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Volume XXI February, 1914 Number 2 


By L. C. KARPINSKI, University of Michigan. 

Popularity of Algebra Among the Arabs. The great interest in the study 
of algebra among the Arabs of the Middle Ages is attested by the numerous works 
written upon this subject by Mohammedan mathematicians. The period of 
this popularity extends from the ninth to the fifteenth century. The students 
of algebra included poets, philosophers, and kings. The popular Omar Khayyam 
was too excellent a mathematician and too fine a poet to woo the muse of algebra 
in verse. But his treatise in prose on algebra, edited with French translation by 
Franz Woepcke, is as good a title to fame as his verse. In the library of the 
Escurial is preserved a poem written six hundred years ago by a native of Granada, 
Mohammed al-Qasim, and treating of algebra. Needless to say the content does 
not compare with the prose of Omar Khayyam. The King of Saragossa Jusuf 
al-Mutamin (reigned 1081-1085) was a devoted student of the mathematical 
sciences. The title of one of his works would seem to indicate its algebraical 
nature. Even in the study of law a knowledge of algebra seems to have been 
necessary, for various questions of inheritance were treated by this science. On 
the whole, however, in that happy day the question of practical applications of 
mathematics did not loom so large upon the horizon of the Arabic scientist. 
That fortunate individual felt entirely free to pursue this subject as one of the 
natural and inevitable activities of the human intellect. 

The First Great Arabic Writer on Algebra. The first systematic treatise on 
algebra was the product of an Arabic writer, Mohammed ibn Musa, al-Khowariz- 
mi (c. 825 A. D.). Commentaries were written upon this work by Arabic authors 
and two different Latin translations were made of it. One of these translations 
was published by Libri in his Histoire des sciences mathematiques en Italie, Vol. I, 
253-297, Paris, 1838. This work is out of print. While the authorship of this 
version is uncertain the composition may safely be placed in the twelfth century, 
as several writers of that century appear to have studied this work and use the 
terminology of this translation. Manuscript copies are somewhat common in 



European libraries. French, German, and Italian writers of the fifteenth and 
sixteenth centuries drew inspiration, and copy, from these pages. The other 
Latin translation was effected in 1144 A. D. (1183 of the Spanish era) by an 
Englishman, Robert of Chester, the first translator of the Koran. This version 
is found, so far as known, only in three manuscripts. 1 The text with English 
translation and notes is being published in Vol. XI, University of Michigan- 
Studies, Humanistic Series. The Arabic text with English translation was pub- 
lished by F. Rosen, London, 1831, but unfortunately this last is out of print. 
Adequate treatment of this initial work in the science of algebra is found in 
Cantor's Vorlesungen iiber Geschichte der Mathematik, volume I. 

The Second Great Arabic Writer on Algebra. The name and fame of Al- 
Khowarizmi is known to most students of mathematics. But fate has not dealt 
so kindly with the second, in point of time, of the great Arabic writers in this 
field. Abu Kamil Shoja' ben Aslam 2 wrote in the late ninth or early tenth century 
a much more extensive treatise on algebra than that of Al-Khowarizmi. Two 
commentaries of the tenth century have been noted by Arabic historians. In 
addition to these a commentary in Arabic was prepared by a Spanish Arab, 
Al-Khoreshi, and a Spanish translation by an unknown Christian Spaniard. 3 
This was followed by a Hebrew translation, which is preserved in manuscript, 
by Mordechai Finzi (c. 1473) of Mantua. Not alone by these translations and 
commentaries was the influence of Abu Kamil exerted upon later writers in this 
science; for two prominent algebraists made large use of his treatise without, 
however, mentioning his name. These two men are Al-Karkhi (died c. 1029), 
whose treatise entitled Al-fakhri was analyzed by Woepcke (Extrait du Fakhri, 
Paris, 1853), and Leonard of Pisa (1202). Nor must it be understood that there 
is any suggestion here of plagiarism, for undoubtedly at the time that Al-Karkhi 
wrote, Abu KamiPs methods were so well known that any writer could employ 
his results as common property, while the phraseology of Leonard of Pisa, in the 
statement of many problems drawn from Abu Kamil, is plainly intended to carry 
the idea of a citation. Al-Karkhi reproduced not only many problems given by 
Abu Kamil but also the geometrical solutions devised by this writer, while 
Leonard of Pisa drew copiously from Abu Kamil for the problems in the section of 
his Liber Abbaci entitled "Expliciunt introductiones algebre et almuchabale; 
Incipiunt questiones eiusdem." 4 

Abu Kamil's Works. Abu Kamil Shoja' ben Aslam ben Mohammed ben 
Shoja, the reckoner from Egypt, was an excellent and learned arithmetician. 
He wrote : The book of fortune, The book of the key to fortune, The book on 
algebra, The book of extracts, The book of omens (by the flight of birds), On 

1 Karpinski, " Robert of Chester's Translation of the Algebra of Al-Khowarizmi," Bibliotheca 
mathematica, third series, XI, 125-131. 

