 must be added to the \hbox{2}~{t\atop 1}, so the equation becomes {m\atop 1}=\hbox{2}~{t\atop 4}. This is a type 6 equation, so one should next follow the rule at 216.11 to solve it.

link to parallel text 223.7 To simplify {m\atop 1}~\ell ~{t\atop 3}=\hbox{24}~\ell ~{t\atop 5} one restores first. Adding both t\atop 3 and t\atop 5 to both sides gives {m\atop 1}~{t\atop 5}=\hbox{24}~{t\atop 3}. Now the t\atop 5 is confronted with the t\atop 3 to get {m\atop 1}~{t\atop 2}=\hbox{24}. This is a type 4 equation, and the rule for its solution is given above at 214.7.

Another way to do it is to confront the deleted t\atop 3 with the deleted t\atop 5 on the other side to get a deleted t\atop 2: {m\atop 1}=\hbox{24}~\ell ~{t\atop 2}. Now the 24 is restored and t\atop 2 is added to the other side to get {m\atop 1}~{t\atop 2}=\hbox{24}.

link to parallel text 223.14 The next example is to simplify {m\atop 1}~\ell ~\hbox{10}={m\atop 1}~\ell ~{t\atop 2{1\over 2}}. Adding \hbox{10}~{t\atop 2{1\over 2}} to both sides gives {m\atop 1}~{t\atop 2{1\over 2}}={m\atop 1}~\hbox{10}; confronting the two {m\atop 1}s results in the equation \hbox{10}={t\atop 2{1\over 2}}. Al-Hawārī cannot begin this one by confronting the {m\atop 1}s, since it would leave terms deleted from nothing on both sides, which would be impossible.

link to parallel text 223.17 In the equation {m\atop 1}~\hbox{10}=\hbox{51}~\ell ~{t\atop 4}, adding the t\atop 4 to both sides gives {m\atop 1}~{t\atop 4}~\hbox{10}=\hbox{51}, and confronting the 10 with the 51 results in the simplified equation {m\atop 1}~{t\atop 4}=\hbox{41}. Al-Hawārī does not mention the confrontation. The other equations he gives are simplified similarly.

link to parallel text 225.1 Section II.2.4. Multiplication and knowing the power and the term.

The power of a term in modern algebra is easy to determine, since it is the exponent. The power of x^3 is 3, the power of x^{12} is 12, etc. Knowing the power is necessary for multiplying or dividing terms, such as figuring that x^2\cdot x^6= x^8 or 6x^3\div 2x^2= 3x. In Arabic algebra the terms are expressed with the names “thing”, “māl”, and “cube”, so the exponent must be calculated from these names. Ibn al-Bannāʾ tells us that the power of the things is 1, of māls is 2, and of cubes is 3, and that for higher powers one collects these numbers together. For example, the power of a māl cube māl māl is 2+3+2+2=9. From 225.2 to the end of the chapter the text has “[a] power (uss)” with the implied indefinite article. We translate it as “the power” to make the reading easier, and because the power of a term is unique.

link to parallel text 226.1 To find a name of the term from the exponent, one need only come up with some combination of māl and cube so the powers add up right. There was no standard way of writing the powers greater than the fourth in Arabic, as we already knew from variations offered by such authors as Abū Kāmil and al-Karajī. Al-Hawārī makes this terminological flexibility explicit.

link to parallel text 226.9 To multiply two monomials, multiply their numbers (i.e., coefficients) and add the powers. If one of the terms is a simple number, then the power of the product is the power of the other term.

link to parallel text 226.12 Al-Hawārī explains his examples well. In notation they are t\atop 5 by t\atop 7 gives m\atop 35, t\atop 10 by m\atop 6 gives c\atop 60, t\atop 1 by c\atop 1 gives mm\atop 1, 6 by m\atop 4 gives m\atop 24, and 7 by mc\atop 3 gives mc\atop 21.

link to parallel text 227.10 In Arabic algebra there was no standard way of making the plurals for species expressed with more than one term (māl, cube). Al-Hawārī makes the first term plural when there is more than one, but because the plural and singular forms of Arabic nouns are the same for numbers greater than ten, we often cannot tell if the second term is intended to be plural. What appears to be “three māls cube” could be “three māls cubes” if we take the “three māls” to be the number of cubes, and this number is greater than ten. See our remarks above at 70.8 for a similar situation with “thousands”. In our translation we leave these potentially plural forms singular. Perhaps Ibn al-Bannāʾ can make all the terms plural at 227.14 because he has not specified the number of the term.

link to parallel text 227.14 Here Ibn al-Bannāʾ explains how to reduce the degree of an equation. This passage seems to be misplaced. It fits better in the section below on division.

If every term in an equation has degree of at least one, or, as al-Hawārī says, “you do not have a number”, then drop the power of each term by the lowest power in the equation. For a modern example, if we had 4x^3-2x^2=x^5, the smallest power is 2, so we drop all powers by two to get 4x-2=x^3. Zero cannot be a solution in medieval algebra, so for al-Hawārī this gives an equivalent equation. Again, al-Hawārī explains his three examples well. In notation they are: {mm\atop 3}~=~{c\atop 4}~{m\atop 10} reduces to {m\atop 3}~=~{t\atop 4}~\hbox{10}, {c\atop 3}~=~{m\atop 10}~{t\atop 20} reduces to {m\atop 3}~=~{t\atop 10}~\hbox{20}, and {c\atop 1}~{m\atop 10}~=~{t\atop 39} reduces to {m\atop 1}~{t\atop 10}~=~\hbox{39}.

link to parallel text 228.8 Before Italian algebraists in the sixteenth century found solutions to irreducible cubic and quartic equations, the only equation types that could be solved generally were those that simplify, or could in some way be reduced, to one of the six types of degree 1 or 2.

link to parallel text 228.9 When multiplying polynomials, some terms are appended and some are deleted. For example, in “a hundred and a māl less twenty things” ( \hbox{100}~{m\atop 1}~\ell ~{t\atop 20}) the “hundred” and the “māl” are appended, while the “twenty things” is deleted. Be careful not to interpret “appended” and “deleted” as “positive” and “negative”. See the discussions at 219.1 and 219.4 above.

Al-Hawārī does not explain why a deleted term multiplied by a deleted term yields an appended term, but the rule was well known. It would have been easy to derive it by geometry, among other possible ways. Abū Kāmil, in fact, gives a geometric proof in the context of the multiplication of “ten dirhams less a thing by ten dirhams less a thing” ( \hbox{10}~\ell ~{t\atop 1} by \hbox{10}~\ell ~{t\atop 1}).[109]

link to parallel text 228.15 The example in notation is to multiply \hbox{8}~\ell ~{t\atop 2} by \hbox{7}~\ell ~{m\atop 4}. First al-Hawārī multiplies the terms that will give appended amounts: 8 by 7 gives 56, and t\atop 2 by m\atop 4 gives c\atop 8. The other two products yield deleted terms: 8 by m\atop 4 gives m\atop 32 and t\atop 2 by 7 gives t\atop 14. The product is then {c\atop 8}~\hbox{56}~\ell ~{t\atop 14}~\ell ~{m\atop 32}.

link to parallel text 229.1 Section II.2.5. Division.

link to parallel text 229.2 Ibn al-Bannāʾ first explains how to divide a monomial by a monomial of lower degree. Al-Hawārī’s first example is to divide m\atop 10 by t\atop 2. The 10 is divided by 2 to get 5. The power of the māls is 2, and the power of the things is 1, so the power of the quotient is their difference, 1. The answer is then “five things”. His other example is to divide c\atop 15 by t\atop 3, from which he gets m\atop 5.

link to parallel text 229.8 Note the wording: “The resulting five is then five māls”. In this term the “māls” is the kind, and the “five” tells how many there are. The two words are the two aspects of a single number or value, like saying “five dollars”. When there is only one of them, the notation must still specify the number. If, instead of “ {m\atop 1}~{t\atop 5}” we write “ \hbox{m}~{t\atop 5}”, it would be like saying “dollars and five pesos”. This leaves unanswered how many dollars/māls there are. For this reason, throughout premodern algebra a “1” was always written in notation when there is one of a term. By contrast, the 5 and the x^2 in our 5x^2 are both numbers, one known and the other not, that are multiplied together. Our x^2 itself, without any coefficient, already designates a value, and so does not need a 1 in front.[110]

link to parallel text 229.12 In the case that the powers of the two terms are the same, the result is a number. Al-Hawārī’s example is to divide m\atop 12 by m\atop 3 to get 4.

link to parallel text 229.17 Dividing by a number gives the same power as the dividend. The example is to divide t\atop 12 by 4 to get t\atop 3. This and the previous rule are given separately because the power of a number was not regarded as being zero. Zero was a place holder in Indian numerical notation, and not a number.

link to parallel text 230.1 If the dividend is diminished, divide both terms by the divisor. By this rule, dividing {c\atop 12}~\ell ~{m\atop 3} by t\atop 2 gives {m\atop 6}~\ell ~{t\atop {1\over 2}1}. A second example is dividing {m\atop 10}~\ell ~{t\atop 3} by 2 to get {m\atop 5}~\ell ~{t\atop {1\over 2}1}. Our authors do not mention that the same rule works if “less” is replaced by “and”, but this may be because the procedure is easily understood in the case of terms gathered together.

link to parallel text 230.11 Ibn al-Bannāʾ writes that one cannot divide a lower power by a higher power. But one can at least eliminate common divisors. The example is to divide m\atop 6 by c\atop 3 to get “two divided by a thing”. The division is still unresolved, but at least the common divisors have been dealt with. Ibn al-Bannāʾ explains the same rule in his book on algebra. In one example he writes “divide eight cubes by three māls māl”. The answer is given as “two and two thirds divided by a thing”.[111]

When working with polynomials, the three operations of addition, subtraction, and multiplication can all be performed to yield another polynomial. But the division of a polynomial by a polynomial often cannot be performed, which is why an algebraist would sometimes admit an unresolved division in an equation. But writing the result of a division in the form “ A divided by B” was generally avoided in algebraic problem-solving. Ibn al-Bannāʾ’s problem (I.10) from his book on algebra is a good example. He solves this particular problem five different ways, each time coming up with a clever way to avoid unresolved divisions. (The enunciation and first solution are translated as problem [6] in Appendix B.) Al-Karajī solves this same problem, but with the number 2{1\over 6} in place of 4{1\over 4}. Only in his fourth and last solution does he set up the equation with divisions: “ten less a thing divided by a thing and a thing divided by ten less a thing, and that equals two dirhams and a sixth”.[112]

In modern notation, al-Karajī’s equation is {10-x\over x}+{x\over 10-x}=2{1\over 6}. We know only one book that shows notation for divisions of algebraic expressions. In place of the division bar, Ibn Ghāzī writes them with an elongated version of the Arabic letter qāf (Too late), the first letter in the word qasama (“divide”). Here is the notational version of his “twenty-four divided by half a thing and four and a half in number”, followed by our transcription (with a “d” for “divided by”), and then the modern version:

The manuscript indicates “things” with only the three dots from the letter Too late. The similarity between the simple bar for fractions and the elongated qāf for algebraic expressions is deliberate, but because divisions are not fractions they are differentiated.

link to parallel text 230.16 One cannot perform a division if the divisor is diminished. To “divide ten māls by three less a thing” can only yield “ten māls divided by three less a thing”. Of course, one cannot perform the division if the divisor consists of two or more appended terms, either. Dividing “ten māls” by “three and a thing” can only give “ten māls divided by three and a thing”. Neither Ibn al-Bannāʾ nor al-Hawārī make note of this.

link to parallel text 231.1 Section [on secret numbers].

It was not uncommon for authors of arithmetic books to include problems on finding secret numbers. Other medieval authors who did so include Ibn al-Yāsamīn (d. 1204), Ibn al-Hāʾim (1389), and Ibn Haydūr (ca. 1400). We have not found the particular problems dictated to al-Hawārī in other books. It is probably by intent that there is one secret number in the first problem, two secret numbers in the second problem, and three secret numbers in the third problem. We describe the conditions using modern algebraic notation, in which the secret numbers are x, y, and z, and a and b are calculated values revealed to the guesser. Keep in mind that there is no algebra (al-jabr wa-l-muqābala) in the questions or solutions.

link to parallel text 231.4 First problem: the person is asked to secretly think of a number, x, and then to calculate x^2 and (10-x)^2. If (10-x)^2<x^2, the remainder a=x^2-(10-x)^2 is revealed. Then x={1\over 2}({a\over 10}+10). If x^2<(10-x)^2 then a=(10-x)^2-x^2 is revealed, and x={1\over 2}(10-{a\over 10}).

link to parallel text 231.13 Second problem: the person is asked to secretly think of two numbers, x and y, that add to 10. This time the revealed number is a=x^2/xy, which of course is the same as x/y. Given the sum and ratio of two numbers, the numbers are also given. Al-Hawārī does not give the rule past this point. We can solve our equations to get x={10a\over 1+a}, but if the person chose whole numbers they are easy to guess without any particular rule. If the revealed number is 9 or 1/9, the numbers are 1 and 9. If they are 4 or 1/4, the numbers are 2 and 8, etc.

link to parallel text 232.1 Third problem: A secret number is divided into two other secret numbers. In notation, suppose z=x+y with x>y. Two numbers are revealed: a=xy-y^2 and b=x^2-xy. Then x-y=\sqrt{b-a} and z=x+y={a+b\over x-y}, so 2x=(x+y)+(x-y) and 2y=(x+y)-(x-y).

link to parallel text 233.2 The date 18 Dhū al-Qaʿda 704 AH corresponds to the Julian date June 12, 1305.

Footnotes

[1]  Translated in (Euclid 1956, vol. 2, 277).

[2]  (al-Fārisī 1994, 71.8).

[3]  (Ibn al-Bannāʾ 1994, 207); (Aballagh 1988, 142-143).

[4]  (Aballagh 1988, 142-143).

[5]  See below at 86.1 for an explanation of the “ \scalebox {-1}[1]{\ell }”.

[6]  Book I, Chapters 7-10 (Nicomachus 1959, 19-28); (Nicomachus 1938, 190-201). Ibn al-Bannāʾ’s terms are precisely those in Thābit ibn Qurra’s late ninth century Arabic translation of Nicomachus’s work (Nicomachus 1959, 20.18). D’Ooge translated Nicomachus’s Greek terms as “even-times even”, “even-times odd”, and “odd-times even”.

[7]  (Nicomachus 1866, 27.12); (Nicomachus 1938, 202, Chapter XII); (Nicomachus 1959, 30.4).

[8]  (Kunitzsch 2003).

[9]  (Lane 1863–1893, 455). Lane gives definitions in third person singular perfect tense. Here and elsewhere I have changed them to the infinitive.

[10]  (Lane 1863–1893, 1275, 646), (Wehr 1994, 640).

[11]  (Saidan 1996, 341ff).

[12]  In the Arithmetical Introduction, Book II, Chapters 20-23 (Nicomachus 1938, 263, 266-70).

[13]  (Abū Kāmil 1986, 218); (Abū Kāmil 2012, 725).

[14]  The first trio of dots on the right should not be there.

[15]  Categories 4b20-24; (Avicenna 2005, 10, 72).

[16]  Most arithmetic books give this definition. We found four arithmetic books that give no definition at all: those of al-Uqlīdisī, Kūshyār ibn Labbān, al-Baghdādī, and al-Qurashī. (Titles of their books are given in Appendix C.) The word for “duplication” is almost always ḍaʿafa.

[17]  (Euclid 1956, vol. 2, 278).

[18]  (Saidan 1971, 124.16).

[19]  (Saidan 1971, 126.10).

[20]  (al-Khwārizmī 2009, 123.6).

[21]  (al-Fārābī 1953, 74.10).

[22]  There were some books that apply algebra to solve problems relating to Book X, but not the other way around.

[23]  (Saidan 1971, 124).

[24]  (Saidan 1968, 707).

[25]  (al-Uqlīdisī 1978, 10).

[26]  For arithmetical proofs in Arabic algebra, see (Oaks 2018a).

[27]  The earliest extant book is Abū l-Wafāʾ’s arithmetic book (Saidan 1971, 113).

[28]  (Nicomachus 1959, 31); (Nicomachus 1938, 203ff).

[29]  Book V, Definition 3 of the Elements (Euclid 1956, vol. 2, 114). The word “size” here is translated from pēlikotēs, which was translated into Arabic as qadr. The meaning of pēlikotēs in the Book V definition is not yet that of numerical size.

[30]  The desire to validate fractions as ratios goes back at least to the tenth century, when Abū l-Wafāʾ also defined fractions as ratios of whole numbers (Saidan 1971, 71).

[31]  (Euclid 1956, vol. 2, 277).

[32]  The “separation” he speaks of may be the preposition “to” (ilā) in the phrase “the ratio of three to six”. Or it may be the mark one makes when writing ratios in notation. Al-Mawāḥidī, another commentator of the Condensed Book, uses the generic “ \cdot \cdot \cdot” which also serves to separate numbers in other contexts. The proportion 3:4::6:8 is written as “8 \cdot \cdot \cdot6 \cdot \cdot \cdot4 \cdot \cdot \cdot3” in (al-Mawāḥidī manuscript, fol. 78a).

