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THE
ALGEBRA
MOHAMMED BEN MUSA.
THE
ALGEBRA
OF
MOHAMMED BEN MUSA.
EDITED AND TRANSLATED
FREDERIC ROSEN.
LONDON:
PRINTED FOR THE ORIENTAL TRANSLATION FUND
AN D SOLO BY
J. MURRAY, ALBEMARLE STREET;
PARBURY, ALLEN, & CO., LEADENHALL STREET;
THACKER & CO., CALCUTTA; TREUTTEL & WUERTZ, PARIS;
AND E. FLEISCHER, LEIPZIG.
1831.
PRINTED BY
J. L. COX, GREAT QUEEN STREET,
LONDON.
us
PREFACE.
In the study of history, the attention of the
observer is drawn by a peculiar charm towards
those epochs, at which nations, after having
secured their independence externally, strive
to obtain an inward guarantee for their power,
by acquiring eminence as great in science and
in every art of peace as they have already at-
tained in the field of war. Such an epoch was,
in the history of the Arabs, that of the Caliphs
Al Mansur, Harun al Rashid, and Al
Mam UN, the illustrious contemporaries of
Charlemagne; to the glory of which era,
in the volume now offered to the public, a new
monument is endeavoured to be raised.
Abu Abdallah Mohammed ben Musa,
of Khowarezm, who it appears, from his pre-
face, wrote this Treatise at the command of the
Caliph Al Mamun, was for a long time consi-
dered as the original inventor of Algebra. * ' Hcbc
ars olim ^Mahomete, Mosis Arabisjilio, initi-
umsumsit: etenim hiijus ret locuples testis Leo-
0Aft09*>
( vi )
NARDUs PisANUs." Sucli are the words with
which HiERONYMUs Cardanus commences
his Ars Magna, in which he frequently refers
to the work here translated, in a manner to
leave no doubt of its identity.
That he was not the inventor of the Art, is
now well established ; but that he was the first
Mohammedan who wrote upon it, is to be found
asserted in several Oriental writers. Haji
Khalfa, in his bibliographical work, cites the
initial words of the treatise now before us,* and
* I am indebted to the kindness of my friend Mr. Gus-
TAv Fluegel of Dresden, for a most interesting extract
from this part of Haji Khalfa's work. Complete ma-
nuscript copies of the ^y^\ 4^o*X are very scarce. The
only two which I have hitherto had an opportunity of exa-
mining (the one bought in Egypt by Dr. Ehrenberg,
and now deposited in the Royal Library at Berlin — the other
among Rich's collection in the British Museum) are only
abridgments of the original compilation, in which the quo-
tation of the initial words of each work is generally omitted.
The prospect of an edition and Latin translation of the
complete original work, to be published by Mr. Fluegel,
under the auspices of the Oriental Translation Committee,
must under such circumstances be most gratifying to all
friends of Asiatic literature.
( vii )
states, in two distinct passages, that its author,
Mohammed ben Musa, was the first Mussul-
man who had ever written on the solution of
problems by the rules of completion and reduc-
tion. Two marginal notes in the Oxford ma-
nuscript — from which the text of the present
edition is taken — and an anonymous Arabic
writer, whose Bibliotheca Philosophorum is fre-
quently quoted byCAsiRi,* likewise maintain
that this production of Mohammed ben Musa
was the first work written on the subjectf by a
Mohammedan.
* -^U.^^1 ^j^ , written in the twelfth century. Casiri
Bibliotheca Ardbica Escurialensis, t. i. 426. 428.
+ The first of these marginal notes stands at the top of
the first page of the manuscript, and reads thus : Jj! ijjb
^liillj^l ti J^iJl ^^ ^ i:;^ J^ " This is the first
book written on (the art of calculating by) completion
and reduction by a Mohammedan : on this account the
author has introduced into it rules of various kinds, in
order to render useful the very rudiments of Algebra."