2 Karpinski, " The Algebra of Abu Kamil Shoja'ben Aslam," Bibliotheca mathematica, third 
series, XII (1912), 40-55. To this article the reader is referred for more complete bibliographical 
references and for citations from the Latin text. 

3 H. Suter, in a personal communication to the author. 

*Scritti di Leonardo Pisano, published by Prince Boncompagni, Rome, 1857, I, 410-459. 


completion and diminution, The book on the rule of double false position, The 
book of surveying and geometry, The book of the adequate (possibly in arith- 
metic). 1 Such is the account of the life and works of Abu Kamil as given in the 
Kitab al-Fihrist (987 A. D.) of An-Nadim, who includes this writer in his list 
of the later (?'. e„ nearly contemporary with An-Nadim) reckoners and arith- 
meticians. Aside from the works included in this list a treatise on the mensura- 
tion of the pentagon and the decagon exists in Latin and Hebrew translations, 
and a further arithmetical work is preserved in the original Arabic as well as in 
Latin and Hebrew. The work on the pentagon and decagon has been published 
in Italian by Sacerdote, 2 and in German by Suter. 3 The arithmetical work, in 
which he deals with first degree indeterminate equations, has been translated by 
Suter. 4 Other information about the life and activity of this great Arab we do 
not have. The titles mentioned indicate that, like so many of the later scientists 
in Europe, Abu Kamil was interested in magic, doubtless including astrology, and 
in omens. 

The Sources of Information. The manuscript upon which I have based 
my study of Abu Kamil's algebra is the Paris manuscript 7377A of the Biblio- 
theque nationale. 5 The treatise covers folios 71 T -93 T , and contains between 35,000 
and 50,000 words. 6 

In the opening sentence the author refers to his famous predecessor in this 
field, Mohammed ibn Musa al-Khowarizmi, and a little later again refers to the 
algebra of the same writer. Several references are made to Euclid but aside from 
these two names no others have been noted in this algebra. 

Quotations from the Manuscript. The following translation of selected 
passages from the algebra is a very free one, as an attempt is made to preserve 
the mathematical significance. Additions to clarify the meaning, and modern 
notations, are put in parentheses. 

The first thing which is necessary for students of this science is to understand the three species 
which are noted by Mohammed ibn Musa al-Khowarizmi in his book. These are roots, squares 
and number. A root is anything which can be multiplied by itself, composed of one, and numbers 
above one, and fractions. A square is that which results from the multiplication by itself of root, 
composed of units, and units and fractions. A number is a quantity by itself which does not have 
the name of root or square, but is proportional to the number of units which it contains. These 
three species are proportional to each other by turn in twos. Thus, squares equal to roots, squares 
equal to number, and roots equal to number (ax 2 = bx, ax 2 = n, bx = n). 

An illustration of squares equal to roots is this: 

A square is equal to five of its roots (x 2 = 5x). The explanation of this is that a square is 
five times the root of it. The root of the square is always the same as the number of the roots 
to which the square is equal. In this question this root is five and the square is twenty-five 

1 1 translate from the German translation of the passages in the Fihrist dealing with mathe- 
matical scientists, by Suter, in Abhandl. z. Geschichte d. math. Wissen., VI, 1-87. 

2 Festschrift Steinschneiders, Leipzig, 1896, 169-194. 

8 Bibliotheca mathematica, X, third series, 15-42. 

4 Das Buch der Seltenheiten der Rechenkunst, in Bibl. math., 3d ser., XI, 100-120. 

6 1 am indebted to the librarian for permission to have photographic reproductions made. 

6 To give an idea of the extent of this work, if the Latin text were printed in the Monthly 
it would take between 75 and 100 pages. 


which is the same as five of its roots. That the root of the square is the same as the number of 

the root this I will explain. 

We place for the square a square surface abgd whose sides are 06, bg, gd, and da (see Fig. 1) . 