[33]  We also find intelligible mathematical objects in al-Khayyām (Omar Khayyam), who followed Aristotle closely. He wrote that while one-, two-, and three-dimensional algebraic powers correspond to geometric magnitudes that exist in sensible things, the zero degree term and powers higher than three are not associated with any sensible objects. These objects are instead “imagined in the continuous magnitudes”, or are “taken as abstracted in the intellect from material things”. See (Oaks 2011a); (Rashed and Vahabzadeh 2000, 171).

[34]  (Avicenna 2005, 122).

[35]  (Ibn al-Bannāʾ 1994, 270.3-4).

[36]  (al-Uqlīdisī 1978, 11).

[37]  (al-Khwārizmī 2009, 147.15, 253.19).

[38]  Our authors use this word to mean “irrational”. See below at 163.2.

[39]  (al-Khwārizmī 2009, 169.14).

[40]  (Saidan 1986, 181.12,14).

[41]  (al-Khwārizmī 2009, 265.7).

[42]  (Abū Kāmil 2012, 471.5).

[43]  (al-Khwārizmī 2009, 169.13).

[44]  Ibn al-Bannāʾ’s example 15\over 24 at 137.1 is first stated in terms of “fourths of a sixth”, and the version in terms of “twenty-fourths” is only added to show the denominator.

[45]  (Zemouli 1993, 171). On Ibn al-Yāsamīn’s way of writing numbers, see (Abdeljaouad 2005b).

[46]  (Mueller 1981, 59).

[47]  (Zemouli 1993, 250).

[48]  Here “surplus (faḍla) between” could be translated as “difference between”, but the notion of a surplus or excess would be lost.

[49]  (Ibn al-Bannāʾ 1994, 285.10).

[50]  (Ibn al-Bannāʾ 1994, 285.15).

[51]  We have found the descriptions in three commentaries on Euclid’s Book X: Anonymous (attributed to al-Māhānī, #82 [M2], ninth century), Sulaymān ibn ʿIṣma al-Samarqandī (#181 [M1], ca. 900), and al-Ahwāzī (#193 [M1], tenth century); in Ibn Sīnā’s epitome of Euclid’s Elements ([M1], ca. 1000); and in two works dealing with algebra: al-Karajī’s Marvelous [Book] of Arithmetic (al-Badīʿ) ([M3], early eleventh century. He gives a minor variation in the descriptions of the second and third apotomes) and Ibn al-Bannāʾ’s Book on the Fundamentals and Preliminaries in Algebra ([M6], before 1300. Only for binomials). Gregg De Young writes that Ibn Sīnā’s book “was heavily influenced by, if not based completely upon, the al-Ḥajjāj translation tradition” (De Young 1991, 665). His evidence comes partly from Ibn Sīnā’s proofs to propositions X.64-66 (De Young 1991, 661). From this, and the fact that their wording is different in the Isḥāq-Thābit translation, we cautiously attribute the wording of these phrases to the translation of al-Ḥajjāj.

[52]  Definitions are given in Propositions X.36 and X.73, and the classifications are given in two sets of definitions presented after Propositions X.47 and X.84.

[53]  By contrast, commensurable roots can be added to make a single root, like the example at 179.16: \sqrt{2} and \sqrt{8} together are \sqrt{18}.

[54]  (Ibn Qunfudh manuscript, 197.24, 198.12).

[55]  (Ibn Qunfudh manuscript, 198.12-13).

[56]  We enclose in square brackets English words implied but not present in the Greek, and were thus not translated into Arabic.

[57]  (Euclid 1956, vol. 3, 85).

[58]  This is Heath’s explanation in (Euclid 1956, vol. 3, 55).

[59]  (Euclid 1956, vol. 3, 87).

[60]  Our translation. Heath translates it as “the side of a rational plus a medial area” (Euclid 1956, vol. 3, 129).

[61]  (Euclid 1956, vol. 3, 206).

[62]  We have corrected misplaced quotation marks here.

[63]  Our translation. Heath translates it as “the side of the sum of two medial areas” (Euclid 1956, vol. 3, 130).

[64]  (Euclid 1956, vol. 3, 209).

[65]  (Saidan 1986, 530.24).

[66]  For Abū Kāmil, see (Oaks 2011b); for al-Baghdādī, see (al-Baghdādī 1985, 200.5).

[67] See (Oaks 2017, 153).

[68]  (Ibn al-Bannāʾ 1994, 293.3-16).

[69]  (Ibn al-Bannāʾ 1994, 293.5).

[70]  (Ibn al-Bannāʾ 1994, 293.7).

[71]  “And the third is harmonic proportion, which is described by the author of the Arithmetic…” (Ibn al-Bannāʾ 1984, 293.7). He then paraphrases (Nicomachus 1959, 102.1). Harmonic proportion is covered by Nicomachus in Book II Chapter 25, starting at (Nicomachus 1959, 101.20); (Nicomachus 1938, 274).

[72]  Translated in (Nicomachus 1938, 284). Both in Greek and in Arabic the words for “harmonic” and “harmony” have the same root.

[73]  Translated in (Nicomachus 1938, 270).

[74]  (Ibn al-Bannāʾ 1994, 293.17). One could argue that he meant “the first two”, since he explicitly said that the third, harmonic proportion, “is composed of the geometric and the arithmetic”. In any case, it is geometric proportion that is in “the four [proportional] numbers”.

[75]  (Ibn al-Bannāʾ 1994, 214.12).

[76]  (Ibn al-Bannāʾ 1994, 297.18).

[77]  (Saidan 1986, 557.7). Ibn al-Bannāʾ’s first of two solutions by algebra is: “You make one of the parts a thing, so the other is ten less a thing. We take a third of a thing, giving a third of a thing, and you confront it with a fourth of ten less a thing, and that is two and a half less a fourth of a thing [this sets up the equation we write as {1\over 3}x=2{1\over 2}-{1\over 4}x]. You restore and confront [i.e., simplify the equation], resulting in the third type [of equation, which we would write as {7\over 12}x=2{1\over 2}], which gives the thing is four and two sevenths, which is the part whose third we took, and the other part is the remainder from the ten.” In the second solution he switches the roles of “a thing” and “ten less a thing”.

[78]  See (Katz 2017).

[79]  (Caianiello 2018).

[80]  (al-Khwārizmī 2009, 97.9).

[81]  (Oaks and Alkhateeb 2005).

[82]  Problems (24) and (28) in (al-Khwārizmī 2009, 185, 191); most of the 20 geometry problems in (Abū Kāmil 2012, 523-577).

[83]  Definitions are taken from (Wehr 1994, 519, 943, 1047). See (Oaks 2010).

[84]  See (Abdeljaouad 2002) and (Oaks 2012a) for a description of the notation.

[85]  (Ibn Ghāzī manuscript, fol. 83a).

[86]  (Ibn Ghāzī manuscript, fol. 84a).

[87]  (Oaks and Alkhateeb 2005, §4.5).

[88]  For more on the word “equation” see (Oaks 2010, §5.2).

[89] For arithmetical proofs, see (Oaks 2018a).

[90]  The books of these authors are named in Appendix C.

[91]  (Ibn al-Hāʾim 2003, 199.22).

[92]  (al-Ḥaṣṣār manuscript, fol. 68b).

[93]  (al-Mawāḥidī manuscript, fol. 77a).

[94]  Occasionally the “and” will be written in the notation, usually to counterbalance a “less” in another part of the expression.

[95]  (Sibṭ al-Māridīnī 1983, 59.8).

[96]  Or, in modern notation, 10+\sqrt{2{1\over 2}}-\sqrt{2{1\over 2}+\sqrt{1000}} (Saidan 1986, 581.6).

[97]  (Ibn Qunfudh manuscript, 272.3). Here Ibn Qunfudh denotes “things” using only the three dots from the letter Too late. The equation is stated verbally as “three things and a root of two māls equal eighteen and a root of seventy-two in number”. The example is given to explain that the term “a root of two māls” is of the order of things, since it comes from the “multiplication of a root of two by a thing”. For more on how medieval algebraists dealt with irrational numbers in algebra, see (Oaks 2017).

[98]  (Saidan 1986, 118.5).

[99]  (al-Qalaṣādī 1999, 215.9, French 248.6).

[100]  (Oaks and Alkhateeb 2007, §3.3).

[101]  (Abū Kāmil 2012, 321.11).

[102]  (Saidan 1986, 120.14).

[103]  (Zemouli 1993, 229.1).

[104]  (Ibn al-Bannāʾ 1994, 309.3).

[105]  (al-Khwārizmī 1895, 201.3). The “what it [i.e., the left side] is equal to” is the “five roots” on the other side of the equation.

[106]  (Ibn al-Bannāʾ 1994, 309.6).

[107]  (Ibn al-Bannāʾ 1994, 309.7).

[108]  (al-Baghdādī 1985, 221.1).

[109]  (Abū Kāmil 2012, 293.7).

[110]  See (Oaks 2009) and (Oaks 2017) for more on this difference.

[111]  (Saidan 1986, 541.2).

[112]  (Saidan 1986, 213.2).

Appendices

A. Conspectus of problems

For the following list of calculations and problems in al-Hawārī’s book, we have adopted our transcription of the Arabic notation. This is explained in our commentary, especially at 219.1. We had introduced the algebraic notation in our commentary at 211.2, and notation for fractions is explained in our commentary in the chapter on fractions, beginning at 135.1. Here is a quick guide to the notation:

The notation is a transcription of how it appears, or in some cases would have appeared, in the manuscripts. With the exception of the figures for double false position, we put the notation in red, as it is in many Arabic manuscripts. It should be read right to left.

The reversed letter “ \scalebox {-1}[1]{\ell }” stands for “less”, and indicates that the number to its left is removed from the number to its right. So, an apotome that we would write as \sqrt{5}-\sqrt{3} is shown as \leavevmode {\color {red}\sqrt{3}~\scalebox {-1}[1]{\ell }~\sqrt{5}}. See our commentary at 86.1. Binomials have no sign for “+”, so the modern 5+\sqrt{21} is shown as \leavevmode {\color {red}\sqrt{21}~5}.

Instead of the letter jīm above a number to indicate square root, we use the modern “ \leavevmode {\color {red}\sqrt{~~}\,}” which functions similarly.

We write “=” for the “equals” in algebraic equations, and for no other purpose. This sign functions like the elongated lām, its counterpart in Arabic manuscripts.

Because the transcribed notation for algebra is entirely different from our notation, we include the versions of the calculations in modern notation as well, even if it sometimes makes little sense (as in the calculations from through 222.1).

An “L” is placed after the reference for examples taken from Ibn al-Bannāʾ’s Lifting the Veil.

Part 1. On known numbers

Chapter 1. On whole numbers

Passage
Example

65.2
Examples of whole numbers: 15, 18, etc.

65.2
Examples of fractions: \leavevmode {\color {red}1\over 2}, \leavevmode {\color {red}1~\,3\over 2~\,8}, \leavevmode {\color {red}{1\over 4}~{1\over 9}}, \leavevmode {\color {red}7~\bullet ~6\over 8~\bullet ~7}, \leavevmode {\color {red}{1\over 9}~\scalebox {-1}[1]{\ell }~{5\over 6}}

65.6,10,17, 66.1
Even: 10, 50; evenly-even: 32; evenly-odd: 14; evenly-evenly-odd: 28

66.7
Odd prime: 11, 29; Oddly-odd: 15, composed of 3 by 5

66.17
Even square: 36 is 6 by 6

67.3
Even, composed of two unequal numbers (a surface): 18 is 3 by 6, and also 2 by 9

67.3
Even, composed of three unequal numbers (a surface): 24 is 3 by 4 by 2

67.12
Even cube: 64 is 4 by 4 by 4

67.19
Odd square: 25 is 5 by 5

68.1
Odd, composed of two unequal numbers (a surface): 35 is 5 by 7; 105 is 3 by 5 by 7

68.8
Odd cube: 27 is composed of 3 by 3 by 3

70.8, 70.23
Sample figures for numbers: 9367184225 and 84725

71.6
143 has three places

71.16
The rank of 10000000 is 8

72.4
The name of 1000000000 is thousands thousands of thousands

74.17
Adding from the units place: add 4043 to 2685 to get 6728

75.9
Adding from the highest place: add 978 to 456 to get 1434

75.20
The most one can get by adding is one extra place: add 9 to 9 to get 18

76.15
Chessboard: add the first sixteen squares to get 65535

77.11
Chessboard: add the first eight squares with 4 as the first square to get 1020

78.4
Adding five numbers with a ratio of \leavevmode {\color {red}2\over 3}: add 16, 24, 36, 54, 81 to get 211

79.4
Adding six numbers starting at 10, with a difference of three: add 10, 13, 16, 19, 22, 25 to get 105

79.13
Add 1, 2, 3, …, 10 to get 55

79.18
Add the squares of 1, 2, 3, …, 10 to get 385

80.1
Add the cubes of 1, 2, 3, …, 10 to get 3025

80.5
Add 1, 3, 5, 7, 9 to get 25

80.10
Add the squares of 1, 3, 5, 7, 9 to get 165

80.15
Add the cubes of 1, 3, 5, 7, 9 to get 1225

80.20
Add 2, 4, 6, 8, 10 to get 30

81.4
Add the squares of 2, 4, 6, 8, 10 to get 220

81.9
Add the squares of 2, 4, 6, 8, 10, 12 to get 364

81.15
Add the cubes of 2, 4, 6, 8, 10 to get 1800

83.16
Subtracting from the highest place: subtract 4968 from 5035 to get 67

84.13
Subtracting from the units place: subtract 3469 from 6543 to get 3074

85.8
The most one can get by subtracting is one fewer place: subtract 1 from 10 to get 9

86.9 L
\leavevmode {\color {red}2~\scalebox {-1}[1]{\ell }~5~\scalebox {-1}[1]{\ell }~7~\scalebox {-1}[1]{\ell }~8~\scalebox {-1}[1]{\ell }~10} is 6

87.19
Casting out nines from 6435 gives nothing

88.5
Casting out eights from 5393 gives 1

89.4
Casting out sevens from 23786435 gives 1

90.7
Casting out sevens from 58064 gives 6

91.4
Add 43 to 64 to get 107 Cast out sevens to check: add 1 to 1 to get 2

91.16
Subtract 74 from 96 to get 22 Cast out sevens to check: subtract 4 from 5 to get 1

92.10
Multiply 12 by 16 to get 192 Cast out sevens to check: multiply 5 by 2 to get 3

92.17 L
Multiply \leavevmode {\color {red}{1\over 3}} by \leavevmode {\color {red}{1\over 4}14} to get \leavevmode {\color {red}{3\over 4}4} Cast out sevens to check: multiply \leavevmode {\color {red}1\over 3} by \leavevmode {\color {red}1\over 4} to get \leavevmode {\color {red}1~0\over 3~4}

93.10
Divide 1488 by 12 to get 124 Cast out sevens to check: multiply 5 by 5 to get 4

93.15 L
Divide \leavevmode {\color {red}{3\over 4}~{5\over 6}} by \leavevmode {\color {red}1\over 2} to get \leavevmode {\color {red}{1\over 6}3}. Cast out sevens to check: multiply \leavevmode {\color {red}5\over 6} by \leavevmode {\color {red}1\over 2}, then convert to 4ths of 6ths to get \leavevmode {\color {red}3~0\over 4~6}

94.1
Denominate 11 with 15 to get \leavevmode {\color {red}2~\,3\over 3~\,5}. Cast out sevens to check: multiply 4 by 1 to get 4

94.5 L
Denominate \leavevmode {\color {red}{2~\,2\over 3~\,6}} with \leavevmode {\color {red}1~\,5\over 3~\,8} to get \leavevmode {\color {red}2\over 3}. Cast out sevens to check: multiply 2 by 2, adjust for the denominators to get 3

95.10 L
Meanings of multiplication: 3 men, each has 5 dirhams; 5 dirhams, how many thirds?

96.3
Multiplication by shifting: multiply 43 by 54 to get 2322

97.4
Vertical multiplication: multiply 42 by 37 to get 1554

98.15
Multiplication by half-shifting: multiply 463 by itself to get 214369

100.5
Lattice multiplication: multiply 435 by 287 to get 124845

102.1
Vertical multiplication (no shifting): multiply 183 by 347 to get 63501

104.10
Sleeper multiplication (no shifting): multiply 253 by 987 to get 249711

107.6
Multiply 444 by 333 to get 147852

108.13
Multiplication by excess: multiply 12 by 15 to get 180

109.1
Multiply 13 by 17 to get 221

109.10
Multiplication by denomination: multiply 6 by 12 to get 72

110.6
Another method of multiplication by denomination: multiply 24 by 8 to get 192

110.12
Multiply 12 by 15 to get 180

111.1
Multiply 3 by 15 to get 45

111.15
Multiplication by nines: multiply 444 by 999 to get 443556

112.10
Another method of multiplication by nines: multiply 999 by 9354 to get 9344646

113.1
Multiplication by squaring: multiply 17 by 19 to get 323

113.9
Another squaring method: multiply 25 by 15 to get 375

113.19
Another squaring method: multiply 36 by 14 to get 504

114.8
Multiplication with zeros: multiply 30 by 140 to get 4200

117.16 L, 118.1 L
Meanings of division: Divide 15 dirhams among 3 men; Divide a piece of wood of 15 spans by a piece of wood of 3 spans