The other scholium stands farther on : it is the same to
which I have referred in my notes to the Arabic text,
p. 177.
( viii )
From the manner in which our author, in
his preface, speaks of the task he had under-
takeuv we cannot infer that he claimed to be the
inventor. He says that the Caliph Al Mamun
encouraged him to write a popular work on Al-
gebra: an expression which would seem to
imply that other treatises were then already
extant. From a formula for finding the circum-
ference of the circle, which occurs in the work
itself (Text p. 51, Transl. p. 72), I have, in a
note, drawn the conclusion, that part of the in-
formation comprised in this volume was derived
from an Indian source ; a conjecture which is
supported by the direct assertion of the author
of the Bihliotheca Philosophorum quoted by Ca-
siRi (1.426, 428). That Mohammed ben Musa
was conversant with Hindu science, is further
evident from the fact* that he abridged, at Al
Mamun's request — but before the accession of
that prince to the caliphat — the Sindhind, or
* Related by Ebn al Ad ami in the preface to his astro-
nomical tables. Casirj, i. 427, 428. Colebrookej Dis-
sertation, &c. p. Ixiv. Ixxii.
( ix )
astronomical tables, translated by Moham-
med BEN Ibrahim al Fazari from the
work of an Indian astronomer who visited the
court of Almansur in the 156th year of the
Hejira (A.D. 773).
The science as taught by Mohammed ben
MusA, in the treatise now before us, does not
extend beyond quadratic equations, including
problems with an affected square. These he
solves by the same rules which are followed by
DioPHANTUs*, and which are taught, though
less comprehensively, by the Hindu mathemati-
cians!. That he should have borrowed from
DioPHANTUs is not at all probable ; for it does
not appear that the Arabs had any knowledge
ofDioPHANTus' work before the middle of the
fourth century after the Hejira, when Abu'l-
WAFA BuzjANi rendered it into ArabicJ. It
* See DioPHANTus, Introd. § ii. and Book iv. pro-
blems 32 and 33.
+ Lilavatl, p. 29, Vijaganita, p. 347, of Mr. (ole-
brooke's translation.
I Casiui Bibl. Arab. Escur. i. 433. Colebrooke's
Dissertation, &c. p. Ixxii.
b
( X )
is far more probable that the Arabs received
their first knowledge of Algebra from the
Hindus, who furnished them with the decimal
notation of numerals, and with various im-
portant points of mathematical and astrono-
mical information.
But under whatever obligation our author
may be to the Hindus, as to the subject matter
of his performance, he seems to have been in-
dependent of them in the manner of digest-
ing and treating it : at least the method which
he follows in expounding his rules, as well as
in showing their application, differs considerably
from that of the Hindu mathematical writers.
BnASKARAand Brahmagupta give dogmati-
cal precepts, unsupported by argument, which,
even by the metrical form in which they are
expressed, seem to address themselves rather
to the memory than to the reasoning faculty
of the learner: Mohammed gives his rules
in simple prose, and establishes their accuracy
by geometrical illustrations. The Hindus give
comparatively few examples, and are fond of
investing the statement of their problems in
( xi >
rhetorical pomp : the Arab, on the contrary,
is remarkably rich in examples, but he intro-
duces them with the same perspicuous simpli-
city of style which distinguishes his rules. In
solving their problems, the Hindus are satisfied
with pointing at the result, and at the principal
intermediate steps which lead to it : the Arab
shows the working of each example at full
length, keeping his view constantly fixed upon
the two sides of the equation, as upon the two
scales of a balance, and showing how any
alteration in one side is counterpoised by a cor-
responding change in the other.
Besides the few facts which have already
been mentioned in the course of this preface,
little or nothing is known of our Author's life.