Of this square any side multiplied by one in number is the same as the area of a surface whose 

length is the side or root of the square. The side ab multiplied by one which is be gives the surface 
ae which is the root of the square ag. The surface ag is the same as five roots 
or five times the root of itself. Therefore, it is five times the surface ae. 
Divide then, the surface ag by equidistant lines into five equal parts, that is 
the surfaces ae, ck, sr, nh, and mg. The lines be, eh, kr, rh and hg are equal. 
Moreover the line be is one. Therefore the line bg will be five which is the 
root of the square and the square itself is twenty-five which is the surface ag. 
This is what we wished to explain. 

FiQ- 1. If it is proposed that one half a square should equal 10 roots (Jx 2 = lOz) 

then the whole square is 20 roots. The root of the square is 20 and the square 
400. Similarly if it is proposed that 5 squares equal 20 roots (5x 2 = 20z) then the root of the 
square is 4 and the square is 16. And whether more than a square or less than a square (is 
proposed) you reduce to 1 square. You operate in the same manner with that which is con- 
nected with the square, the roots. 

A square which is equal to a number is illustrated: 

A square equals 16 (z 2 = 16). Therefore, it itself is 16 and the root of it is 4. Similarly if 
given 5 squares equal 45 drachmas (5z 2 = 45), then 1 square is a fifth of 45 which is 9 and the root 
of it is 3. . . . So whether more or less than a square (is given) it is reduced to 1 square and 
similarly is treated that which is joined or equal to them of numbers. 

Roots which are equal to numbers are illustrated: 

A root equals 4 (x = 4). Therefore, the root itself is 4 and the square is 16. Similarly again 
if it is proposed that 5 roots should equal 30 (5x = 30) then the root is equal to 6 and the square 
36. . . . 

We construct now these three species, that is roots, squares and numbers, joined by twos and 
proportional to the third. This gives, a square and roots equal to number, a square and number 
equal to roots, and roots and number equal to squares. 

(In modern notation ax 2 + bx — n, ax 2 + n = bx, bx + n — ax 2 .) 

First Type of Quadratic Equation. An illustration of squares and roots 
equal to number is the following: 

A square and 10 roots equal to 39 drachmas (x 2 + 10z = 39). The explanation of this is 
that there is some square to which if you add the same as 10 of its roots the number of it results 
in 39 drachmas. In this question there are two ways (methods) of which one leads you to know 
the root of the square and the other to know the square of the root. We will set forth and explain 
each of these by a geometrical figure which will be understood by those who understand the book 
of Euclid, that is the Elements. The method which leads us to know the root of the square has 
been narrated by Mohammed ibn Musa al-Khowarizmi in his book. This is that you divide the 
roots in half, that is in two equal parts. In this question this gives 5. You multiply this by itself, 
giving 25, which you add to 39. This gives 64 of which number you take the root which is 8; 
from this you take the half of the roots, that is 5, and there remains 3, which is the root of the 
square. The square is 9. 

The method indeed which leads you to know the square is that you multiply 10 roots by itself, 
giving 100. This you multiply by 39, which is equal to the square and roots, giving 3,900. Now 
divide 100 in halves, giving 50, which you multiply by itself, giving 2,500. This add to 3,900, 
giving 6,400, and of this number take the root, which is 80. This you subtract from 50, which is 
the half of 100, and from 39, which is equal to the square and the roots, together giving 89, and 9 
remains, which is the square. 1 

1 The text to this point is found in the Paris MS. on folio 7r-72 D ; given in my article in the 
Bibliotheca Mathematica, he. cit., p. 42-44. 


In modern notation the steps of this solution are indicated as follows, in which 
both the general case and this special problem are given in parallel columns. 

(1) a: 2 + bx = n, x 2 + lOx = 39, 

(2) bV + Vx = nV, 100a; 2 + 1000a; = 3,900, 

/ ft 2 \ 2 ft 4 

(3) 6 2 x 2 + Vx + ( |- J = nb* + ~, 100a; 2 + 1000a; + 2500 = 6,400, 

(4) bx + | = + -yj nV + j, 10a; + 50 = 80. 

Only the positive root is taken, as the negative root leads to the negative solution of the equation. 
Subtracting (4) from (1), member for member, 


v r b* 

- <L = n - yj nV + -r, x* - 50 = 39 - 80, 

Z 2 = n + ¥■ - y nV- + j, x* = 50 + 39 - 80, 

a; 2 = 9. 
Following this the problem is proposed: 

2a; 2 + 10a; = 48. 