119.1
Divide 245 by 12 to get \leavevmode {\color {red}{1~\,2\over 2~\,6}\,20}

120.5
Divide 44 among 11 men to get 4

120.10
Divide 96 by 12 to get 8

120.16
Divide 35 by 15 to get \leavevmode {\color {red}{1\over 3}2}

121.4
Apportionment. Wealth of donors: 4, 5, 6 dinars, amount of loan: 10 dinars

122.5
Apportionment. Wealth of donors: \leavevmode {\color {red}{1\over 3}4}, \leavevmode {\color {red}{1\over 4}5}, \leavevmode {\color {red}{1\over 6}6} dinars, amount of loan: 12 dinars

123.22
Common denomination: denominate 11 with 15 to get \leavevmode {\color {red}{2~\,3\over 3~\,5}}

124.12,14,17
Other denominations: denominate 4 with 12 to get \leavevmode {\color {red}1\over 3}; denominate 9 with 15 to get \leavevmode {\color {red}3\over 5}; denominate 10 with 16 to get \leavevmode {\color {red}5\over 8}

124.20ff
Finding divisors: 50, 36, 66, 42, 64, 68, 14, 26, 81, 39, 123, 77, 221

129.6
Restore 8 to 19; reduce 50 to 6

129.8
Restore 3 to 6: divide 6 by 3 to get 2

129.12
Reduce 8 to 3. Denominate 3 with 8 to get \leavevmode {\color {red}3\over 8}

Part 1, Chapter 2. On fractions

Passage
Example

134.2 L
Language of parts: \leavevmode {\color {red}1\over 11}, \leavevmode {\color {red}1\over 17}

134.8
Numerator denominator: \leavevmode {\color {red}1\over 4}, \leavevmode {\color {red}2\over 4}, \leavevmode {\color {red}3\over 4}; and \leavevmode {\color {red}1\over 7} through \leavevmode {\color {red}6\over 7}

135.1
Fractions with two or more names: \leavevmode {\color {red}1~\,2\over 7~\,8}; \leavevmode {\color {red}1~\,2~\,4~\,8\over 3~\,6~\,7~\,9}

135.10 L
A related fraction: \leavevmode {\color {red}2~\,4~\,5\over 3~\,5~\,6}

136.8 L
A distinct fraction: \leavevmode {\color {red}{4\over 5}\,{5\over 6}}

137.1 L
A portioned fraction: \leavevmode {\color {red}5~\bullet ~3\over 6 ~\bullet ~4}

137.13
The numerator of \leavevmode {\color {red}1\over 7} is 1

138.5
The numerator of \leavevmode {\color {red}2~\,3~\,4~\,5\over 3~\,5~\,7~\,8} is 596

139.2
The numerator of \leavevmode {\color {red}{4\over 6}~{1~\,5\over 2~\,7}} is 122

139.11
The numerator of \leavevmode {\color {red}3~\bullet ~5~\bullet ~7\over 10\,\:\bullet \,\:6\,\:\bullet \,\:9\,} is 105

140.8
The numerator of \leavevmode {\color {red}{1\over 9}~\scalebox {-1}[1]{\ell }~{6\over 8}} is 46

141.1
The numerator of \leavevmode {\color {red}{1\over 3}~\scalebox {-1}[1]{\ell }~{1~\,6\over 2~\,7}} is 26

141.11
The numerator of the connected fraction \leavevmode {\color {red}{1\over 5}~\scalebox {-1}[1]{\ell }~{1\over 7}~\scalebox {-1}[1]{\ell }~{1\over 4}~\scalebox {-1}[1]{\ell }~{1\over 3}\,5}, also written \leavevmode {\color {red}{1\over 5}\,{1\over 7}\,{1\over 4}\,\scalebox {-1}[1]{\ell }\,{1\over 3}\,5}, is 912

142.6
The numerator of the distinct fractions \leavevmode {\color {red}{1\over 5}~\scalebox {-1}[1]{\ell }~{1\over 7}~\scalebox {-1}[1]{\ell }~{1\over 4}~\scalebox {-1}[1]{\ell }~{1\over 3}\,5} is 1991

143.5
The numerator of \leavevmode {\color {red}{3~5\over 4~6}\,5} is 143

143.13
The numerator of \leavevmode {\color {red}10\,{6\over 8}\,{4\over 7}} is 740

144.10
The numerator of \leavevmode {\color {red}{3\over 6}\,5\,{4\over 9}} is 147

145.4
The numerator of \leavevmode {\color {red}{4\over 7}\,7\,{2\over 3}} is 106

147.4
Add \leavevmode {\color {red}{6\over 8}\,{4\over 5}\,3} to \leavevmode {\color {red}{1~3~\,\,4\,\over 2~8~10}} to get \leavevmode {\color {red}{1~7~\,\,9\,\over 2~8~10}4}

147.15
Subtract \leavevmode {\color {red}{1\over 3}~\scalebox {-1}[1]{\ell }~2\,{7\over 10}} from \leavevmode {\color {red}{5\,\bullet \, 3\over 6\,\bullet \, 4}\,4} to get \leavevmode {\color {red}{1~3~\,\,5\,\over 2~6~10}3}

149.8
Multiply \leavevmode {\color {red}{1\over 3}\,{3\over 4}} by \leavevmode {\color {red}{1~\,4~\,3\over 5~\,6~\,9}} to get \leavevmode {\color {red}{1~\,0~\,0~\,4\over 4~\,5~\,6~\,9}}

150.2
Multiply \leavevmode {\color {red}{1\over 8}\,4\,{1\over 3}} by \leavevmode {\color {red}10\,{2\,\bullet \, 1\over 3\,\bullet \, 5}} to get \leavevmode {\color {red}{5\over 6}1}

151.4
Divide \leavevmode {\color {red}{1\over 3}\,6} by \leavevmode {\color {red}3\,{7\,\bullet \, 4\over 8\,\bullet \, 5}} to get \leavevmode {\color {red}{1~\,0\over 7~\,9}\,3}

151.14
Denominate \leavevmode {\color {red}{2\over 9}~\scalebox {-1}[1]{\ell }~{1\over 4}\,3} with \leavevmode {\color {red}{3\over 5}\,{2\over 8}\,6} to get \leavevmode {\color {red}{5~\,\,50\,\over 9~137}}

152.11
Divide \leavevmode {\color {red}{2~\,\,9\,\over 3~\,10}\,8} by \leavevmode {\color {red}{1~\,\,5\,\over 3~\,10}} to get \leavevmode {\color {red}{1~\,6\over 2~\,8}\,16}

153.3
Denominate \leavevmode {\color {red}{1\over 3}\,2} with \leavevmode {\color {red}{2\over 3}\,6} to get \leavevmode {\color {red}{\,1~~3~\over 2~\,10}}

153.12 L
Divide 5 by \leavevmode {\color {red}5\over 6} to get 6; denominate \leavevmode {\color {red}5\over 6} with 5 to get \leavevmode {\color {red}1\over 6}

154.6ff
Restore \leavevmode {\color {red}1\over 2} to \leavevmode {\color {red}9\over 10}; \leavevmode {\color {red}1~\,2\over 2~\,7} to \leavevmode {\color {red}{1\over 2}5}; \leavevmode {\color {red}5~\bullet ~2\over 7~\bullet ~3} to 10; 5 to \leavevmode {\color {red}{4\over 6}10}; \leavevmode {\color {red}{1~\,3\,\over 2~10}4} to 8; \leavevmode {\color {red}{1\over 3}~\scalebox {-1}[1]{\ell }~{3\over 5}3} to \leavevmode {\color {red}{3\over 5}12}

155.11ff
Reduce \leavevmode {\color {red}7\over 10} to \leavevmode {\color {red}1\over 3}; 8 to \leavevmode {\color {red}{1\over 2}2}; 10 to \leavevmode {\color {red}3\over 4}; \leavevmode {\color {red}{1\over 4}7} to \leavevmode {\color {red}{4\over 6}3}; \leavevmode {\color {red}{4~\,9\,\over 7~10}11} to 5; \leavevmode {\color {red}{1\over 3}2} to \leavevmode {\color {red}1\over 9}

157.2 L
Convert \leavevmode {\color {red}{3\over 4}~{5\over 6}} to tenths. Answer: \leavevmode {\color {red}{5~\,5\,\over 6~10}1}

157.12 L
How many tenths are in \leavevmode {\color {red}{3\over 4}~{5\over 6}}? Again, it is \leavevmode {\color {red}{5~\,5\,\over 6~10}1}

158.1 L
How many tenths are in 5? Answer: 50

158.11
How many ninths are in \leavevmode {\color {red}{4\over 10}~{6\over 8}}? Answer: \leavevmode {\color {red}{4~\,\,\,3\,~\,1\over 8~\,10~\,9}\,1}

Part 1, Chapter 3. On roots

Passage
Example

163.4 L
Examples: \leavevmode {\color {red}\sqrt{10}}, \leavevmode {\color {red}\sqrt{1\over 2}}, \leavevmode {\color {red}\sqrt{{1\over 2}10}}, \leavevmode {\color {red}\sqrt{\sqrt{10}}}

166.13
\leavevmode {\color {red}\sqrt{625}} is 25

167.8
\leavevmode {\color {red}\sqrt{20}} is approximately \leavevmode {\color {red}{1\over 2}4}

167.14
\leavevmode {\color {red}\sqrt{54}} is approximately \leavevmode {\color {red}{1~\,2\over 2~\,7}7}

168.1
\leavevmode {\color {red}\sqrt{92}} is approximately \leavevmode {\color {red}{3\over 5}9}

168.16
\leavevmode {\color {red}\sqrt{92}} is approximately \leavevmode {\color {red}{1~\,5~\,\,\,5\,\over 2~\,6~\,10}\,9}

169.4
\leavevmode {\color {red}\sqrt{12}} is approximately \leavevmode {\color {red}{1~\,3\over 4~\,7}3}

169.10,17 L
\leavevmode {\color {red}\sqrt{625}} is 25; \leavevmode {\color {red}\sqrt{729}} is 27

170.6
\leavevmode {\color {red}\sqrt{100}} is 10

170.16
\leavevmode {\color {red}\sqrt{1~\,4\over 6~\,6}} is \leavevmode {\color {red}{5\over 6}}

171.1
\leavevmode {\color {red}\sqrt{{1\over 4}12}} is \leavevmode {\color {red}{1\over 2}3}

171.6
\leavevmode {\color {red}\sqrt{3~\,4\over 6~\,9}} is \leavevmode {\color {red}{1~\,5~\,3~\,13\over 2~\,6~\,9~\,19}}

171.15
\leavevmode {\color {red}\sqrt{{1~\,7\over 2~\,8}10}} is approximately \leavevmode {\color {red}{1~\,2~\,16\over 4~\,8~\,53}\,3}

172.6
\leavevmode {\color {red}\sqrt{{1~\,7\over 2~\,8}10}} is approximately \leavevmode {\color {red}{4\over 13}3}

172.12
\leavevmode {\color {red}\sqrt{1~\,4\over 2~\,7}} is approximately \leavevmode {\color {red}{3~\,5~\,\,\,8\,\over 4\,~7~\,11}}

173.13 L
Binomials: \leavevmode {\color {red}\sqrt{3}~5}; \leavevmode {\color {red}\sqrt{3}~\sqrt{5}}

173.16 L
Apotomes: \leavevmode {\color {red}\sqrt{3}~\scalebox {-1}[1]{\ell }~5}; \leavevmode {\color {red}\sqrt{3}~\scalebox {-1}[1]{\ell }~\sqrt{5}}

174.5
1st & 4th binomials: \leavevmode {\color {red}\sqrt{21}~5}, \leavevmode {\color {red}\sqrt{2}~2}

174.8
2nd & 5th binomials: \leavevmode {\color {red}\sqrt{45}~5}, \leavevmode {\color {red}\sqrt{72}~5}

174.11
3rd & 6th binomials: \leavevmode {\color {red}\sqrt{18}~\sqrt{10}}, \leavevmode {\color {red}\sqrt{8}~\sqrt{7}}

175.11
\leavevmode {\color {red}\sqrt{\sqrt{60}~8}} is \leavevmode {\color {red}\sqrt{3}~\sqrt{5}}

175.19
\leavevmode {\color {red}\sqrt{\sqrt{60}~\scalebox {-1}[1]{\ell }~8}} is \leavevmode {\color {red}\sqrt{3}~\scalebox {-1}[1]{\ell }~\sqrt{5}}

176.6
\leavevmode {\color {red}\sqrt{\sqrt{55}~8}} is \leavevmode {\color {red}\sqrt{{1\over 2}2}~\sqrt{{1\over 2}5}}; \leavevmode {\color {red}\sqrt{\sqrt{55}~\scalebox {-1}[1]{\ell }~8}} is \leavevmode {\color {red}\sqrt{{1\over 2}2}~\scalebox {-1}[1]{\ell }~\sqrt{{1\over 2}5}}

176.10
\leavevmode {\color {red}\sqrt{\sqrt{112}~7}} is \leavevmode {\color {red}\sqrt{\sqrt{{3\over 4}1}}~\sqrt{\sqrt{{3\over 4}85}}}; \leavevmode {\color {red}\sqrt{\sqrt{112}~\scalebox {-1}[1]{\ell }~7}} is \leavevmode {\color {red}\sqrt{\sqrt{{3\over 4}1}}~\scalebox {-1}[1]{\ell }~\sqrt{\sqrt{{3\over 4}85}}}

176.15
\leavevmode {\color {red}\sqrt{\sqrt{14}~\sqrt{32}}} is \leavevmode {\color {red}\sqrt{\sqrt{1\over 2}}~\sqrt{\sqrt{{1\over 2}24}}}; \leavevmode {\color {red}\sqrt{\sqrt{14}~\scalebox {-1}[1]{\ell }~\sqrt{32}}} is \leavevmode {\color {red}\sqrt{\sqrt{1\over 2}}~\scalebox {-1}[1]{\ell }~\sqrt{\sqrt{{1\over 2}24}}}

176.20
\leavevmode {\color {red}\sqrt{\sqrt{30}~7}} is \leavevmode {\color {red}\sqrt{\sqrt{{3\over 4}4}~\scalebox {-1}[1]{\ell }~{1\over 2}3}~\sqrt{\sqrt{{3\over 4}4}~{1\over 2}3}}; \leavevmode {\color {red}\sqrt{\sqrt{30}~\scalebox {-1}[1]{\ell }~7}} is \leavevmode {\color {red}\sqrt{\sqrt{{3\over 4}4}~\scalebox {-1}[1]{\ell }~{1\over 2}3}~\scalebox {-1}[1]{\ell }~\sqrt{\sqrt{{3\over 4}4}~{1\over 2}3}}

177.5
\leavevmode {\color {red}\sqrt{\sqrt{20}~3}} is \leavevmode {\color {red}\sqrt{\sqrt{{1\over 4}~{1\over 2}2}~\scalebox {-1}[1]{\ell }~\sqrt{5}}~\sqrt{\sqrt{{1\over 4}~{1\over 2}2}~\sqrt{5}}}; \leavevmode {\color {red}\sqrt{3~\scalebox {-1}[1]{\ell }~\sqrt{20}}} is \leavevmode {\color {red}\sqrt{\sqrt{{1\over 4}~{1\over 2}2}~\scalebox {-1}[1]{\ell }~\sqrt{5}}~\scalebox {-1}[1]{\ell }~\sqrt{\sqrt{{1\over 4}~{1\over 2}2}~\sqrt{5}}}

177.11
\leavevmode {\color {red}\sqrt{\sqrt{11}~\sqrt{10}}} is \leavevmode {\color {red}\sqrt{{1\over 2}~\scalebox {-1}[1]{\ell }~\sqrt{{3\over 4}2}}~\sqrt{\sqrt{{3\over 4}2}~{1\over 2}}}; \leavevmode {\color {red}\sqrt{\sqrt{10}~\scalebox {-1}[1]{\ell }~\sqrt{11}}} is \leavevmode {\color {red}\sqrt{{1\over 2}~\scalebox {-1}[1]{\ell }~\sqrt{{3\over 4}2}}~\scalebox {-1}[1]{\ell }~\sqrt{\sqrt{{3\over 4}2}~{1\over 2}}}

179.7,11
Add \leavevmode {\color {red}\sqrt{3}}to \leavevmode {\color {red}\sqrt{27}} to get \leavevmode {\color {red}\sqrt{48}}

179.16
Add \leavevmode {\color {red}\sqrt{2}} to \leavevmode {\color {red}\sqrt{8}} to get \leavevmode {\color {red}\sqrt{18}}

179.20
Add half of \leavevmode {\color {red}\sqrt{20}} to two \leavevmode {\color {red}\sqrt{5}}s to get \leavevmode {\color {red}\sqrt{45}}

180.10
Add \leavevmode {\color {red}\sqrt{3}} to \leavevmode {\color {red}\sqrt{15}} to get \leavevmode {\color {red}\sqrt{15}~\sqrt{3}}