He lived and wrote under the caliphat of Al
Mamun, and must therefore be distinguished
from Abu Jafar Mohammed ben Musa*,
* The father of the latter, Musa ben Shaker, whose
native country I do not find recorded, had been a robber
or bandit in the earlier part of his life, but had after-
wards found means to attach himself to the court of the
Caliph Al-Mamun ; who, after Musa's death, took care of
( xii )
likewise a mathematician and astronomer, who
flourished under the Caliph Al Motaded
(who reigned A.H. 279-289, A.D. 892-902).
the education of his three sons, Mohammed, Ahmed, and
Al Hassan. (Abilfaragii Histor. Dyn. p. 280. Casiri,
I. 386. 418). Each of the sons subsequently distinguished
himself in mathematics and astronomy. We learn from
Abulfaraj (/. c. p. 281) and from T-bn Khallikan
(art. ^ ^^ '^i^^) that Thabet ben Korrah, the well-
known translator of the Almagest, was indebted to Mo-
hammed for his introduction to Al Motaded, and the
men of science at the court of that caliph. Ebn Khalli-
kan's words are: l^ (♦'^'J ^y^j^ tlpJ <J\/^ cT* "^J^
(Thabet ben Korrah) left Harran, and established
himself at Kafratutha, where he remained till Mohammed
BEN MusA arrived there, on his return from the Greek domi-
nions to Bagdad. The latter became acquainted with Thabet
and on seeing his skill and sagacity, invited Thabet to ac-
company him to Bagdad, where Mohammed made him
lodge at his own house, introduced him to the Caliph, and
procured him an appointment in the body of astronomers."
Ebn Khallikan here speaks of Mohammed ben Musa as
of a well-known individual : he has however devoted
no special article to an account of his life. It is possible
( xiii )
The manuscript from whence the text of the
present edition is taken — and which is the only
copy the existence of which I have as yet been
able to trace — is preserved in the Bodleian col-
lection at Oxford. It is, together with three
other treatises on Arithmetic and Algebra,
contained in the volume marked cmxviii.
Hunt. 214, foL, and bears the date of the
transcription A.H. 743 (A. D. 1342). It is
written in a plain and legible hand, but unfor-
tunately destitute of most of the diacritical
points : a deficiency which has often been very
sensibly felt ; for though the nature of the sub-
ject matter can but seldom leave a doubt as to
the general import of a sentence, yet the true
reading of some passages, and the precise in-
terpretation of others, remain involved in ob-
scurity. Besides, there occur several omissions
of words, and even of entire sentences ; and
also instances of words or short passages writ-
that the tour into the provinces of the Eastern Roman Em-
pire here mentioned, was undertaken in search of some
ancient Greek works on mathematics or astronomy.
( xiv )
ten twice over, or words foreign to the sense in-
troduced into the text. In printing the Arabic
part, I have included in brackets many of those
words which I found in the manuscript, the
genuineness of which I suspected, and also
such as I inserted from my own conjecture, to
supply an apparent hiatus.
The margin of the manuscript is partially filled
with scholia in a very small and almost illegible
character, a few specimens of which will be found
in the notes appended to my translation. Some
of them are marked as being extracted from a
commentary (^j--») by Al Mozaihafi*, pro-
bably the same author, whose full name is Je-
MALEDDIN AbU AbDALLAH MoHAMMED BEN
Omar al Jaza'i-}- al Mozaihafi, and whose
" Introduction to Arithmetic," (c->L^l ^ i«jJU)
is contained in the same volume with Moham-
med's work in the Bodleian library.
Numerals are in the text of the work always
* Wherever I have met with this name, it is written
without the diacritical points j^Aac^^l , and my pronuncia-
tion rests on mere conjecture.
+ ^\A'^ ( J )
( XV )
expressed by words : figures are only used in
some of the diagrams, and in a few marginal
notes.
The work had been only briefly mentioned in
Uris' catalogue of the Bodleian manuscripts.