This is a problem which is also given by Al-Khowarizmi and the same is true of 
many further problems given by Abu Kamil. The latter author gives the method 
of solution presented by Al-Khowarizmi and also the method, above illustrated, 
leading to the value of x 2 directly. Further than this he presents the geometrical 
solution for the first method as given by his more famous predecessor. Abu 
Kamil adds the geometrical solution corresponding to his own method, the second 
of those given above. This seems to be worth presenting here. 

In order to show you the method (the door, literally) which leads to a knowledge of the square 
we place for the square the line ah (see Fig. 2). To this we add 10 roots of itself which are repre- 
sented by the line bg, giving the line ag, which is 39. We wish to know then the value of the line 
ah. We construct then upon the line bg a square surface, which is the surface degb. This then 
will be 100a; 2 (in modern notation), that is, 100 times the line ah. . . . We construct then the 
surface ah the same as the square surface be, that is equal to the 
square, which is the same as the multiplication of the line ah in 
one of the units which it contains taken 100 times. Then we 
complete the surface an which is 3,900, since the line ag is 39 and 
am is 100, which lines contain this surface. Moreover the surface 
ah is the same as the surface be. Therefore, the surface dn is 
3,900, which is produced by the multiplication of the line ne by 
eg, as the line eg is the same as the line ed and the line gn is 100 
since it is the same as the line am. Therefore we divide the line Fig. 2. 

gn in two equal parts at the point I, to which line so divided the 

line ge is added in length. Therefore, the multiplication of the whole line ne by eg together with 
the product of the line gl by itself is the same as the product of the line le by itself, as Euclid 
says in the second book 1 of his Elements. The product of ne by eg is 3,900 and the product of gl 

1 Euclid's Elements, II, 6, " If a straight line is bisected and a straight line be added to it in a 
straight line, the rectangle contained by the whole with the added straight line and the added 
straight line, together with the square on the half, is equal to the square on the straight line 
made up of the half and the added straight line." 

e 9 










by itself is 2,500, which being combined make 6,400, which is the same as the product of the line 
le by itself. Therefore, the product of the line le by itself is 6,400, of which the root is 80, which 
is the line le. But the line ge is the same as the line gb, and so the line Ig and the line gb make 80. 
When we take away the line Ig and the line dg, which are 80, from the lines ag, 39, and gl, 50, 
which make 89, the line ah remains, which is 9. This is the square and that is what we desired to 
explain. 1 

Second Type of Quadratic Equations. The first problem which Abu Kamil 
presents of the second type of composite quadratic equations is again the same 
as that given by Al-Khowarizmi, 

x 2 + 21 = 10a;. 

For this type of equation also Abu Kamil presents two solutions, the one leading 
directly to the value of the root of the equation and the other to the value of the 
square of the root. Further the author notes that there are two solutions for 
both the root and the square in this type of quadratic equation. The reason that 
this was noted by both of these Arabic writers is that in this type of equation the 
two roots are positive, whereas in the other two types the one root is positive and 
the other root is negative. Negative quantities as such are not accepted by 
these Arabic mathematicians and this, of course, accounts for the three types of 
quadratic equations. Express mention is also made of the fact that if the square 
of half the coefficient of the roots is less in magnitude than the constant then the 
problem is impossible, from the Arabic standpoint, or contradictory. Further 
also, the author mentions that when this square is equal to the given constant 
the root of the equation is the same as one half the given coefficient. 
In the above problem, 

* 2 + 21 = 10*, 

the analytical solution leading to the value of the square of the unknown quantity 
proceeds in substance as follows: 

Multiply 10 by 10, giving 100. Multiply this by 21, giving 2,100. Then take half of 100, 
giving 50, which you multiply by itself, giving 2,500. From this subtract 2,100, leaving 400 of 
which the root is 20, which you subtract from 50, the half of 100, leaving 30. From this you sub- 
tract 21, leaving 9, which is the square. And if you wish, add 20 to 50, giving 70, from which you 
subtract 21, leaving 49, which is the square and the root of it is 7. 2 

Al-Khowarizmi gave the geometrical figure corresponding only to the smaller 
root of the equation. Thus in the equation, * 2 + n = bz, in which 

b |6 2 

one of these roots, when both are real, is necessarily less than b/2 and the other 
is greater. Abu Kamil gives the geometrical explanation corresponding to both, 
thus completing the work of Al-Khowarizmi. Thus he says that when you 
assume that the square is less (in numerical value) than the number which is 
given, then the solution appears by subtraction, and when you assume the same 

!In the manuscript folio 72 v -73 r ; in Bib. Math., loc. cit., p. 46. 
2 Manuscript folio, 73 r ; Bib. Math., loc. cit., p. 47. 



greater than the constant the solution appears by addition. Further he gives a 
geometrical solution for the type in which the square of the coefficient of x is 
equal to the given constant. The figure in this case consists simply of two equal 
squares with a side in common. We follow with a demonstration, original with 
Abu Kamil, leading to value of x 2 in the equation, x 2 + 21 = 10a:. 