180.15
Add half of \leavevmode {\color {red}\sqrt{\sqrt{80}}} to \leavevmode {\color {red}{1~\bullet ~1\over 4~\bullet ~3}} of \leavevmode {\color {red}\sqrt{684}} to get \leavevmode {\color {red}\sqrt{\sqrt{{1~4\over 2~\,8}\,22}}~\sqrt{\sqrt{5}}}

181.6
Subtract \leavevmode {\color {red}\sqrt{8}} from \leavevmode {\color {red}\sqrt{32}} to get \leavevmode {\color {red}\sqrt{8}}

181.12,16
Subtract \leavevmode {\color {red}\sqrt{12}} from \leavevmode {\color {red}\sqrt{27}} to get \leavevmode {\color {red}\sqrt{3}}

182.4
Subtract \leavevmode {\color {red}\sqrt{8}} from \leavevmode {\color {red}\sqrt{10}} to get \leavevmode {\color {red}\sqrt{8}~\scalebox {-1}[1]{\ell }~\sqrt{10}}

183.4
Multiply \leavevmode {\color {red}\sqrt{8}} by \leavevmode {\color {red}\sqrt{9}} to get \leavevmode {\color {red}\sqrt{72}}

183.7
Multiply \leavevmode {\color {red}\sqrt{\sqrt{5}}} by \leavevmode {\color {red}\sqrt{\sqrt{7}}} to get \leavevmode {\color {red}\sqrt{\sqrt{35}}}

183.11
Multiply \leavevmode {\color {red}\sqrt{\sqrt{\sqrt{3}}}} by \leavevmode {\color {red}\sqrt{\sqrt{\sqrt{8}}}} to get \leavevmode {\color {red}\sqrt{\sqrt{\sqrt{24}}}}

183.15
Multiply 3 by \leavevmode {\color {red}2~\scalebox {-1}[1]{\ell }~\sqrt{7}} to get \leavevmode {\color {red}6~\scalebox {-1}[1]{\ell }~\sqrt{63}}

184.1
Multiply 3 by \leavevmode {\color {red}\sqrt{7}} to get \leavevmode {\color {red}\sqrt{63}}

184.4
Multiply 2 by \leavevmode {\color {red}\sqrt{\sqrt{3}}} to get \leavevmode {\color {red}\sqrt{\sqrt{48}}}

184.11
Multiply 2 by two \leavevmode {\color {red}\sqrt{7}}s to get \leavevmode {\color {red}\sqrt{112}}

184.18
Multiply 5 by three \leavevmode {\color {red}\sqrt{\sqrt{2}}}s to get \leavevmode {\color {red}\sqrt{\sqrt{101250}}}

185.6
Multiply \leavevmode {\color {red}2\over 3} by half of \leavevmode {\color {red}\sqrt{20}} to get \leavevmode {\color {red}\sqrt{{2\over 9}2}}

185.12
Multiply \leavevmode {\color {red}\sqrt{5}} by half of \leavevmode {\color {red}\sqrt{\sqrt{40}}} to get \leavevmode {\color {red}\sqrt{{1\over 2}62}}

186.1
Duplicate \leavevmode {\color {red}\sqrt{3}} twice to get \leavevmode {\color {red}\sqrt{12}}

186.4
Duplicate \leavevmode {\color {red}\sqrt{7}} five times to get \leavevmode {\color {red}\sqrt{175}}

186.8
Half of \leavevmode {\color {red}\sqrt{10}} is \leavevmode {\color {red}\sqrt{{1\over 2}2}}

186.11
\leavevmode {\color {red}4~\bullet ~1\over 8~\bullet ~3} of \leavevmode {\color {red}\sqrt{\sqrt{60}}} is \leavevmode {\color {red}\sqrt{\sqrt{1~2~0\over 2~6~9}}}

187.4
Divide \leavevmode {\color {red}\sqrt{20}} by \leavevmode {\color {red}\sqrt{3}} to get \leavevmode {\color {red}\sqrt{{2\over 3}6}}

187.7
Denominate \leavevmode {\color {red}\sqrt{3}} with \leavevmode {\color {red}\sqrt{8}} to get \leavevmode {\color {red}\sqrt{3\over 8}}

187.10
Divide \leavevmode {\color {red}\sqrt{\sqrt{6}}} by \leavevmode {\color {red}\sqrt{\sqrt{2}}} to get \leavevmode {\color {red}\sqrt{\sqrt{3}}}

187.14
Denominate \leavevmode {\color {red}\sqrt{\sqrt{18}}} with \leavevmode {\color {red}\sqrt{\sqrt{32}}} to get \leavevmode {\color {red}\sqrt{\sqrt{1~4\over 2~8}}}

188.6
Divide \leavevmode {\color {red}\sqrt{\sqrt{14}}} by \leavevmode {\color {red}\sqrt{2}} to get \leavevmode {\color {red}\sqrt{\sqrt{{1\over 2}3}}}

188.11
Divide two \leavevmode {\color {red}\sqrt{15}}s by 2 to get \leavevmode {\color {red}\sqrt{15}}

188.15
Divide half of \leavevmode {\color {red}\sqrt{24}} by \leavevmode {\color {red}\sqrt{2}} to get \leavevmode {\color {red}\sqrt{3}}

189.1
Divide 12 by \leavevmode {\color {red}\sqrt{3}~5} to get \leavevmode {\color {red}\sqrt{9~~9\over 11~\,11}~\scalebox {-1}[1]{\ell }~{8\over 11}2}

189.11
Divide 10 by \leavevmode {\color {red}\sqrt{7}~\scalebox {-1}[1]{\ell }~ 3} to get \leavevmode {\color {red}\sqrt{175}~15}

Part 2. Finding unknown numbers

Chapter 1. Solving problems by proportion

Passage | Example
195.16 | Example of four proportional numbers: 3:6::4:8
198.7 L | Too late
199.1 | Too late
200.1 | Too late
201.1 | Too late
202.3 | Too late
203.1 | Too late
203.11 L | Too late
205.5 L | Too late
207.6 L | Too late

Part 2, Chapter 2. Solving problems by algebra

Passage
Example

211.15
Simple equations: \leavevmode {\color {red}{t\atop 7}={m\atop 3}}; \leavevmode {\color {red}\hbox{20}={m\atop 5}}; \leavevmode {\color {red}\hbox{12}={t\atop 3}} ( 3x^2=7x; 5x^2=20; 3x=12)

212.6
Composite equations: \leavevmode {\color {red}\hbox{24}={t\atop 10}~{m\atop 1}}; \leavevmode {\color {red}{t\atop 5}=\hbox{4}~{m\atop 1}}; \leavevmode {\color {red}\hbox{5}~{t\atop 4}={m\atop 1}} ( x^2+10x=24; x^2+4=5x; x^2=4x+5)

213.7
Solve \leavevmode {\color {red}{t\atop 15}={m\atop 3}} to get \leavevmode {\color {red}t\atop 1} is 5 and \leavevmode {\color {red}m\atop 1} is 25 ( 3x^2=15x\Rightarrow x=5, x^2=25)

213.13
Solve \leavevmode {\color {red}\hbox{18}={m\atop 2}} to get \leavevmode {\color {red}m\atop 1} is 9 and \leavevmode {\color {red}t\atop 1} is 3 ( 2x^2=18\Rightarrow x^2=9, x=3)

214.1
Solve \leavevmode {\color {red}\hbox{20}={t\atop 5}} to get \leavevmode {\color {red}t\atop 1} is 4 and \leavevmode {\color {red}m\atop 1} is 16 ( 5x=20\Rightarrow x=4, x^2=16)

214.9
Solve \leavevmode {\color {red}\hbox{15}={t\atop 2}~{m\atop 1}} to get \leavevmode {\color {red}t\atop 1} is 3 and \leavevmode {\color {red}m\atop 1} is 9 ( x^2+2x=15\Rightarrow x=3, x^2=9)

215.6
Solve \leavevmode {\color {red}{t\atop 6}=\hbox{8}~{m\atop 1}} to get \leavevmode {\color {red}t\atop 1} is 4 and \leavevmode {\color {red}m\atop 1} is 16, or \leavevmode {\color {red}t\atop 1} is 2 and \leavevmode {\color {red}m\atop 1} is 4 ( x^2+8=6x\Rightarrow x=4, x^2=16 or x=2, x^2=4)

215.14
Solve \leavevmode {\color {red}{t\atop 6}=\hbox{9}~{m\atop 1}} to get \leavevmode {\color {red}m\atop 1} is 9 and \leavevmode {\color {red}t\atop 1} is 3( x^2+9=6x\Rightarrow x^2=9, x=3)

216.13
Solve \leavevmode {\color {red}\hbox{3}~{t\atop 2}={m\atop 1}} to get \leavevmode {\color {red}t\atop 1} is 3 and \leavevmode {\color {red}m\atop 1} is 9 ( x^2=2x+3\Rightarrow x=3, x^2=9)

217.10
\leavevmode {\color {red}\hbox{36}={t\atop 6}~{m\atop 2}} simplifies to \leavevmode {\color {red}\hbox{18}={t\atop 3}~{m\atop 1}} ( 2x^2+6x=36\Rightarrow x^2+3x=18)

218.1
\leavevmode {\color {red}\hbox{6}={t\atop 2}~{m\atop {1\over 2}}} simplifies to \leavevmode {\color {red}\hbox{12}={t\atop 4}~{m\atop 1}} ( {1\over 2}x^2+2x=6\Rightarrow x^2+4x=12)

219.2
Add \leavevmode {\color {red}m\atop 1}, \leavevmode {\color {red}t\atop 6}, and \leavevmode {\color {red}\hbox{10}} to get \leavevmode {\color {red}\hbox{10}~{t\atop 6}~{m\atop 1}} ( 6x+x^2+10\Rightarrow x^2+6x+10)

219.5 L
Add \leavevmode {\color {red}{t\atop 1}~\scalebox {-1}[1]{\ell }~{m\atop 1}} to \leavevmode {\color {red}\hbox{10}} to get \leavevmode {\color {red}{t\atop 1}~\scalebox {-1}[1]{\ell }~\hbox{10}~{m\atop 1}} ( (x^2-x)+10\Rightarrow x^2+10-x)

219.7 L
Add \leavevmode {\color {red}{m\atop 1}~\scalebox {-1}[1]{\ell }~{m\atop 2}} to \leavevmode {\color {red}\hbox{10}} to get \leavevmode {\color {red}\hbox{10}~{m\atop 1}} ( (2x^2-x^2)+10\Rightarrow x^2+10)

219.10 L
Add \leavevmode {\color {red}{t\atop 2}~\scalebox {-1}[1]{\ell }~{m\atop 1}} to \leavevmode {\color {red}t\atop 10} to get \leavevmode {\color {red}{t\atop 8}~{m\atop 1}} ( (x^2-2x)+10x\Rightarrow x^2+8x)

219.12 L
Add \leavevmode {\color {red}{t\atop 5}~\scalebox {-1}[1]{\ell }~{m\atop 1}} to \leavevmode {\color {red}t\atop 10} to get \leavevmode {\color {red}{t\atop 5}~{m\atop 1}} ( (x^2-5x)+10x\Rightarrow x^2+5x)

220.1
Subtract \leavevmode {\color {red}t\atop 1} from \leavevmode {\color {red}m\atop 1} to get \leavevmode {\color {red}{t\atop 1}~\scalebox {-1}[1]{\ell }~{m\atop 1}} ( x^2-x\Rightarrow x^2-x)

220.3
Subtract \leavevmode {\color {red}\hbox{10}} from \leavevmode {\color {red}{t\atop 3}~{m\atop 2}} to get \leavevmode {\color {red}\hbox{10}~\scalebox {-1}[1]{\ell }~{t\atop 3}~{m\atop 2}} ( (2x^2+3x)-10\Rightarrow 2x^2+3x-10)

220.9
Subtract \leavevmode {\color {red}{t\atop 1}~\hbox{2}} from \leavevmode {\color {red}{t\atop 3}~\scalebox {-1}[1]{\ell }~{m\atop 1}} to get \leavevmode {\color {red}\hbox{2}~\scalebox {-1}[1]{\ell }~{t\atop 4}~\scalebox {-1}[1]{\ell }~{m\atop 1}} ( (x^2-3x)-(2+x)\Rightarrow x^2-4x-2)

221.1
Subtract \leavevmode {\color {red}{t\atop 5}~\scalebox {-1}[1]{\ell }~\hbox{52}} from \leavevmode {\color {red}\hbox{30}~{c\atop 2}} to get \leavevmode {\color {red}{t\atop 22}~\scalebox {-1}[1]{\ell }~{t\atop 5}~{c\atop 2}} ( (2x^3+30)-(52-5x)\Rightarrow 2x^3+5x-22)

221.13
Subtract \leavevmode {\color {red}{t\atop 4}~\scalebox {-1}[1]{\ell }~\hbox{12}} from \leavevmode {\color {red}{t\atop 2}~\scalebox {-1}[1]{\ell }~{m\atop 3}} to get \leavevmode {\color {red}\hbox{12}~\scalebox {-1}[1]{\ell }~{t\atop 2}~{m\atop 3}} ( (3x^2-2x)-(12-4x)\Rightarrow 3x^2+2x-12)

222.1
Subtract \leavevmode {\color {red}{m\atop 2}~\scalebox {-1}[1]{\ell }~{c\atop 1}} from \leavevmode {\color {red}{t\atop 4}~\scalebox {-1}[1]{\ell }~\hbox{30}} to get \leavevmode {\color {red}{t\atop 4}~\scalebox {-1}[1]{\ell }~{c\atop 1}~\scalebox {-1}[1]{\ell }~{m\atop 2}~\hbox{30}} ( (30-4x)-(x^3-2x^2)\Rightarrow 30+2x^2-x^3-4x)

223.2
\leavevmode {\color {red}{t\atop 1}~\hbox{2}={t\atop 3}~\scalebox {-1}[1]{\ell }~{m\atop 1}} simplifies to \leavevmode {\color {red}{t\atop 4}~\hbox{2}={m\atop 1}} ( x^2-3x=2+x\Rightarrow x^2=2+4x)

223.7
\leavevmode {\color {red}{t\atop 5}~\scalebox {-1}[1]{\ell }~\hbox{24}={t\atop 3}~\scalebox {-1}[1]{\ell }~{m\atop 1}} simplifies to \leavevmode {\color {red}\hbox{24}={t\atop 2}~{m\atop 1}} ( x^2-3x=24-5x\Rightarrow x^2+2x=24)

223.14
\leavevmode {\color {red}{t\atop {1\over 2}2}~\scalebox {-1}[1]{\ell }~{m\atop 1}=\hbox{10}~\scalebox {-1}[1]{\ell }~{m\atop 1}} simplifies to \leavevmode {\color {red}{t\atop {1\over 2}2}=\hbox{10}} ( x^2-10=x^2-2{1\over 2}x\Rightarrow 10=2{1\over 2}x)

223.17
\leavevmode {\color {red}{t\atop 4}~\scalebox {-1}[1]{\ell }~\hbox{51}=\hbox{10}~{m\atop 1}} simplifies to \leavevmode {\color {red}\hbox{41}={t\atop 4}~{m\atop 1}} ( x^2+10=51-4x\Rightarrow x^2+4x=41)

224.1
\leavevmode {\color {red}{t\atop 1}~\scalebox {-1}[1]{\ell }~{m\atop 2}~\hbox{10}={t\atop 5}~{m\atop 1}} simplifies to \leavevmode {\color {red}\hbox{10}~{m\atop 1}={t\atop 6}} ( x^2+5x=10+2x^2-x\Rightarrow 6x=x^2+10)

224.4
\leavevmode {\color {red}{t\atop 7}~\scalebox {-1}[1]{\ell }~\hbox{7}~{m\atop 3}={t\atop 2}~\scalebox {-1}[1]{\ell }~\hbox{10}~{m\atop 2}} simplifies to \leavevmode {\color {red}{t\atop 5}~\hbox{3}={m\atop 1}} ( 2x^2+10-2x=3x^2+7-7x\Rightarrow x^2=3+5x)

225.8
The power of the māl māl is 4; of the māl cube is 5; of the māl māl māl is 6; of the māl cube māl cube is 10

225.13
The power of the māl cube māl māl is 9; of the cube māl cube cube māl māl is 15

226.1
A term for 4 is a māl māl; for 7 is a cube māl māl; for 6 is a māl māl māl or a cube cube

226.3
A term for 8 is a māl māl māl māl or a cube māl cube, or a cube cube māl, or a māl cube cube, etc.

226.7
A term for 9 is a cube cube cube or a cube māl māl māl, etc.

226.12
Multiply \leavevmode {\color {red}t\atop 5} by \leavevmode {\color {red}t\atop 7} to get \leavevmode {\color {red}m\atop 35} ( 5x\times 7x\Rightarrow 35x)

226.17
Multiply \leavevmode {\color {red}t\atop 10} by \leavevmode {\color {red}m\atop 6} to get \leavevmode {\color {red}c\atop 60} ( 10x\times 6x^2\Rightarrow 60x^3)

227.4
Multiply \leavevmode {\color {red}t\atop 1} by \leavevmode {\color {red}c\atop 1} to get \leavevmode {\color {red}mm\atop 1} ( x\times x^3\Rightarrow x^4)