Mr. H. T. Cole BROOKE first introduced it to
more general notice, by inserting a full account
of it, with an English translation of the direc-
tions for the solution of equations, simple and
compound, into the notes of the " Dissertation'''
prefixed to his invaluable work, '* Algebra, with
Ar^ithmetic and Mensuration, from the Sanscrit
of Brahmegupta and Bhascara," (London, 1817,
4to. pages Ixxv-lxxix.)
The account of the work given by Mr. Cole-
BROOKE excited the attention of a highly dis-
tinguished friend of mathematical science, who
encouraged me to undertake an edition and
translation of the whole : and who has taken the
kindest interest in the execution of my task.
He has with great patience and care revised
and corrected my translation, and has furnished
the commentary, subjoined to the text, in the
form of common algebraic notation. But my
( xvi )
obligations to him are not confined to this only ;
for his luminous advice has enabled me to over-
come many difficulties, which, to my own limit-
ed proficiency in mathematics, would have been
almost insurmountable.
In some notes on the Arabic text which are
appended to my translation, I have endeavoured,
not so much to elucidate, as to point out for
further enquiry, a few circumstances connected
with the history of Algebra. The comparisons
drawn between the Algebra of the Arabs and
that of the early Italian writers might perhaps
have been more numerous and more detailed ;
but my enquiry was here restricted by the
want of some important works. Montucla,
CossALi, HuTTON, and the Basil edition of
Card AN us' Ars magna, were the only sources
which I had the opportunity of consulting.
THE AUTHOR'S PREFACE
In the Name of God, gracious and merciful!
This work was written by Mohammed ben Musa, of
Khowarezm. He commences it thus :
Praised be God for his bounty towards those who
deserve it by their virtuous acts : in performing which,
as by him prescribed to his adoring creatures, we ex-
press our thanks, and render ourselves worthy of the
continuance (of his mercy), and preserve ourselves from
change : acknowledging his might, bending before his
power, and revering his greatness ! He sent Moham-
med (on whom may the blessing of God repose !) with
the mission of a prophet, long after any messenger
from above had appeared, when justice had fallen
into neglect, and when the true way of life was sought
for in vain. Through him he cured of blindness, and
saved through him from perdition, and increased
( 2 )
through him what before was small, and collected
through him what before was scattered. Praised be
God our Lord ! and may his glory increase, and may
all his names be hallowed — besides whom there is no
God; and may his benediction rest on Mohammed
the Prophet and on his descendants !
The learned in times which have passed away, and
among nations which have ceased to exist, were con-
stantly employed in writing books on the several de-
partments of science and on the various branches of
knowledge, bearing in mind those that were to come
after them, and hoping for a reward proportionate to
their ability, and trusting that their endeavours would
meet with acknowledgment, attention, and remem-
brance — content as they were even with a small degree
of praise; small, if compared with the pains which they
had undergone, and the difficulties which they had
encountered in revealing the secrets and obscurities of
science.
(2) Some applied themselves to obtain information which
was not known before them, and left it to posterity ;
others commented upon the difficulties in the works
left by their predecessors, and defined the best method
(of study), or rendered the access (to science) easier or
( 3 )
placed it more within reach ; others again discovered
mistakes in preceding works, and arranged that which
was confused, or adjusted what was irregular, and cor-
rected the faults of their fellow-labourers, without arro-
gance towards them, or taking pride in what they did
themselves.
That fondness for science, by which God has distin-
guished the Imam al Mamun, the Commander of the
Faithful (besides the caliphat which He has vouchsafed
unto him by lawful succession, in the robe of which He
has invested him, and with the honours of which He
has adorned him), that affability and condescension
which he shows to the learned, that promptitude with
which he protects and supports them in the elucida-
tion of obscurities and in the removal of difl&culties,
— has encouraged me to compose a short work on Cal-
culating by (the rules of) Completion and Reduction,
confining it to what is easiest and most useful in arith-
metic, such as men constantly require in cases of
inheritance, legacies, partition, law-suits, and trade,
and in all their dealings with one another, or where
the measuring of lands, the digging of canals, geo-
metrical computation, and other objects of various
sorts and kinds are concerned — relying on the good-
( 4 )
ness of my intention therein, and hoping that the
learned will reward it, by obtaining (for me) through
their prayers the excellence of the Divine mercy :
in requital of which, may the choicest blessings and
the abundant bounty of God be theirs ! My confi-
dence rests with God, in this as in every thing, and
in Him I put my trust. He is the Lord of the Sub-
lime Throne. May His blessing descend upon all the
prophets and heavenly messengers !