. . . Place the line ab to represent x 2 (Fig. 3), and we add to it the drachmas which accom- 
pany it, 21, and let it be the line bg. Therefore, the line ag is 10 roots of the line ab. We construct 
then upon the line ag, a square surface, the square aged and this is 100 times the line ab . . . since 
the line ag is 10 roots of the line ab, and 10 roots multiplied by itself gives 100 squares (x 2 ). We 
oonstruct then the surface ah equal to the square aged and let one side of it be the line ab. There- 
fore, the other side bh is 100. Then we complete the surface an and therefore the surface bn is 
2,100 since the line bg is 21 and the line gn 100. We construct then the surface my equal to the 
surface ae, but the surface ae is equal to the surface ah. Take away, then, the surface th and 
there remains the area en equal to the area ca. Adding then the surface by, the whole surface ay 
is equal to the surface bn, which is 2,100. Therefore the area ay is 2,100 which is the product of 
gy by yn, since yn is equal to yt, since the area in is a square. We divide then the line gn in two 
equal parts at the point I. Therefore the line ng is divided into two equal parts by the point I 
and into two unequal parts by the point y. Therefore the product of gy by yn together with the 
square upon ly is the same as the product of In by itself. 1 But In by itself makes 2,500, since it 
is 50, and the product of gy by yn is 2,100. The square on the line ly is then left as 400 and the 
line ly 20. But In is 50 and as it (ly) is to be subtracted the line yn is left as 30. Since the line 
bg is_21 the remaining line ab is 9 which is the square. This is what we wished to explain. 2 

9 y 

Fig. 3. 

9 y 

Fig. 4. 

Having given the explanation "by subtraction," that is to say, when the 
square root of 2,500 — 2,100 is subtracted from 50, Abu Kamil proceeds to explain 
"by addition." gn'is again divided into two equal parts but the point of division 
falls between y and n (Fig. 4). This corresponds to the assumption that the 
square is greater than the number which accompanies it. This has also been 
noted above. 

The product of gy by yn together with the square of yl equals the product of Ig by itself. 
Ig multiplied by itself gives 2,500 and the line yn by yg gives 2,100. Therefore the square of ly 
is 400 and ly is 20. Since In is 50, the whole line yn is 70. But yn is the same as yt and yt equals 
ag. Therefore ag is 70. But bg is 21 whence ab is 49 which is the square. This is what we desired 
to explain. 3 

In the third type of quadratic equations Abu Kamil again follows Al-Kho- 
warizmi in taking the equation 3x + 4 = x 2 . Three explanations are presented 
for finding the root of the square. The first is equivalent to the demonstration 
presented by Al-Khowarizmi but the figure is not completed. Instead of this 
the proof is made to depend upon the second book of Euclid. The second gives 

^This is by Euclid VI, 5, " If a straight line be cut into equal and unequal segments, the 
rectangle contained by the unequal segments of the whole, together with square on the straight 
line between the points of section, is equal to the square on the half." 

2 Manuscript folio, 74 r ; Bib. Math., loc. cit., pp. 48-49. 

8 Manuscript folio, 74 r ; Bib. Math., loc. cit., p. 49. 



the completed figure and explains as in the algebra of Al-Khowarizmi. In his 
third figure (Fig. 5) the square abdg is placed to represent x 2 which by the con- 
ditions of the problem equals 3x + 4. Then the lines ah and yd are taken If 
units in length and the rectangles completed cutting off from the original square 
3x less the little square, If on a side, which these two rectangles have in common. 
Hence the square which remains hnyz is 4 + (If) 2 or 6|. It follows that gh is 
2f and ay (or x) is 4. 

9 h 
V n 


Fig. 5. 

Fig. 6. 