227.7
Multiply 6 by \leavevmode {\color {red}m\atop 4} to get \leavevmode {\color {red}m\atop 24} ( 6\times 4x^2\Rightarrow 24x^2)

227.10
Multiply 7 by \leavevmode {\color {red}mc\atop 3} to get \leavevmode {\color {red}mc\atop 21} ( 7\times 3x^5\Rightarrow 21x^5)

227.17
\leavevmode {\color {red}{m\atop 10}~{c\atop 4}={mm\atop 3}} simplifies to \leavevmode {\color {red}\hbox{10}~{t\atop 4}={m\atop 3}} ( 3x^4=4x^3+10x^2\Rightarrow 3x^2=4x+10)

228.1
\leavevmode {\color {red}{t\atop 20}~{m\atop 10}={c\atop 3}} simplifies to \leavevmode {\color {red}\hbox{20}~{t\atop 10}={m\atop 3}} ( 3x^3=10x^2+20x\Rightarrow 3x^2=10x+20)

228.4
\leavevmode {\color {red}{t\atop 39}={m\atop 10}~{c\atop 1}} simplifies to \leavevmode {\color {red}\hbox{39}={t\atop 10}~{m\atop 1}} ( x^3+10x^2=39x\Rightarrow x^2+10x=39)

228.11
Multiply \leavevmode {\color {red}t\atop 5} by \leavevmode {\color {red}{t\atop 4}~\scalebox {-1}[1]{\ell }~\hbox{13}} to get \leavevmode {\color {red}{m\atop 20}~\scalebox {-1}[1]{\ell }~{t\atop 65}} ( 5x\times (13-4x)\Rightarrow 65x-20x^2)

228.15
Multiply \leavevmode {\color {red}{t\atop 2}~\scalebox {-1}[1]{\ell }~\hbox{8}} by \leavevmode {\color {red}{m\atop 4}~\scalebox {-1}[1]{\ell }~\hbox{7}} to get \leavevmode {\color {red}{m\atop 32}~\scalebox {-1}[1]{\ell }~{t\atop 14}~\scalebox {-1}[1]{\ell }~\hbox{56}~{c\atop 8}} ( (8-2x)\times (7-4x^2)\Rightarrow 8x^3+56-14x-32x^2)

229.4
Divide \leavevmode {\color {red}m\atop 10} by \leavevmode {\color {red}t\atop 2} to get \leavevmode {\color {red}t\atop 5} ( 10x^2\div 2x\Rightarrow 5x)

229.8
Divide \leavevmode {\color {red}c\atop 15} by \leavevmode {\color {red}t\atop 3} to get \leavevmode {\color {red}m\atop 5} ( 15x^3\div 3x\Rightarrow 5x^2)

229.13
Divide \leavevmode {\color {red}m\atop 12} by \leavevmode {\color {red}m\atop 3} to get 4 ( 12x^2\div 3x^2\Rightarrow 4)

229.17
Divide \leavevmode {\color {red}t\atop 12} by 4 to get \leavevmode {\color {red}t\atop 3} ( 12x\div 4\Rightarrow 3x)

230.4
Divide \leavevmode {\color {red}{m\atop 3}~\scalebox {-1}[1]{\ell }~{c\atop 12}} by \leavevmode {\color {red}t\atop 2} to get \leavevmode {\color {red}{t\atop {1\over 2}1}~\scalebox {-1}[1]{\ell }~{m\atop 6}} ( (12x^3-3x^2)\div 2x\Rightarrow 6x^2-1{1\over 2}x)

230.8
Divide \leavevmode {\color {red}{t\atop 3}~\scalebox {-1}[1]{\ell }~{m\atop 10}} by 2 to get \leavevmode {\color {red}{t\atop {1\over 2}1}~\scalebox {-1}[1]{\ell }~ {m\atop 5}} ( (10x^2-3x)\div 2\Rightarrow 5x^2-1{1\over 2}x)

230.13
Divide \leavevmode {\color {red}m\atop 6} by \leavevmode {\color {red}c\atop 3} to get \leavevmode {\color {red}2\over {t\atop 1}} ( 6x^2\div 3x^3\Rightarrow {2\over x})

230.17
Divide \leavevmode {\color {red}m\atop 10} by \leavevmode {\color {red}{t\atop 1}~\scalebox {-1}[1]{\ell }~\hbox{3}} to get \leavevmode {\color {red}{m\atop 10}\over {{t\atop 1}~\scalebox {-1}[1]{\ell }~\hbox{3}}} ( 10x^2\div (3-x)\Rightarrow {10x^2\over 3-x})

B. Some sample problems solved in other books

[1] A problem of Ibn al-Bannāʾ’s solved by “the four proportional numbers”.

In some problems, the method of “the four proportional numbers” is applied directly. This example is from the chapter on business transactions (muʿāmalāt) in Ibn al-Bannāʾ’s Essays on Arithmetic. It asks for the price of a certain amount of a commodity given the price of another amount of the same commodity.[1]

If someone said to you, “Five amdāʾ[2] for ten dinars.[3] How much do fourteen mudy cost?”

These are four proportional amounts: the first is ten dinars, the second is five amdāʾ, the third is the unknown dinars, and the fourth is fourteen mudy. The third in this problem is the unknown. You multiply the first by the fourth, and you divide the outcome by the second, giving you the third, which is the cost of fourteen mudy, the unknown, and that is twenty-eight dinars.

[2] A problem of Ibn al-Bannāʾ’s solved by single false position.

This problem, also from Ibn al-Bannāʾ’s Essays on Arithmetic, shows how the method of “the four proportional numbers” is used in single false position. In this method a convenient but probably incorrect value is posited for the solution, here 60, and the calculations are performed on this value. The answer is then found from a proportion.[4]

If someone said to you, “A quantity (māl): you add its half and a third of what remains, and a fourth of what remains, and a fifth of what remains, to get twenty dirhams. How much is the quantity?”

You can take the third and the fourth and the fifth of sixty. Then take its half, giving thirty. And take a third of what remains, giving ten, which leaves twenty. Take its fourth, giving five. There remains fifteen. Take its fifth, giving three. Then gather the thirty with the ten and the five and the three, and you add them all, so the sum is forty-eight.

The ratio of this forty-eight with respect to the sixty is as the ratio of the twenty with respect to the unknown quantity. So you work it out as described above to get the quantity, which is fifteen.

Like al-Hawārī’s problem at 200.1, this enunciation requires some explanation. If we use modern notation and call the unknown A, and we make the remainders B=A-{1\over 2}A, C=B-{1\over 3}B, and D=C-{1\over 4}C, then the problem asks to add together {1\over 2}A+{1\over 3}B+{1\over 4}C+{1\over 5}D. It would not be efficient to solve this problem by algebra, either premodern or modern. If we let the unknown be x, then the equation would be set up as {1\over 2}x+{1\over 3}(x-{1\over 2}x)+{1\over 4}(x-{1\over 2}x-{1\over 3}(x-{1\over 2}x))+{1\over 5}(x-{1\over 2}x-{1\over 3}(x-{1\over 2}x)-{1\over 4}(x-{1\over 2}x-{1\over 3}(x-{1\over 2}x)))=20. Even if you manage to simplify this to {1\over 2}x+{1\over 6}x+{1\over 12}x+{1\over 20}x=20, it is much easier to pick a value like 60 and work through the operations with it.

For another problem solved by single false position, see problem [7] below.

[3] Ibn Qunfudh’s problem (14), solved by algebra.

Problems [3], [4], and [5] are simple examples that show how the rules illustrated by al-Hawārī in the chapter on algebra are applied in the algebraic solutions to problems. For each problem, we give the breakdown according to the three stages. These stages are not indicated in the texts.

Ibn Qunfudh completed his commentary on Ibn al-Bannāʾ’s Condensed Book in 1370. It is titled Lowering the Veil from the Faces of Arithmetical Operations. This book shows the algebraic notation.[5]

[Enunciation] Problem. Ten: you divide it into two parts, and you multiply one of them by the other, resulting in twenty-two and two ninths.

[Stage 1] Make one of the parts a thing and the other ten less a thing, and multiply one of them by the other, resulting in ten things less a māl, and it is equal to twenty-two and two ninths, and it is this: {t\atop 10}~\ell ~{m\atop 1}=22{2\over 9}.

[Stage 2] So you restore by adding the excluded [amount] to each [side], so the problem becomes: ten things equal a māl and twenty-two and two ninths, and it is this: {t\atop 10}={m\atop 1}~~22{2\over 9}.

[Stage 3] This results in the fifth type [of equation]. So work it out, which is that you halve the roots and you square the half to get twenty-five. Subtract the number from it, leaving two and seven ninths. Take its root, to get one and two-thirds. And if you add it to the half, which is five, it results in the thing by addition, and if you subtract it, it results [in the thing] by subtraction. One of them is [one] part, and the remainder from the ten is the other part. So you find one of the parts to be six and two thirds, and the other three and a third, and that’s it.

The enunciation of the problem is an arithmetic question which is solved using algebra. Ibn Qunfudh applies the rule for multiplying polynomials in stage 1, a restoration (al-jabr) to simplify the equation in stage 2, and the rule for solving the type 5 equation in stage 3. The type 5 equation here has two solutions, and in this problem they lead to the two parts of ten. The two equations in notation are shown as Too late and Too late in the manuscript. They correspond to the modern 10x-x^2=22{2\over 9} and 10x=x^2+22{2\over 9}, respectively. In this problem Ibn Qunfudh shows “things” in notation with only the three dots from the letter Too late, though he shows the whole letter in other examples.

[4] A geometry problem of al-Karajī’s solved by algebra.

This next problem is from al-Karajī’s Sufficient Book on Arithmetic, written in the early eleventh century. There is no need for stage 2 because the equation that is set up at the end of stage 1 is already simplified. Its solution in stage 3 is trivial.[6]

[Enunciation]If someone said, “A rectangle: its length is twice its width, and its area is equal to its perimeter. How much are its area and its sides?”

[Stage 1]Make its length two things and its width a thing, and multiply the length by the width to get two māls. This is the area, and it equals the perimeter, which is six things [ {m\atop 2}={t\atop 6}].

[Stage 3]So the one thing equals three, which is the width.

[5] ʿAlī al-Sulamī’s problem (I.36), solved by algebra.

This author’s tenth-century book Sufficient Introduction on Calculation by Algebra and What One Can Learn from Its Examples contains many problems, among which is the following[7]:

[Enunciation]Two men encounter a horse. The first says to the second, if you offer a fourth and a fifth of what you have in addition to what I have, then I will have the price of the horse. And the second says to the first, if you offer a sixth and a seventh of what you have in addition to what I have, then I will have the price of the horse.

[Stage 1]So make what the first has a thing, and what the second has a number that has a fourth and a fifth, and that is twenty.[8] If the first takes from the second a fourth and a fifth of what he has, he will have a thing and nine dirhams, which is the price of the horse.

Then the second takes from the first a sixth and a seventh of a thing, so he will have twenty and thirteen parts of forty-two of a thing, which is also the price of the horse. Then confront what they have, [i.e., establish the equation {t\atop 1}~\hbox{9}=\hbox{20}~{t\atop {13\over 42}}]

[Stage 2]and cast away like terms, leaving twenty-nine parts of forty-two equals eleven dirhams [ {t\atop {29\over 42}}=\hbox{11}].

[Stage 3]Complete the thing, by which you multiply everything you have by forty-two and divide it by twenty-nine, to get: the thing is fifteen and twenty-seven[9] parts of twenty-nine [ 15{27\over 29}], which is what the first has, and the second has twenty dirhams.

If you want the results to be whole numbers, multiply what you have by twenty-nine, so the first has four hundred sixty-two and the second has five hundred eighty.

[6] Ibn al-Bannāʾ’s problem (I.10), solved by algebra.

Many authors, Ibn al-Bannāʾ included, condensed stages 2 and 3 of their solutions. Here is the enunciation and the first of five different algebraic solutions to problem (I.10) from his Book on the Fundamentals and Preliminaries in Algebra. Note how he manipulates the operations to avoid unresolved divisions in his equation, and his justification for this manipulation after the solution.[10]

[Enunciation]Ten: you divide it into two parts. You divide each of them by the other, so it results in a total of four and a fourth.

[Stage 1]You make one of the parts a thing, so the other is ten less a thing. Then you square each of them and you add the squares to get a hundred and two māls less twenty things [ \hbox{100}~{m\atop 2}~\ell ~{t\atop 20}]. Keep this in mind. Then you multiply the surface of the parts by the result of the divisions, which is four and a fourth, to get forty-two things and half a thing less four māls and a fourth of a māl. This equals the sum of the remembered squares [ {t\atop 42{1\over 2}}~\ell ~{m\atop 4{1\over 4}}=\hbox{100}~{m\atop 2}~\ell ~{t\atop 20}].

[Stage 2]Then you restore and confront, to get the fifth type [of equation: {t\atop 62{1\over 2}}=\hbox{100}~{m\atop 6{1\over 4}}].

[Stage 3]So you look for the thing, finding two by subtraction, which is one of the parts, and eight by addition, which is the other part.

You saw the reason for this way above in the chapter on division, in the first part, which is: for any two numbers, if each one is divided by the other, then the sum of their squares is equal to the product of their surface by the sum of the two quotients.

We can represent the enunciation with modern algebraic notation by naming the two parts of ten a and b so that a+b=10, with the condition that {a\over b}+{b\over a}=4{1\over 4}. Ibn al-Bannāʾ names one of the parts “a thing”, so the other is necessarily “ten less a thing”. The modern versions would be x and 10-x, and we would set up the equation as

and maybe combine the terms on the left to get

after which we would multiply both sides by x(10-x).

Like other Arabic algebraists, Ibn al-Bannāʾ avoids operations in equations. He instead reformulates the problem according to an arithmetical rule he had given earlier in the book, and which was stated rhetorically: for any two numbers a and b, a^2+b^2=ab\left({a\over b}+{b\over a}\right). This way, the equation is set up without operations: “forty-two things and half a thing less four māls and a fourth of a māl…equals” “a hundred and two māls less twenty things”. There is no addition, subtraction, or scalar multiplication in this equation. Our modern version 42{1\over 2}x-4{1\over 4}x^2=100+2x^2-20x, however, is built from addition, subtraction, scalar multiplication, and exponentiation. See our commentary above at 219.1, 219.4, and 229.8.

Because stages 2 and 3 follow prescribed rules, he does not need to spell out the steps. Like some other algebraists, Ibn al-Bannāʾ writes “restore and confront” to mean restoration and/or confrontation in stage 2. For this equation two restorations are performed.

[7] A problem of al-Ḥaṣṣār’s solved by single false position, algebra, and double false position.

Al-Ḥaṣṣār wrote his Book of Demonstration and Recollection in the Art of Dust-Board Reckoning in the late twelfth century. He is one of several authors to solve the same arithmetic problem by different methods.[11]

If someone said to you, “A quantity (māl): you add its third and its fourth so it brings back twenty-one dirhams”.

[Solution by single false position]In this problem you can take the third and the fourth from twelve. Then you take from the twelve its third and its fourth, which is seven. The ratio of this seven with respect to twelve is as the ratio of the twenty-one with respect to the required quantity. So multiply the twelve by the twenty-one, and you divide the result by the seven. This gives you thirty-six, which is the quantity.

[Solution by algebra]And if you wish, work it out by algebra (al-jabr), which is that you make the quantity a thing. You take from it its third and its fourth, and that is three sixths of a thing and half a sixth of a thing. This equals the twenty-one [ {t\atop {1~\,3\over 2~\,6}}=\hbox{21}].

So we say to you: restore the three sixths and a half so that it yields a thing. What determines that is one and five sevenths, and we will explain it, God willing, in the chapter on the restoration of fractions.[12] So you multiply one and five sevenths by twenty-one. This gives you thirty-six, which is the quantity.

[Solution by double false position]And if you wish, work it out by the [method of] scales, which is that you take any number you wish for the front [scale], which you are allowed to do. So you make the front scale three. You take its third and its fourth: it is one and three fourths. Confront it with the twenty-one. You find it is smaller. So you made an error in the “three” scale of nineteen and a fourth falling short, and keep this error in mind.

Then you take the other scale and let it be, for example, four. You take its third and its fourth, and that is two and a third. Confront it with the twenty-one. So you made an error in the “four” scale of eighteen and two thirds falling short.

Multiply the error of the first scale, which is nineteen and a fourth, by the second scale, which is four. That gives seventy-seven. Keep it in mind. Then multiply the error of the second scale, which is eighteen and two thirds, by the first scale, which is three. It gives you fifty-six. Subtract it from the seventy-seven, leaving twenty-one. Then you subtract the smaller of the two errors from the greater. The remainder is three sixths and half of a sixth. Divide the twenty-one by it, giving you thirty-six, which is the quantity itself, and we will clarify this ratio when we get to it.

Footnotes

[1]  (Ibn al-Bannāʾ 1984, 212.7).

[2]  The mudy, plural amdāʾ, was a unit of capacity measure (Rebstock 2008). The plural form of Arabic nouns are used for numbers from 3 to 10 only, so fourteen of them are mudy.

[3]  The dinar (dīnār) was the standard gold coin in Islamicate countries.