MOHAMMED BEN MUSA'S
COMPENDIUM
ON CALCULATING BY
COMPLETION AND REDUCTION.
When I considered what people generally want in (^)
calculating, I found that it always is a number.
I also observed that every number is composed of
units, and that any number may be divided into units.
Moreover, I found that every number, which may
be expressed from one to ten, surpasses the preceding
by one unit : afterwards the ten is doubled or tripled,
just as before the units were : thus arise twenty, thirty,
&c., until a hundred ; then the hundred is doubled and
tripled in the same manner as the units and the tens,
up to a thousand ; then the thousand can be thus re-
peated at any complex number ; and so forth to the
utmost limit of numeration.
J observed that the numbers which are required
in calculating by Completion and Reduction are of
three kinds, namely, roots, squares, and simple numbers
relative to neither root nor square.
( 6 )
A root is any quantity which is to be multiplied by
itself, consisting of units, or numbers ascending, or
fractions descending.^
A square is the whole amount of the root multiplied
by itself.
A simple number is any number which may be pro-
nounced without reference to root or square.
A number belonging to one of these three classes
may be equal to a number of another class; you
may say, for instance, " squares are equal to roots," or
" squares are equal to numbers," or " roots are equal to
numbers."!
/,l\ Of the case in which squares are equal to roots, this
is an example. " A square is equal to five roots of the
same ;"J the root of the square is five, and the square
is twenty-five, which is equal to five times its root.
So you say, " one third of the square is equal to four
roots ;"§ then the whole square is equal to twelve
roots; that is a hundred and forty-four; and its root
is twelve.
Or you say, " five squares are equal to ten roots ;" ||
then one square is equal to two roots; the root of
the square is two, and its square is four.
♦ By the word root, is meant the simple power of the
unknown quantity.
f cx'^ — bx cx^ = a bx=a
X x^=5x .*. x-^
:4^ .•.^a:^=l2J? ,\ x=.\q.
5f=.
3
5x2=100; /. 0:2 = 2X -^ x^2
( 7 )
In this manner, whether the squares be many or few,
{i, e. multiplied or divided by any number), they are
reduced to a single square ; and the same is done with
the roots, which are their equivalents ; that is to say,
they are reduced in the same proportion as the squares.
As to the case in which squares are eqtial to numbers ;
for instance, you say, " a square is equal to nine ;"*
then this is a square, and its root is three. Or " five
squares are equal to eighty ; "f then one square is equal
to one-fifth of eighty, which is sixteen. Or "the half
of the square is equal to eighteen ;"J then the square is
thirty-six, and its root is six.
Thus, all squares, multiples, and sub-multiples of
them, are reduced to a single square. If there be only
part of a square, you add thereto, until there is a whole
square; you do the same with the equivalent in numbers.
As to the case in which roots are equal to numbers ;
for instance, " one root equals three in number ; "§ then
the root is three, and its square nine. Or " four roots (5)
are equal to twenty ;" || then one root is equal to five,
and the square to be formed of it is twenty-five.
Or "half the root is equal to ten; "f then the
* x^:=g ar=3
t 5^2=80/. «2^^ = i6
t ^= 18/. x^ = s6 /. *=6
§ x=3
II 4^=20 /. x=5
f -|=io .-. X = 20
( 8 )
whole root is equal to twenty, and the square which is
formed of it is four hundred.