To arrive at the value of x 2 geometrically Abu Kamil places (Fig. 6) ab for 
the square and cuts off gb equal to 4, leaving ag equal to the three roots. Then 
he continues: 

Upon ag a square is constructed aged whose area is 9x 2 or 9 times the line ab. ah is constructed 
equal to the square ae. From this it follows that an is 9 and that gc which equals an is 9. Since 
bg is 4 the area of gh is 36. Take ay equal to an and draw yl parallel to de. The area yg equals ac 
and the area ae equals ah. Hence ye equals cb which is 36. But ya is 9 since an is 9. Divide 
ay into two equal parts by the point m. Now yd is added in length to ay. Hence ad by dy plus 
ym multiplied by itself equals md multiplied by itself, as Euclid says in his second book. But 
ad by dy is ye or 36. ym multiplied by itself is 20|. Adding you obtain 56J. Therefore md 
multiplied by itself gives 56J and md is 7J. But am is 4J and gb is 4. Hence ab is 16 and this is 
what we desired to demonstrate. 1 

Abu Kamil summarizes the results of the algebra up to this point, stating that 
all questions that can be solved by algebra and almuchabala, that is, restoration 
and opposition, must reduce to one or the other of the six types of equations 

Multiplication and Division. In the next section the multiplication and 
division of algebraic quantities is considered. Binomials, involving both addi- 
tion and subtraction, are given particular consideration. The writer uses res 
and radix, indifferently, for the first power of the unknown. This is combined 
by addition or subtraction with a pure number. The second power of the 
unknown is designated by census, but this word is sometimes used simply for 
any unknown quantity as in problems below. Similar usage obtains in the 
work of Leonard of Pisa and in the Liber augmenti et diminutionis by one 
Abraham, published by Libri, Histoire, I, 304-371. 

After a general statement about multiplication which summarizes the work 
presented, Abu Kamil continues with a geometrical explanation of the problems. 
Thus a figure containing four small squares is used to illustrate the fact that 2x 
multiplied by 2x gives 4a; 2 . The next problem is 3a; multiplied by 6 drachmas. 

'Manuscript folio, 75 r ; Bib. Math., loc. cit., pp. 49-50. 


In this section the constant is referred to drachmas although in other places the 
word numerus is also used. Figures are shown for 10 -f- x by x, 10 — x by x, 
10 + x by 10 — x and other similar examples. 

Radical Expressions. The addition and subtraction of radicals, involving 
quadratic irrationalities only, is effected by means of the relations given in modern 

symbols by the equalities V a =*= Vb = * a + b =*= 2 ^ ab. Thus, to subtract 
the square root of 8 from the square root of 18 the rule is simply: "subtract 
24 from 26, as you know, and 2 remains. The root of this is the root of 8 sub- 
tracted from the root of 18." This problem, by the way, is given by Al-Karkhi. 
Leonard of Pisa 1 employs the same method but uses 18 and 32, instead of 8 and 
18, combining them by addition and subtraction in the same way as Abu Kamil. 

Al-Karkhi 2 proceeds to the consideration of cube roots in a similar way but 
our author takes up for a final problem in roots the addition of l/lO to the V2 
which, he states, does not lead to a simple result since 10 : 2 = 5 which is not a 
square anld aso 2 : 10 = \ which is not a square. This section concludes with 
the statements that l/l0+ 1/2= ^/l2+2 ^20 and l/10-V / 2= A /l2-2v / 20. 

Problems. The section dealing with problems is introduced in much the 
same way that Al-Khowarizmi does the corresponding section, and many of the 
problems are taken from the older work. The first problem is to divide ten into 
two parts such that the square of the larger part should equal If times the product 
of the two parts. This leads to the equation 

x 2 = lfz(10 - x), 

2|a; 2 = 15a;. 

Thus the problem leads to the first of the six types of quadratic equations. 

The above problem is the second presented by Leonard of Pisa 3 in the section 
with the heading: Expliciunt introductiones algebre et almuchabale. Incipiunt 
questiones eiusdem. Furthermore this problem does not appear in Al-Karkhi 4 
in the list given by Woepcke. The conclusion is reasonable that Leonard had 
some source of information concerning the algebra of Abu Kamil other than 
Al-Karkhi's work, and even more particularly in view of the long series of prob- 
lems found in Leonard's work which correspond to those given by Abu Kamil. 
The second and third questions here are found to be the same as the following 
two questions in Leonard of Pisa, 5 but with slightly different numbers. Thus in- 
stead of 6|a; 2 = 100, Leonard and Al-Khowarizmi give 2^x 2 = 100 and in place of 
xj (10 — x) = 4 as given by Abu Kamil and Al-Khowarizmi, Leonard sets this 
same fraction equal to 2\. Our space does not permit an examination of all 
similarities of this kind. 