[4]  (Ibn al-Bannāʾ 1984, 206.17).

[5]  (Ibn Qunfudh manuscript, 297.13).

[6]  (al-Karajī 1986, 204.8).

[7]  (ʿAlī al-Sulamī manuscript, fol. 54b.1).

[8]  The problem is indeterminate, so the author is allowed to pick one of the numbers as he pleases.

[9]  The MS mistakenly has “nineteen”.

[10]  (Saidan 1986, 562.4).

[11]  (al-Ḥaṣṣār manuscript, fol. 72r.6).

[12]  Al-Hawārī explains the restoration of fractions at 154.1.

C. Chronological list of mathematicians and other scholars

Below is a list of exceptionally brief notices on some of the different mathematicians and philosophers mentioned in this book. References are given to (Rosenfeld and Ihsanoğlu 2003) and (Lamrabet 2014) where applicable. For example, for al-Ḥaṣṣār we write (#532, M55), which means that he is scholar #532 in Rosenfeld & Ihsanoğlu, and scholar M55 in Lamrabet. Rosenfeld and Ihsanoğlu give numbered lists of individual titles by each author. For example, “[M2]” after an Arabic title indicates book M2 in their list. After Aristotle and Euclid all dates are CE, and with a few AH.

Aristotle (384-322 BC). Three aspects of Aristotle’s thought impacted al-Hawārī’s book, most likely through the works of Ibn Sīnā. One is the distinction between quality and quantity at 92.17, another is the distinction between sensible and intelligible objects at 133.3, and the third is the division of the genus of “quantity” into discrete and continuous, discussed in our commentary at 117.2 and 133.3. For a brief overview of Aristotle’s influence on Arabic mathematics, see Section 4 in our introduction.

Euclid (ca. 300 BC). Euclid wrote his Elements in Greek around 300 BC in Alexandria. The work consists of thirteen “Books”, or what we might call long chapters. Books I through VI cover plane geometry, Books VII to IX discuss number theory, the long Book X deals with the theory of quadratic irrationals in a geometric context, and Books XI to XIII deal with three-dimensional geometry.[1]

    Euclid’s Elements is reported to have been first translated into Arabic by al-Ḥajjāj ibn Yūsuf ibn Maṭar during the reign of the caliph Hārūn al-Rashīd (786-809). Al-Ḥajjāj later produced a second and improved translation during the reign of al-Maʾmūn (813-833). Later in the century Isḥāq ibn Ḥunayn (830-910) translated the Elements, and this was subsequently revised by Thābit ibn Qurra (836-901). While the surviving manuscripts apparently all derive from Thābit’s version, they also contain passages from al-Ḥajjāj. In addition, quotations from other books preserve parts of the earlier translations. The textual history of Euclid’s Elements in Arabic is complex and at present not well understood.

    Euclid’s Data, another work on geometry, was also translated into Arabic and influenced the ways some Arabic mathematicians presented proofs.

    The influence of Euclid’s Elements on al-Hawārī’s book is seen in the definitions of “number” (see our commentary at 65.2), “oddly-odd” (66.7), “multiplication” (95.2), “ratio”, “part”, “parts” (133.1), “medial” (163.4), and “ex-aequali” (195.2). The terms “side”, “surface”, and “square” (66.17) also ultimately come from Euclid, as do the various manipulations of ratios covered at 196.16. The entire theory of quadratic irrationals beginning at 173.4 derives from an arithmetical reading of Elements Book X. Some fragments of al-Ḥajjāj’s translation of this book may have found their way into al-Hawārī’s book. See our comments at 173.10.

Nicomachus of Gerasa (fl. second century). Nicomachus was a Neopythagorean philosopher who wrote his Arithmetical Introduction more as a prolegomenon (introduction) to philosophy than as a treatise on mathematics. He devotes the first six chapters to a philosophical argument for the primacy of arithmetic in the mathematical sciences, and he covers aspects of what we would call elementary number theory in the rest of the book. Ḥabīb ibn Bihrīz translated the Arithmetical Introduction into Arabic from a Syriac version shortly before 822. Thābit ibn Qurra translated Nicomachus’s work into Arabic again some decades later, this time directly from the Greek.[2]

    Nicomachus’s influence is evident in the definitions of “evenly-even”, etc. (65.10), the distinction between disparities of quantity and quality in sequences (73.7, 73.17), the sieve of Eratosthenes (127.10), and in the list of the kinds of proportion that al-Hawārī copied from Ibn al-Bannāʾ’s Lifting the Veil (195.2).

Diophantus (ca. 300). This Alexandrian mathematician wrote his Arithmetica in thirteen “books”. Six survive in Greek and another four in the Arabic translation of Qusṭā ibn Lūqā. In this work Diophantus solves determinate and indeterminate problems by algebra.[3]

al-Khwārazmī, early ninth century (#41). Muḥammad ibn Mūsā al-Khwārazmī worked in Baghdad and authored works on Indian arithmetic, algebra, astronomy, astrology, geography, and the Jewish calendar. The earliest known books on algebra are al-Khwārazmī’s Book of Algebra (Kitāb al-jabr wa l-muqābala, [M3]) and Ibn Turk’s book by the same title. Only a fragment of the latter is extant, while al-Khwārazmī’s book survives in its entirety. Al-Khwārazmī’s Book on Indian Reckoning (Kitāb al-ḥisāb al-hindī, [M1]) covers the rules of calculating with Arabic numerals, and survives only in a medieval Latin reworking of a Latin translation.[4]

Ibn Turk, early ninth century (#59). The only part of ʿAbd al-Ḥamīd Ibn Turk’s Book of Algebra (Kitāb al-jabr wa l-muqābala, [M1]) that is extant is the part giving proofs for the rules for solving simplified equations.[5]

Thābit ibn Qurra, 836-901 (#103). Abū l-Ḥasan Thābit ibn Qurra al-Ḥarrānī was a prolific translator and scholar who worked in all branches of the mathematical sciences. He revised Isḥāq ibn Ḥunayn’s Arabic translation of Euclid’s Elements, and he translated Nicomachus’s Arithmetical Introduction as well as some works of Archimedes into Arabic.

    Among Thābit’s original contributions to mathematics are his calculations of area and volume in the tradition of Archimedes, including the volume of a section of a paraboloid. These are not numerical calculations. Instead, they equate the areas and volumes to other known areas and volumes. Thābit also wrote a short treatise on algebra titled Establishing the Correctness of Algebra Problems by Geometric Proofs (Qawl fī taṣḥīḥ masāʾil al-jabr bi l-barāhīn al-handasiyya, [M19]) in which he proved the rules for solving the three composite quadratic equations in the style of Euclid’s Data.[6]

Qusṭā ibn Lūqā, died ca. 910 (#118). Among the many contributions Qusṭā made to mathematics, the two of interest to the current study are his translation of Diophantus’s Arithmetica and his short treatise in which he proved the rules of double false position via geometry in the style of Euclid’s Data.[7]

Abū Kāmil, late ninth century (#124). Abū Kāmil, whose full name is Shujāʿ ibn Aslam ibn Muḥammad ibn Shujāʿ, is best known for his comprehensive Book on Algebra (Kitāb fī l-jabr wa l-muqābala, [M1]), written as a kind of commentary on al-Khwārazmī’s algebra book. The rules in algebra for finding the māl directly, first mentioned at 214.2 ultimately come from this book.[8]

ʿAlī al-Sulamī, tenth century (#267). This Syrian mathematician is known only for his Sufficient Introduction on Calculation by Algebra and What One Can Learn from its Examples (al-Muqaddima al-kāfiyya fī ḥisāb al-jabr wa l-muqābala wa mā yuʿrafu bihi qiyāsuhū min al-amthila, [M1]), which survives in a single manuscript. The book exhibits borrowings from Abū Kāmil and al-Khwārazmī.[9] His full name is Abū l-Ḥasan ʿAlī ibn al-Muslim ibn Muḥammad ʿAlī al-Fatḥ al-Sulamī.

Ikhwān al-Ṣafāʾ (Brethren of Purity), tenth century (#226). The Brethren of Purity were a group of anonymous mystical scholars centered in the city of Basra. Their highly influential Epistles of the Brethren of Purity (Rasā’il Ikhwān al-Ṣafāʾ, [E1]) is a kind of encyclopedia of all knowledge situated in a universalist philosophical/theological setting. The first of its four parts is on mathematics, where the main influences for arithmetic are the Greek Neopythagoreans, especially Nicomachus, and for geometry Euclid’s Elements.[10]

al-Uqlīdisī, tenth century (#232). Abū l-Ḥasan Aḥmad ibn Ibrāhīm al-Uqlīdisī wrote his Chapters on Indian Arithmetic (al-Fuṣūl fī l-ḥisāb al-Hindī, [M1]) in 952-3. It is the oldest extant Arabic treatise on calculating with Arabic numerals. The nickname “al-Uqlīdisī” can be translated as “the Euclidean”, presumably because he worked on Euclid’s Elements. His book on arithmetic is unrelated to Euclid’s work.[11]

Abū l-Wafāʾ, 940-998 (#256). The earliest extant book dedicated primarily to finger-reckoning is Abū l-Wafāʾ’s Book of What is Necessary for Scribes, Businessmen, and Others in the Science of Arithmetic (Kitāb fīmā yaḥtāju ilayhi al-kuttāb wa l-ʿummāl wa ghayruhum min ʿilm al-ḥisāb, [M2]), written between 961 and 976.[12] Full name: Abū l-Wafā Muḥammad ibn Muḥammad ibn Yaḥyā ibn Ismāʿīl ibn al-ʿAbbās al-Būzjānī.

al-Karajī, died ca. 1025 (#309). Fakhr al-Dīn Abū Bakr Muḥammad ibn al-Ḥasan (or al-Ḥusayn) al-Karajī was a Persian mathematician and engineer who wrote his works on algebra and arithmetic in Baghdad. His three main works, in chronological order, are:

[Book of] al-Fakhrī on the Art of Algebra (al-Fakhrī fī ṣināʿat al-jabr wa l-muqābala, [M2]). This is al-Karajī’s book on algebra, modeled on Abū Kāmil’s Book on Algebra and Diophantus’s Arithmetica. It was probably completed in 401H/1010-1 CE.

The Sufficient [Book] on Arithmetic (al-Kāfī fī l-ḥisāb, [M1]). A work on calculation by finger-reckoning with a section on algebra. It was probably completed in 402H.

Marvelous [Book] of Arithmetic (al-Badīʿ fī l-ḥisāb, [M3]). This book covers various techniques of calculation, beginning with an introduction to arithmetic based in Euclid and Nicomachus, then a section on algebra, and ending with an exposition on techniques of solving indeterminate problems by algebra.[13]

Kūshyār ibn Labbān, ca. 970-1030 (#308). Kūshyār was a Persian mathematician and astronomer who wrote a book on calculating with Arabic numerals, titled Principles of Indian Reckoning (Kitāb fī uṣūl ḥisāb al-Hindī, [M1]).[14] Full name: Abū l-Ḥasan Kūshyār ibn Labbān ibn Bāshahrī al-Jīlī.

Ibn Sīnā, ca. 970-1037 (#317). Abū ʿAlī al-Ḥusayn ibn ʿAbdallāh ibn Sīnā, known in Latin as Avicenna, was the most important philosopher of medieval Islam. He was second only to Aristotle in influence in the Islamic world and in the late medieval and early modern West. For epistemology and ontology he built on the works of Aristotle, and it is most likely through him that Aristotelian ideas are present in al-Hawārī’s book.[15]

al-Baghdādī, died 1038 (#320). Abū Manṣūr ʿAbd al-Qāhir ibn Ṭāhir al-Baghdādī was a Persian mathematician and judge best known for his Completion of Arithmetic (al-Takmila fī l-ḥisāb, [M1]). This book covers calculation with Indian numerals on the dust-board, sexagesimal arithmetic, number theory, and business arithmetic.[16]

Ibn al-Samḥ, early eleventh century (A93). Abū l-Qāsim Aṣbagh ibn Muḥammad ibn Aṣbagh ibn al-Samḥ was a mathematician, astronomer, and physician who taught in Grenada. His Sufficient Book on Mental Reckoning (Risāla kāfiya fī ʿilm al-ḥisāb) covers techniques of finger-reckoning and double false position.[17]

Ibn al-Haytham, 965-1041 (#328). Known also by his Latinized name Alhazen, Ibn ʿAlī al-Ḥasan ibn al-Ḥasan ibn al-Haytham is best known for his groundbreaking work in optics. What interests us is his short work The Arithmetic of Transactions (Ḥisāb al-muʿāmalāt, [M24]).[18] It was common for scientists conducting advanced work to produce elementary texts like this. Naṣīr al-Dīn al-Ṭūsī (below) is another.

al-Bīrūnī, 973-1048 (#348). Abū l-Rayḥān Muḥammad ibn Aḥmad al-Bīrūnī wrote extensively on a wide range of topics including anthropology, linguistics, natural science, mathematics, and astronomy. Book II of his Book of Instruction in the Elements of the Art of Astrology (Kitāb al-tafhīm li-awāʾil ṣināʿat al-tanjīm, [A2]) covers arithmetic.[19] The first part, on number theory (pp. 72-95), is largely taken from Nicomachus, with some influence from Indian sources. The second part (pp. 96-119) covers calculation: multiplication, division, roots, sexagesimal and base ten arithmetic, algebra, double false position, and abjad calculation.

al-Qurashī, 1030-1067. Abū l-Ḥasan ʿAlī ibn al-Khiḍr al-Qurashī hailed from Syria. His Note on the Elements of Calculation and Inheritance (al-Tadhkira bi-uṣūl al-ḥisāb wa l-farāʾiḍ) is a summary of his more extensive, and lost, Book of Sustenance (Kitāb al-Maʿūna). It covers practical arithmetic, with a focus on the division of estates.[20]

al-Khayyām, 1048-1131 (#420). Better known in the West as Omar Khayyam, Ghiyāth al-Dīn Abū l-Fatḥ ʿUmar ibn Ibrāhīm al-Khayyāmī (Khayyām) was a poet, mathematician, and astronomer best known to historians of mathematics for his Treatise on the Proofs of Algebra Problems (Risāla fī l-barāhīn ʿalā masāʾil al-jabr wa l-muqābala, [M2]), in which he classifies and solves the twenty-five equations of degree three and less. Al-Khayyām saw algebra as a tool for geometric problem-solving, so he gives constructions for solving each equation type. He also gave numerical solutions where he could, but rules for solving irreducible cubic equations would not be found until the sixteenth century. Al-Khayyām was able to justify the appropriation of algebra for solving problems in geometry by regarding the numbers of the algebraists as the dimensionless measures of continuous magnitudes.[21]

al-Samawʾal, died ca. 1175 (#487). Al-Samaw-al ibn Yaḥyā ibn ʿAbbās al-Maghribī was a native of Baghdad who later worked in Iran. He built on the works of al-Karajī and others to develop computational rules for polynomials (“composite numbers”) in his The Dazzling [Book] on the Science of Calculation (al-Bāhir fī ,ilm al-ḥisāb, [M1]).[22]

al-Ḥaṣṣār, died before 1194 (#532, M55). Abū Bakr Muḥammad ibn ʿAbdallah ibn ʿAyyāsh al-Ḥaṣṣār’s Book of Demonstration and Recollection in the Art of Dust-Board Reckoning (Kitāb al-bayān wa l-tadhkār fī ṣanʿat ʿamal al-ghubār, [M1]) is the earliest extant Arabic book showing the notation of fractions with the division bar, among other innovations. He also wrote a more comprehensive book on arithmetic, the Complete [Book] on the Art of Number (al-Kāmil fī ṣināʿa al-ʿadad, [M2]). Al-Ḥaṣṣār worked in al-Andalus and Morocco.[23]

Ibn al-Yāsamīn, died 1204 (#521, M83). Originally from Fez in Morocco, Abū Muḥammad ʿAbdallah ibn Muḥammad ibn Hajjāj ibn al-Yāsamīn al-Adrīnī composed his Grafting of Opinions of the Work on Dust Figures (Talqīḥ al-afkār fī l-ʿilm bi-rushūm al-ghubār, [M3]) in Seville. This book is a hybrid between a textbook on Indian arithmetic and one on finger reckoning. It is the earliest book we know to show the Arabic algebraic notation, though it is clear by the way he presents it that it was not his invention. Ibn al-Yāsamīn is also well known for his 54-line Poem on Algebra (Urjūza fī l-jabr, [M1]), which inspired as many commentaries as Ibn al-Bannāʾ’s Condensed Book.[24]

    Ibn al-Bannāʾ copied many parts of Ibn al-Yāsamīn’s Grafting of Opinions nearly word-for-word into his Condensed Book. Much of the chapter on the addition of whole numbers (73.2 through 79.13), the paragraph on multiplication at 98.4, the rules (mostly) from finger-reckoning from 109.7 through 114.4 (except 113.6), and the first couple of sentences on division at 117.2-5 were taken from it.

Ibn Munʿim, twelfth-thirteenth centuries (#556, M89). Abū Jaʿfar Aḥmad ibn Ibrāhīm ibn Munʿim al-ʿAbdarī was a Moroccan mathematician who wrote his book Understanding Calculation (Fiqh al-ḥisāb, [M1]) at a more theoretical level than the textbooks of people like al-Ḥaṣṣār or Ibn al-Yāsamīn.[25]

Ibn Fallūs, 1194-1239 (#584). Shams al-Dīn Abū l-Ṭāhir Ismāʿīl ibn Ibrāhīm ibn Ghāzī al-Māridīnī was known by the name Ibn Fallūs. It was during a pilgrimage to Mecca that he wrote his treatise on algebra with a rhyming title: Kitāb niṣāb al-ḥabr fī ḥisāb al-jabr (Preparation for writing on calculation in algebra, [M3]).[26] He also wrote a textbook on finger-reckoning, Direction to Reckoners Showing the Right Path in Revealing the Science of Arithmetic (Irshād al-ḥussāb fī al-maftūḥ min ʿilm al-ḥisāb, [M2]).

Ibn Badr, ca. thirteenth century (#587). Abū ʿAbdallah Muḥammad ibn ʿUmar Ibn Badr al-Balansī wrote his Brief Book on Algebra (Kitāb fīhi ikhtiṣār al-jabr wa l-muqābala, [M1])[27] with borrowings from Abū Kāmil and al-Khwārazmī.

Naṣīr al-Dīn al-Ṭūsī, 1201-1274 (#606). One of the most prolific scholars of medieval Islam, Naṣīr al-Dīn Abū Jaʿfar Muḥammad ibn Muḥammad al-Ṭūsī wrote on philosophy, theology, and a variety of scientific topics. In the sciences he is best known for his work on planetary astronomy, but we are more interested in his relatively insignificant textbook Gathering of Arithmetic by Means of Board and Dust (Jāmiʿ al-ḥisāb bi l-takht wa l-turāb, [M17]).[28]

al-Fārisī, died 1319 (#674). Kamāl al-Dīn al-Fārisī was a Persian mathematician who wrote his scientific examination of the rules of finger reckoning, Foundation of the Rules on Elements of Benefits (Asās al-qawāʿid fī uṣūl al-fawāʾid, [M2]), in the form of a commentary on a textbook of his teacher Ibn al-Khawwām.[29]

Ibn al-Bannāʾ, 1256-1321 (#696, M134). See section 1.2 in the Introduction for bio-bibliographic details.

al-Hawārī, early fourteenth century (#747, M148, M150). See section 1.2 in the Introduction for bio-bibliographic details.

Ibn Qunfudh, 1339-1407 (#780, M193). Abū l-ʿAbbās Aḥmad ibn Ḥasan ibn ʿAlī ibn al-Khaṭīb ibn Qunfudh al-Qustantīnī completed his commentary on Ibn al-Bannāʾ’s Condensed Book, titled Lowering the Veil from the Faces of Arithmetical Operations (Ḥaṭṭ al-niqāb ʿan wujūh aʿmāl al-ḥisāb, [M1]), in 1370. He also wrote Foundations for Beginning the Commentary on Ibn al-Yāsamīn’s Poem (on algebra) (Mabādīʾ al-sālikīn fī sharḥ urjūzat Ibn al-Yāsamīn). Both books show the Arabic algebraic notation.[30]

al-Mawāḥidī, second half of the fourteenth century (M176). Abū ʿAbd al-Raḥmān Yaʿqūb ibn Ayyūb al-Mawāḥidī al-Jazūlī wrote the popular commentary titled Achieving the Desire of Commenting on Ibn al-Bannāʾ’s Condensed [Book] (Taḥṣīl al-munā fı sharḥ Talkhīṣ Ibn al-Bannāʾ), ca. 1382.[31]

al-ʿUqbānī, died 1408 (#781, M195). Abū ʿUthmān Saʿīd ibn Muḥammad al-Tujībī al-Tilimsānī al-ʿUqbānī wrote his Commentary on the Condensed [Book] (Sharḥ al-Talkhīṣ)[32] with a focus on providing Euclidean-style proofs to the rules in Ibn al-Bannāʾ’s book. He was a native of Tlemcen, and spend his career between cities now located in Algeria and Morocco.

al-Ḥanbalī, died 1409 (#782). Taqī al-Dīn ibn ʿIzz al-Dīn al-Ḥanbalī copied numerical examples from al-Hawārī’s book into his own Commentary on the Condensed [Book] (Sharḥ al-Talkhīṣ).[33]

Ibn al-Hāʾim, ca. 1355-1412 (#783). Abū l-ʿAbbās Shihāb al-Dīn Aḥmad ibn Muḥammad ibn Imād al-Dīn ibn ʿAlī was known by the name Ibn al-Hāʾim. Among his many books on arithmetic and algebra are his 1387 Commentary on the Poem of al-Yāsamīn (Sharḥ al-Urjūza al-Yāsmīnīyya, [M13]), his 1389 Guidebook for the Science of Aerial Calculation (al-Maʿūnah fī ʿilm al-ḥisāb al-hawāʾī, [M1]), and his Abridgement [of the] Condensed [Book] of Ibn al-Bannāʾ (Mukhtaṣar Talkhīṣ Ibn al-Bannāʾ, [M17]).[34]

al-Kāshī, died 1429 (#802). Giyāth al-Dīn Jamshīd ibn Masʿūd al-Kāshī (or al-Kāshānī) was a mathematician and astronomer in the court of Ulugh Beg in Samarqand. He is best known for his work in planetary astronomy and for his calculations of 2\pi and the sine of one degree, each accurate to the equivalent of 16 decimal places. What interests us is his 1427 book Key to Calculation (Miftāḥ al-ḥisāb, [M1]).[35]

Ibn al-Maghribī, fifteenth century (#910). We mention his Poem on Reckoning with Finger-Joints (Manẓūma fī l-ḥisāb al-ʿuqūd, [M1]) in our Introduction.[36] His full name is Abū l-Ḥasan ʿAlī ibn al-Maghribī.

al-Qalaṣādī, died 1486 (#865, M229). Abū l-Ḥasan ʿAlī ibn Muḥammad al-Qurashī al-Basṭī al-Qalaṣādī was a native of Baza in al-Andalus. His Commentary on the Condensed [Book] on the Operations of Arithmetic (Sharḥ Talkhīs aʿmāl al-Ḥisāb, [M7]) shows the Arabic algebraic notation.[37]

Sibṭ al-Māridīnī, 1423-1506 (#873). Muḥammad ibn Muḥammad ibn Aḥmad Abū ʿAbdallah Badr [Shams] al-Dīn al-Miṣrī al-Dimashqī was the timekeeper at the al-Azhar mosque in Cairo, and authored dozens of works on arithmetic and astronomy. We cite two works in particular: his Student’s Guide to the Way of Arithmetic (Irshād al-ṭullāb ilā wasīlat al-ḥisāb, [M7]) and The Light of al-Māridīnī on Commentary on Ibn al-Yāsamīn (al-Lamʿah al-Māridīnīyah fī sharḥ al-Yāsamīnīyah, [M10]).[38]

Ibn Ghāzī, 1437-1513 (#913, M246). Abū ʿAbdallah Muḥammad ibn Aḥmad ibn Muḥammad Ibn Ghāzī al-ʿUthmānī al-Miknāsī al-Fāsī further condensed Ibn al-Bannāʾ’s Condensed Book into a 461-line poem titled Desire of Reckoners (Munyat al-ḥussāb, [M1]). Later, in 1483, he wrote a commentary on his poem titled Aim of the Students in Commentary on Desire of Reckoners (Bughyat al-ṭullāb fī sharḥ munyat al-ḥussāb, [M2]). This book shows the Arabic algebraic notation, including an entire worked-out problem.[39]

Luca Pacioli, ca. 1446-1517. Italian Renaissance author whose massive, printed, and highly influential 1494 Summa de Arithmetica Geometria Proportioni & Proportionalita presents most of the mathematical knowledge of his time. Much of it ultimately comes from Arabic sources, often via Fibonacci.[40]

Footnotes

[1]  The standard English translation of the Elements is (Euclid 1956). The Greek text with the English translation by Richard Fitzpatrick is available online: http://farside.ph.utexas.edu/Books/Euclid/Elements.pdf (accessed May 1, 2019). The extant Arabic translation has not been published.

[2]  Thābit’s version is published in (Nicomachus 1959). Ḥabīb ibn Bihrīz’s translation survives only in a Hebrew translation from 1317 of a redaction by al-Kindī (Freudenthal and Zonta 2007). The Greek text is published in (Nicomachus 1866), and an English translation from the Greek is published in (Nicomachus 1938).

[3]  The extant Greek books are published in (Tannery 1893–1895), and the Arabic books in (Sesiano 1982); (Diophantus 1984). See (Christianidis and Oaks 2013) for a comparison between the method of Diophantus and Arabic algebra.

[4]  The Book of Algebra is published with an English translation in (al-Khwārizmī 2009). This is translated from the French edition of 2007. The Book on Indian Reckoning is published in (al-Khwārizmī 1997). See also (al-Khwārizmī 1992).

[5]  The Arabic text is published with English and Turkish translations in (Sayili 1962).

[6]  Thābit’s translation of Nicomachus is published in (Nicomachus 1959). The treatise on algebra is published in (Luckey 1941); (Thābit ibn Qurra 2009, 159-169), the former with a German translation and the latter with a French translation. An English translation appears in (al-Khwārizmī 2009, 34-35, 38, 41-42).

[7]  The four extant books of Qusṭā’s translation of Diophantus have been edited and translated twice: (Sesiano 1982), into English, and (Diophantus 1984), into French. His treatise on double false position is edited and translated into German in (Suter 1908–1909).

[8]  A facsimile of the Istanbul manuscript of the Book on Algebra is published in (Abū Kāmil 1986). An edition with a German translation is published in (Abū Kāmil 2004), and an edition with a French translation appears in (Abū Kāmil 2012).

[9]  (ʿAlī al-Sulamī manuscript).

[10]  English translations of the mathematical portions of the Epistles have been published in (Goldstein 1964) and (El-Bizri 2012), the latter also containing the Arabic text.

[11]  The Arabic text is published in (al-Uqlīdisī 1984), and Saidan’s English translation is published in (al-Uqlīdisī 1978).

[12]  The Arabic text is edited in (Saidan 1971). See (Saidan 1974) for a description of the contents in English.

[13]  The Arabic texts of these three books are published, in order, in (Saidan 1986), (al-Karajī 1986), and (al-Karajī 1964).

[14]  A facsimile of the Istanbul manuscript with an English translation is published in (Kūshyār ibn Labbān 1965).

[15]  For more on Ibn Sīnā, see his entry in the online Stanford Encyclopedia of Philosophy, https://plato.stanford.edu/entries/ibn-sina/ (accessed July 29, 2018).

[16]  The Arabic text is published in (al-Baghdādī 1985).

[17]  A facsimile of the Escurial manuscript with a Spanish translation is published in (Ibn al-Samḥ 2006).

[18]  Published in (Rebstock 1998).

[19]  A facsimile with an English translation is published in (al-Bīrūnī 1934).

[20] A facsimile of the Medina manuscript, with a German translation and commentary, is published in (al-Qurashī 2001). For a description of the book in English, see (Rebstock 2002).

[21]  For the Arabic text with a French translation of al-Khayyām’s mathematical works see (Rashed and Vahabzadeh 1999). An English translation of this book, but without the Arabic text, is published in (Rashed and Vahabzadeh 2000). Earlier editions and translations have also been published. See (Oaks 2011a) for the ontological foundation of his algebra.

[22]  This book is published in (al-Samawʾal 1972).

[23]  A facsimile of a late twelfth-century manuscript is available online: (al-Ḥaṣṣār manuscript).

[24]  T. Zemouli’s edition of the Arabic text of Grafting of Opinions appears in (Zemouli 1993). We lack pages 103-116 of this book, but we have a typeset version covering these pages from the edition that Zemouli will publish at some time in the future, (Zemouli n.d.). The Poem on Algebra is published with an English translation in (Abdeljaouad 2005a).

[25]  The Arabic text is published in (Ibn Munʿim 2005).

[26]  (Ibn Fallūs manuscript).

[27]  The Arabic text is published in (Saidan 1986).

[28]  The Arabic text is published in (Saidan 1967).

[29]  The Arabic text is published in (al-Fārisī 1994).

[30]  A facsimile of a fifteenth-century manuscript of the first work is available online: (Ibn Qunfudh manuscript).

[31]  We consulted (al-Mawāḥidī manuscript), which is available online.

[32]  The Arabic text is published in (Harbili 1997).

[33]  MSS Paris 2463/1; Tunis 16448/1.

[34]  The first is edited with a partial French translation in (Ibn al-Hāʾim 2003), the second is edited in (Ibn al-Hāʾim 1988), and the third is unpublished.

[35]  The Arabic text is published in (al-Kāshī 1969).

[36]  Partially translated into English in (Saidan 1968).

[37]  Published with a French translation in (al-Qalaṣādī 1999).

[38]  The Arabic texts are published in (Sibṭ al-Māridīnī 2004) and (Sibṭ al-Māridīnī 1983), respectively.

[39]  For a facsimile of the Library of Congress manuscript see (Ibn Ghāzī manuscript). The Arabic text is published in (Ibn Ghāzī 1983).

[40]  (Pacioli 1494).

D. Glossary

For each Arabic word, we give references to the text, and thus also to the translation. References to explanations of the word or concept in our commentary are preceded by a “C”. For words that were used in both technical and non-technical ways, we include references to both. The listing follows the alphabetical order of the roots of the words.

|                       ا
Too late | āḥād: units, 65.2, 65.8, 68.19, 69.3,16. C65.2.
Too late | uss: 1. index, 70.15, 71.12, 101.19, 107.12. C68.18.
| 2. power, 225.1,8, 226.3, 227.15, 229.19. C68.18.
Too late | illā: less, 76.19, 86.11, 140.2, 173.18, 219.5. C141.7, C173.10, C219.4.
Too late | imām: denominator, 94.12, 120.1, 135.12, 138.1, 170.10, 206.6. C93.6.
Too late | awwal: prime, 66.7. C66.7.

|                       ب
Too late | basaṭa: to numerate, 93.3, 134.1, 138.5, 145.9. C93.10.
Too late | basṭ: numerator, 93.17, 137.13, 139.10, 142.8, 170.10. C93.10.
Too late | mabsūṭ: numerator, 144.6, 145.3, 151.2, 158.7. C93.10.
Too late | basīṭ (and related forms): simple, 66.9, 67.17, 69.15, 134.2, 217.8. C66.7.
Too late | mubʿʿiḍ: portioned, 137.1, 139.10.
Too late | baqiya: to leave, remain, 72.2, 85.5, 88.10, 113.4, 165.1, 229.19. C83.1.
Too late | bāqī: remainder, 72.7, 84.12, 90.9, 141.6, 176.2, 201.4, 216.2, 220.2, 231.6. C83.1.
Too late | baqiya: residue, 89.2, 90.5, 91.13, 92.6. C83.1.
Too late | bāqiya: residual, 92.1,12, 93.12. C83.1.
Too late | mutabāyna: 1. different (i.e., relatively prime), 122.13, 123.11. C122.2.
| 2. incommensurable, 180.11, 182.5. C173.10.

|                       ث
Too late | mustathnā (and related forms): excluded, 137.12, 140.1, 142.11, 183.17, 220.5, 230.1. C140.1, C173.10, C219.4.
Too late | mustathnā minhu (and related forms): diminished, 140.14, 141.3, 230.1. C140.1, C173.10, C219.4.

|                       ج
Too late | jabara: to restore, 129.6, 154.7, 217.3, 223.9. C129.1, C211.2, C219.4, C223.1.
Too late | majbūr: number to be restored, 129.8, 154.2.
Too late | majbūr ilayhi: restored number, 129.8, 154.2.
Too late | al-jabr: 1. restoration, 129.1, 154.1, 211.1,2, 217.5, 220.11. C129.1, C211.1, C220.5.
| 2. algebra, 183.19, 189.3, 211.8. C191.1, C211.1.
Too late | al-jabr wa l-muqābala: algebra, 191.4, 209.2. C191.1, C211.1, C223.1.
Too late | jadhr: 1. root (square root), 66.18, 68.12, 163.5, 173.5, 180.7, 187.11, 213.8,11, 214.12, 216.22. C213.7.
| 2. root (unknown in algebra), 211.9,15, 212.8, 214.14, 215.12, 216.20. C213.7.
Too late | majdhūr: has a root, 66.17, 164.2. C66.13.
Too late | majdhūra: has a root, 163.14, 166.14. C66.13.
Too late | jazaʾa: to partition, 184.10, 186.8. C65.2, C179.1, C186.8.
Too late | juzʾ: 1. part, 61.9, 94.11, 95.8, 117.2, 121.1, 127.2, 131.1, 134.6, 152.6, 172.2. C131.1. See also ajzāʾ ṣumm and ajzāʾ muḥāṣṣa.
| 2. portion, 198.5, 199.4, 200.5, 202.1, 203.5.
Too late | ajzāʾ: parts (from Euclid), 133.1. C131.1.
Too late | jamaʿa: to add, 74.9, 86.15, 100.2, 134.8, 179.2, 201.11, 219.5. C73.2.
Too late | jamʿ: addition, 73.1, 74.1, 80.20, 147.4, 176.11, 219.1.
Too late | majmūʿ: 1. sum, 71.20, 80.15, 100.3, 173.8, 198.17, 226.10. C73.2.
| 2. addend, 74.18, 75.15, 76.5, 91.9, 179.11, 180.7.
Too late | majmūʿ ilayhi: augend, 74.18, 91.9, 147.6, 180.7.
Too late | jumla: sum, 79.6, 109.8. C73.2.
Too late | jins: species, 104.16, 105.4, 219.2, 220.1, 221.16; type, 117.9, 134.6; kind, 195.14, 211.3,6. C68.18.