1 Scritti di Leonardo Pisano, published by Boncompagni, Vol. I : II liber abbaci, Roma, 1857, 
p. 363-365. 

2 F. Woepcke, Exlrait du Fakhri, Paris, 1853, p. 57-59. 

3 Liber abbaci, p. 410. 

4 Woepcke, loc. cit., p. 75-137. 

5 Liber abbaci, p. 410. 


A problem given on folio 81* is of particular interest because of its similarity 
to a problem of Al-Khowarizmi's 1 which has given rise to discussion. The 
problem is to divide 50 among some number of men and then to add three men 
and again distribute 50 (drachmas) among them (equally). In the second 
instance each man receives 3f j (meaning 3 1/2 + J) drachmas less than in the 
first. Leonard states the corresponding problem as follows : 

Distribute 60 among (a certain number of) men and each will receive something. Add two 
men and among them all again distribute 60 and then each man will receive 2J denars less than 
at first. 

The statement of Abu Kamil is : 

And if we tell you, distribute 50 among (a certain number of) men and each receives a certain 
amount (res). Add three men and distribute again 50 drachmas among them. Each then 
receives 3§ J drachmas less than before. 

Both follow with a geometrical explanation. 

Problems Involving Fractions. The notation of fractions employed by this 
translator of Abu Kamil resembles the peculiar system employed later by Leonard. 
Thus on fol. 83 r the square of § and of ^ are given in combinations as in the 
following translation: 

. . . = and - which multiplied by itself, gives -r and — . To this we add 11 drachmas and - 
ji y 4 y y y 

giving 11 drachmas and -r and^ and — and -r-z. Of this we take the root or 3 and - and ^ . . . . 
4 o y 2* y y o y / 

The fraction q-~ stands for -z plus ^r whereas ^ as well as ^ represents simply 

1 5 e 5 

75. So also Leonard 2 uses 5-5 for ^7. This system is explained by Leonard 3 

who writes for 77 the form ^ and 2 „ 1f) for ttc + ™ + ^on • ^he difference is 

only in the order of reading. 

Al-Khowarizmi gives several problems of this nature: 4 

To find a square of which if one-third be added to three dirhems, and the sum be subtracted 
from the square, the remainder, multiplied by itself restores the square. 

The solution involves the use of x (root) to represent the square. So also Leonard 5 
in a similar problem states: "... place for that square x (res)." Our version of 
Abu Kamil has several problems of the same kind : 

And if we tell you there is a quantity (census) from which when you subtract J of itself and 
2 drachmas, the product of 'the remainder by itself gives the quantity and 24 drachmas. 

The solution is obtained by placing x (res) for the quantity (census). 

1 Libri, Histoire des sciences mathematiques en Italie, Vol. 1, Paris, 1838: "Liber Maumeti . . . 
de algebra et almuchabala," p. 286. Rosen, The Algebra of Mohammed ben Musa, London, 1831, 
p. 63-64. Al-Khowarizmi does not present any geometrical explanation. 

2 Liber abbaci, p. 447. 4 Rosen, loc. cit., p. 56, 57. 

3 Liber abbaci, p. 24, 25. 6 Liber abbaci, p. 422. 


The longest discussion of any algebra problem in Leonard is that of the 
following: 1 

Divide 10 in two parts, and divide the larger by the smaller, and the smaller by the larger. 
Sum the results of the division and this equals the square root of 5. 

Leonard presents several solutions. In the first he arrives at the equation 
VEx* + 2a; 2 + 100 = 20a: + 1/500?. The coefficient of x 2 is made equal to 
unity by multiplying through by V5 — 2. The same equation is found in 
Abu Kamil and the treatment is the same. The value of x is 5 — \/225 — ^50000 
w hich is found by both writers. Leonard goes on to find the value of 

-^225 - ^50000 as l/l25 - 10. Al-Khowarizmi and Al-Karkhi do not give 
this problem although they do give one upon which this is based, namely: to 
divide 10 into two parts such that the sum of each divided by the other is 2\. 
Al-Karkhi 2 discusses four solutions. 