|                       ح
Too late | ḥarf: 1. letter, 88.12, 89.1,6. C68.18.
| 2. particle, 142.2, 173.16, 182.5, 220.1. C68.18.
| 3. conjunction, 142.10, 173.14, 180.11, 181.2. C68.18.
Too late | maḥsūsa: sensible, 133.6. C133.3.
Too late | muḥāṣṣa: apportionment, 120.20. C120.20.
Too late | ajzāʾ muḥāṣṣa: apportioned parts, 121.1, 122.2. C120.20.
Too late | ḥaṭṭa: to reduce, 129.7, 155.13, 217.1. C129.1, C217.1.
Too late | ḥaṭṭ: reduction, 129.1, 154.1, 217.5. C129.1, C217.1.
Too late | maḥṭūṭ: number to be reduced, 129.12, 154.4, 155.14.
Too late | maḥṭūṭ ilayhi: reduced number, 129.12, 154.2, 155.14.
Too late | ḥalla: to decompose, 68.15, 117.16, 120.1, 123.4. C68.11.
Too late | ḥall: decomposition, 68.15, 117.2, 122.9.
Too late | ḥamala: to add, 71.18, 84.17, 108.10, 167.12, 200.4, 214.7, 216.11. C73.2.

|                       خ
Too late | mukhtalif: distinct, 136.8, 137.12, 139.1, 142.7, 145.14. C93.15, C136.8, C139.1.
Too late | mukhtalifa: difference, 73.20.

|                       د
Too late | dirham: 1. dirham (unit of currency), 95.10, 117.16, 158.2, 207.7. C211.8.
| 2. dirham (unit in arithmetic/algebra), 219.3, 220.3, 221.13, 223.7, 228.3. C211.8.
Too late | adaqq: finer, 94.11, 135.8, 136.7, 158.10.

|                       ذ
Too late | dhahaba: to take away, 169.12, 199.1. C83.1.

|                       ر
Too late | rabaʿa: to square, 80.5, 112.17, 113.14, 214.7, 215.8, 232.3.
Too late | murabbaʿ: square, 66.17, 112.17, 164.17, 173.4, 211.10. C66.17.
Too late | tarbīʿ: squaring, 79.18, 80.10, 112.16.
Too late | martaba: rank, 65.1, 68.18, 74.22, 174.2, 180.6, 230.14. C68.18, C174.1.
Too late | radda: transform, 143.1, 180.1,6, 184.9, 188.2, 228.8. C179.20.
Too late | rakiba: 1. to compose (numbers), 122.13, 123.5. C66.7, C68.11.
| 2. to combine (a proportion), 121.23, 196.17, 204.2. C196.16.
Too late | murakkaba (and related forms): 1. composed; composite; composition (numbers), 66.12,16, 68.1, 120.1, 127.4,12, 128.2. C66.7.
| 2. composite (equations), 211.14, 212.6, 217.1. C211.13, C219.4.

|                       ز
Too late | zawj: even, 65.6.
Too late | zawj al-zawj: evenly-even, 65.12. C65.10.
Too late | zawj al-fard: evenly-odd, 65.17. C65.10.
Too late | zawj al-zawj wa l-fard: evenly-evenly-odd, 66.1. C65.10.
Too late | zāda: 1. to add, 71.13, 110.2, 113.17, 175.7, 204.19, 220.10. C73.2.
| 2. to exceed, 77.6, 108.9, 169.12. C73.2.
Too late | zāʾid: 1. appended, 87.11, 183.19, 189.3, 211.5, 228.9. C219.4.
| 2. exceeds, 198.15, 200.6, 201.9, 203.5. C219.4.
Too late | ziyāda: added; additional; excess, 77.7, 108.13, 115.5, 200.6.

|                       س
Too late | musaṭṭaḥ: 1. surface (in arithmetic), 67.3, 80.10, 182.5, 225.6, 231.14. C66.17.
| 2. surface (in geometry), 133.7. C66.17.
Too late | saqaṭa: to drop, 76.13, 109.17, 123.4, 140.11, 166.17, 175.13, 199.4, 216.18, 231.4. C83.1.
Too late | ism: 1. name, 59.1, 66.13, 67.16, 69.15, 71.6, 134.1, 157.9, 229.5. C68.18.
| 2. term, 173.5, 174.5, 176.1, 211.12, 217.2, 225.1, 229.15.
Too late | dhū l-ismīn: binomial (lit., “two unified names”), 173.13, 175.11, 180.13, 188.18. C173.10.
Too late | sammā: to denominate, 94.1, 108.9, 121.15, 166.9, 181.16, 187.1. C93.6.
Too late | tasmiya: denomination, 93.6, 109.7, 118.15, 151.1. C118.14.
Too late | musammā: denominated number, 94.3, 124.8, 153.5. C93.6.
Too late | musammā minhu: denominating number, 93.6, 123.18, 153.5, 181.18. C93.6.

|                       ش
Too late | ishtirāk: common divisor, 122.4, 146.3, 153.10, 230.11. C122.2.
Too late | shayʾ: thing, 211.8, 213.7, 219.10, 225.2, 227.20. C211.8.

|                       ص
Too late | ṣaḥīḥ: 1. whole; whole number, 61.10, 65.2, 92.15, 149.2, 154.10, 163.1, 207.10.
| 2. posited number, 198.14, 199.14, 200.10, 201.10. See footnote in translation at 198.14.
Too late | ṣarafa: to convert, 157.10, 158.9. C157.1.
Too late | ṣarf: conversion, 95.8.
Too late | masrūf: fraction to be converted, 158.7,12, 159.1.
Too late | masrūf ilayhi: converted fraction, 158.1,8,12.
Too late | taṣfīr: voiding, 114.7. C74.17.
Too late | ṣifr: zero, 69.9, 75.5, 84.1, 98.22, 112.8, 114.4, 119.13, 170.8. C74.17.
Too late | aṣamm: deaf (i.e., prime), 127.5,14, 128.12. C66.7, C134.2.
Too late | ajzāʾ ṣumm: deaf parts (i.e., prime), 66.9, 126.2, 127.9, 128.12. C134.2.
Too late | ṣanf: kind, 207.8, 208.8. C68.18.

|                       ض
Too late | ḍaraba: to multiply, 71.13, 95.10, 96.3, 129.3, 183.2, 211.11, 226.9.
Too late | ḍarb: 1. multiplication, 92.8, 95.1, 98.4, 103.18, 157.11, 225.1, 227.6. C95.2, C149.2, C183.4, C225.1.
| 2. variety, 65.6, 66.15, 67.17; type, 73.3,10, 83.2, 170.14, 211.1,14, 214.3, 223.6; kind, 95.15, 98.4. C68.18.
Too late | maḍrūb: 1. multiplicand, 92.12, 95.4, 96.4, 99.17, 102.4, 149.4, 226.9.
| 2. multiplication, 93.3, 98.9.
Too late | maḍrūb fīhi: multiplier, 92.12, 95.4, 96.4, 101.1, 103.4, 149.4, 226.9.
Too late | ḍaʿafa: to duplicate, 149.5, 186.1.
Too late | taḍʿīf: duplication; duplicating, 95.2,5, 114.7, 149.2, 184.10, 186.1. C74.17, C95.2, C149.2, C179.1, C186.1.
Too late | ḍilʿ: side, 66.18, 67.2,13, 68.9,12. C66.17.
Too late | ḍāfa: 1. to add, 97.15, 102.6, 105.3, 110.10, 215.18. C73.2.
| 2. to attach, 124.5, 135.1, 144.4.
Too late | muḍāf: combined, 137.14. C137.13.

|                       ط
Too late | ṭaraḥa: 1. to subtract, 76.4, 83.7, 84.2, 147.1, 179.1, 211.3, 220.1, 221.1. C83.1.
| 2. to cast out, 87.18, 88.1, 91.2, 125.16.
Too late | ṭarḥ: 1. subtraction, 83.1, 91.12, 147.3, 176.11, 211.7, 219.1. C83.1,2, C219.4.
| 2. modulus, 91.13, 125.6, 126.3, 127.1.
Too late | maṭrūḥ: subtrahend, 83.6, 86.15, 148.1, 182.1, 220.13.
Too late | maṭrūḥ minhu: minuend, 83.6, 86.4, 91.12, 148.1, 220.10.
Too late | munṭariḥa (and related forms): cast out entirely, 87.22, 88.6, 125.8,9, 126.1, 127.1.
Too late | maṭlūb: required number, 76.14, 101.14, 147.12, 171.3, 198.18, 220.17, 228.13, 231.12.

|                       ع
Too late | ʿadda: to count, 66.8, 127.11, 128.1,12, 166.1. C127.9.
Too late | ʿadad: number, 61.9, 65.2, 68.18, 69.15, 71.13, 73.2, 133.1, 163.3, 170.7, 173.13, 211.8. C65.2, C219.1.
Too late | ʿidda: number, 79.1,3,10.
Too late | ʿadala: to equal, 211.13,15, 216.13, 223.2, 227.14, 227.16, 228.1. C211.13, C213.7.
Too late | muʿādala: 1. equation, 217.2,4, 227.16,18, 228.2. C213.7.
| 2. equalization, 211.5, 220.11, 221.8. C213.7.
Too late | mutaʿādilān: two sides of [the] equation, 220.8, 223.1.
Too late | ʿaqd: power of ten (lit., “bent finger-joint”), 87.17, 109.19, 110.7, 111.3. C110.5.
Too late | maʿqūl: intelligible, 133.6. C133.3.
Too late | ʿilla: cause, 87.11.
Too late | ʿalā: to surpass, 121.6. C73.2.
Too late | maʿnā: meaning, 68.12, 94.12, 95.5,9,11,13, 113.2, 117.9, 211.1, 213.8. C95.3.

|                       غ
Too late | ghirbāl: sieve, 127.10. C127.9.

|                       ف
Too late | fard: odd, 65.6.
Too late | mufrad (and related forms): simple, 109.20, 137.13, 211.14, 213.2. C137.13, C211.13.
Too late | munfaṣila: detached, 141.9, 142.8,13. C141.7.
Too late | munfaṣil: 1. apotome, 173.4, 173.16, 176.4,18, 182.9, 188.18. C173.10.
| 2. discrete, 117.4, 133.8. C117.2, C133.3.
Too late | faḍala: to exceed, 78.1,7. C73.2.
Too late | faḍl mā bayna: difference between (sometimes just mā bayna or faḍl bayna), 78.2, 111.21, 173.9, 197.1, 201.1, 204.14, 229.5.
Too late | tafāḍul (and related forms): disparity, 73.7,15, 76.7, 79.1.

|                       ق
Too late | qabila: 1. to confront, 198.10, 199.5, 207.3, 217.7, 221.19, 223.9. C223.1.
| 2. to match, 99.18.
Too late | muqābala: confrontation, 199.6,12, 211.3, 221.8. C211.1, C220.5.
Too late | qadr: size (of a ratio), 124.6.
Too late | taqrīb: approximation, 166.12, 168.6, 169.1, 173.2.
Too late | qasama: to divide, 61.10, 65.2, 72.1, 93.10, 117.16, 118.1, 119.18, 151.4, 187.2, 189.1, 229.2,12,16, 230.13. C117.2.
Too late | qism: 1. kind, 65.2, 73.6; type, 82.1, 95.3, 157.11; variety, 163.2. C68.18.
| 2. part, 118.8, 139.1, 144.8, 202.4,10.
Too late | qisma: division, 72.1, 117.2, 118.14,15, 120.20, 151.4, 188.18, 229.6. C68.18, C95.2, C117.2,9, C118.14, C230.11.
Too late | maqsūm: dividend, 93.14, 117.8, 119.2, 151.5, 152.14, 187.3, 229.2,14. C93.6.
Too late | maqsūm ʿalayhi: divisor, 93.6, 117.2, 118.4,18, 119.7, 151.5, 181.15, 188.7, 229.3,18. C93.6.
Too late | munqaṭiʿ: disconnected, 140.1,18, 143.1. C140.1.
Too late | qānūn: basic rule, 170.7, 191.1.
Too late | quwwa: area (lit., “power”), 177.8,14. C163.4, C177.5. See also munṭaq fī l-quwwa below.
Too late | maqām: denominator, 122.15, 171.2. C93.6.
Too late | qiyās: rule, 175.18, 183.10, 184.7, 187.18.

|                       ك
Too late | kasr: fraction, 65.2, 93.15, 133.4, 134.1, 135.10, 143.5, 154.6, 157.4, 170.10. C65.2, C131.1, C133.3, C134.2.
Too late | mukaʿʿab: cube, 67.12,15, 68.14,15, 74.1, 80.2. C67.12, C211.8.
Too late | kaʿb: 1. cube root, 67.13,16, 68.9,12. C67.12.
| 2. cube (the third power in algebra), 221.1, 225.2,13, 226.7, 227.10, 229.8. C67.12, C211.8.
Too late | kamm (and kammīya): quantity, 73.17, 93.5. C73.17, C92.17.
Too late | kayf (and kaifīya): quality, 73.7, 93.5. C73.17, C92.17.

|                       ل
Too late | lafẓ: 1. term, 95.5,8,11,13, 184.9, 188.1. C95.3.
| 2. word, 70.3, 73.24, 117.11.

|                       م
Too late | māl: 1. quantity, 199.1,19, 200.1, 201.1, 203.13, 213.8,14, 215.7. C211.8.
| 2. māl (the second power in algebra), 211.8,15, 213.7, 214.9, 215.12, 225.4,8, 226.1. C211.8.

|                       ن
Too late | nazala: to remove, 114.8, 123.1, 142.2, 146.3, 221.5. C83.1.
Too late | manzila: place, 69.1,17, 71.8,19, 76.18, 83.10, 90.4, 112.6, 164.1. C68.18.
Too late | nisba: 1. ratio, 73.8, 78.4, 108.10, 113.10, 117.5, 123.21, 124.6, 133.3, 163.3, 195.7. C131.1, C133.3, C193.1.
| 2. proportion, 121.23, 191.4, 195.2. C193.1, C195.2.
Too late | nisba ʿadadiyya: arithmetic progression, 73.17, 74.5.
Too late | nisba handasiyya: geometric progression, 73.7, 78.13. C73.17.
Too late | muntasib: related, 135.10, 138.1, 139.4. C135.1,10.
Too late | munṭaq: rational (lit., “expressible” in speech), 163.2, 167.9, 172.10, 174.2, 177.8, 183.10. C134.2.
Too late | ghayr munṭaq: surd (lit., “inexpressible”), 163.2,4,6. C163.2.
Too late | munṭaq fī l-quwwa: rational in square (lit., “expressible in power”), 163.7, 180.7, 188.7. C163.4.
Too late | naqaṣa: to subtract, 112.9, 168.14, 173.6, 183.16, 204.10, 215.4, 232.8. C83.1.
Too late | nāqiṣ: 1. deleted, 87.11, 189.3, 211.5, 220.10, 228.9. C219.4.
| 2. falls short, 198.12, 199.7, 202.7, 205.19. C219.4.
Too late | tanqīl (and related forms): shifting, 95.16, 97.3, 98.8, 99.14, 157.7.
Too late | muntahā ilayhi: upper extreme, 79.13, 80.5,20, 81.2, 82.2.
Too late | nāfa: excess, 108.9.
Too late | nawʿ: species, 65.10, 66.7, 71.13, 211.3,8, 220.5, 226.9; kind, 66.16, 67.18, 157.2, 158.1, 195.2; type, 92.15, 97.1, 99.14, 118.14, 207.9. C68.18.

|                       و
Too late | wa: and, C219.1.
Too late | muwassaṭ (and mutawassit): medial, 163.8, 176.14, 180.8, 187.10, 188.7. C163.4.
Too late | muwassaṭayn: bimedial, 176.13,18, 177.14. C176.10.
Too late | waṣala: to join (i.e., form a binomial), 174.15, 175.1,3,5,7,9. C173.10.
Too late | muttaṣil: 1. continuous, 117.6. C117.2, C133.3.
| 2. connected, 140.14,16, 141.9, 143.1. C140.14.
| 3. joining, 177.9,15. C173.10.
| 4. binomial (lit., “joining”), 188.19,20, 189.12,14. C173.10.
Too late | muttaṣila: connected, 142.2,8,11. C140.14.
Too late | mawdhiʿ: place, 108.12, 109.5. C68.18.
Too late | wafiqa: to reconcile (i.e., cancel common factors), 120.3,17, 123.1,5. C119.18.
Too late | ʿalā tawālī: consecutive; consecutively, 79.13,15, 80.2,5, 81.16, 89.6.

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