Leonard of Pisa occasionally solves incorrectly, if this expression be permissible. 
Thus the problem: 3 To find a number such that, if the square root of 3 be added 
to it and then the square root of 2 be added, the product of the two sums will 
be 20. Algebraically (x + V?>) (x + V% = 20. The product of the two 
binomials is given as x 2 + 6 + V\2x 2 + VSx 2 . This leads to the incorrect 
value -\/19 + ]/ 24 — VZ — V2 for a:. Our manuscript of Abu Kamil gives 

the correct value, ^21| + ^lf.- ^6- V\- V\, for x. The problems 4 
following this in the algebra of Leonard, with the exception of the two final ones, 
are all taken from Abu Kamil and in the same order in which they occur in the 
work of the Arabic writer. 

The Italian writer frequently uses an expression which might be supposed to 
refer to Abu Kamil : " Operate according to algebra," 6 . . . " Operate therefore in 
this according to algebra, etc." However our manuscript employs a similar ex- 
pression. Thus in a problem arriving at 16a; 4 = 256x 2 , this Latin version of Abu 
Kamil reads: Fac secundum algebra in eis et erunt 16 census census equales 256 
censibus. census census (x 1 ) ergo equatur . 16 censibus et census equatur 16 
dragmis, ergo res equatur 4 dragmis . . ." (fol. 91 r ). In the discussion of the 
same problem Leonard concludes one form of solution with the words: "Age ergo 
in eis secundum algebra, et inuenies, census census equari 16 censibus: quare 
census est 16, et radix eius est 4, ut dictum est." Both writers present several 
solutions, including a geometrical one, of the problem in question which is to 
divide 10 into two parts such that if from the larger you subtract two of its roots 
and to the smaller add two of its roots the quantities are then equal. The ex- 
pression secundum algebra refers to the Book of Algebra and Almucabala by 
Mohammed ben Musa. 

1 Liber abbaci, p. 434-438. 

2 Woepcke, loc. cit., p. 91, 92. 

3 Liber abbaci, p. 445. 

4 Liber abbaci, 445-459. 
6 Liber abbaci, 438. 


Treatise on the Pentagon and Decagon. The algebra terminates with the 
first eight lines on fol. 93 T , and is followed by the work on the pentagon and 
decagon by the same author. Although this treatise on the pentagon and decagon 
by Abu Kamil is geometrical in its nature yet the treatment and the solutions 
are algebraical, including a fourth degree equation (x i = 8000a; 2 — V51 200 000) 
as well as mixed quadratics with irrational coefficients. The twelfth problem is 
in the Latin: "Et si dicemus tibi trianguli equilateri et equianguli mensura est 
cum perpendiculari ipsius est 10 ex numero, quanta sit perpendicularis?" 1 This 
suggests the similar problems in Greek of unknown date and author presented by 
Heiberg and Zeuthen, 2 as well as the similar problems given by Diophantos. 
In these Greek problems also lines and areas are summed quite contrary to ancient 
Greek usage. A further point of interest is that in the equation x 2 + 75 = 75x, 
to which the solution of the thirteenth problem leads, Abu Kamil gives only one 
solution whereas in the algebra he recognizes that this equation has two positive 
roots. The opening sentence of this treatise on the pentagon and decagon makes 
reference to the algebra as immediately preceding it, which indeed is the fact in 
the Hebrew and Latin manuscripts that are preserved. 

Conclusion. The algebra terminates with a general statement to the effect 
that by the methods taught in this book many more problems can be easily 
solved. In true Arabic fashion, the closing words are: " Whence praise and glory 
be to the only Creator." 

Let us summarize the results of our study. The most important conclusion 
of this investigation of Abu Kamil's algebra is that Al-Karkhi and Leonard 
of Pisa drew extensively from this Arabic writer. Through them this man, though 
himself comparatively unknown to modern writers, exerted a powerful influence 
on the early development of algebra. Abu Kamil deserves somewhat the same 
recognition from modern mathematicians and historians of science as that which 
Leonard of Pisa and Al-Khowarizmi have received. We may hope that the 
future will be more just than the past in according to Abu Kamil a prominent 
place among the mathematicians of the middle ages. 


By A. J. KEMPNER, University of Illinois. 
It is well known that the series 

diverges. The object of this Note is to prove that if the denominators do not 

1 "If we say to you that an equilateral and equiangular triangle, together with its altitude, is 
measured by 10, what is the altitude?" 

2 Einige griechische Aufgaben der unbestimmten Analytik, in Biblioth. mathem., VIII, third 
series, 118-134.