I found that these three kinds ; namely, roots,
squares, and numbers, may be combined together, and
thus three compound species arise ;* that is, " squares
and roots equal to numbers ;'* " squares and numbers
equal to roots ,*" "roots and numbers equal to squares."
Roots and Squares are equal to Numbers ;\ for in-
stance, " one square, and ten roots of the same, amount
to thirty-nine dirhems ;" that is to say, what must be
the square which^ when increased by ten of its own
roots, amounts to thirty-nine? The solution is this : you
halve the numberj of the roots, which in the present
instance yields five. This you multiply by itself;
the product is twenty-five. Add this to thirty-nine;
the sum is sixty- four. Now take the root of this, which
is eight, and subtract from it half the number of the
roots, which is five ; the remainder is three. This is
the root of the square which you sought for; the
square itself is nine.
* The three cases considered are,
1st. cx^-^bx=a
2d. cx'-\- a —bx
3d. cx^ — bx-\-a
f 1 St case : cx"^ -^bx=za
Example x' + 1 ox = 39
= ^64 — 5
= 8-5 = 3
X i. e. the coefficient.
( 9 )
The solution is the same when two squares or three,
or more or less be specified ;* you reduce them to one
single square, and in the same proportion you reduce
also the roots and simple numbers which are connected
therewith.
For instance, " two squares and ten roots are equal
to forty-eight dirhems ;"f that is to say, what must be '^
the amount of two squares which, when summed up and
added to ten times the root of one of them, make up a
sum of forty-eight dirhems ? You must at first reduce
the two squares to one ; and you know that one square
of the two is the moiety of both. Then reduce every
thing mentioned in the statement to its half, and it will
be the same as if the question had been, a square and
five roots of the same are equal to twenty-four dirhems;
or, what must be the amount of a square which, when
added to five times its root, is equal to twenty-four dir-
hems ? Now halve the number of the roots; the moiety
is two and a half. Multiply that by itself; the pro-
duct is six and a quarter. Add this to twenty-four ; the
sum is thirty dirhems and a quarter. Take the root of
this ; it is five and a half. Subtract from this the moiety
of the number of the roots, that is two and a half; the
* ex' -\-bx=a is to be reduced to the form x^-^—x.
t 2aP+iox=4S
^ = ^/[(|)''+24]-|
= 5i - 2i = 3
c
( 10 )
remainder is three. This is the root of the square, and
the square itself is nine.
The proceeding will be the same if the instance be,
" half of a square and five roots are equal to twenty-eight
dirhems ;"* that is to say, what must be the amount of
a square, the moiety of which, when added to the equi-
valent of five of its roots, is equal to twenty-eight dir-
hems ? Your first business must be to complete your
square, so that it amounts to one whole square. This
you effect by doubling it. Therefore double it, and dou-
ble also that which is added to it, as well as what is equal
to it. Then you have a square and ten roots, equal to
fifty-six dirhems. Now halve the roots ; the moiety is
five. Multiply this by itself; the product is twenty-five.
Add this to fifty-six ; the sum is eighty-one. Extract
the root of this; it is nine. Subtract from this the
moiety of the number of roots, which is five ; the re-
mainder is four. This is the root of the square which
you sought for ; the square is sixteen, and half the
(*7) square eight.
Proceed in this manner, whenever you meet with
squares and roots that are equal to simple numbers : for
it will always answer.
* 1-1-50:^28
x--|-iox=56
10\2 I ^fi-i 1
= v^ 25 -}- 56 - 5
= n/8i - 5
-9-5 = 4
( 11 )
Squares and Numbers are equal to Roots;* for
instance, " a square and twenty-one in numbers are
equal to ten roots of the same square." That is to say,
what must be the amount of a square, which, when
twenty-one dirhems are added to it, becomes equal to
the equivalent of ten roots of that square? Solution :
Halve the number of the roots; the moiety is five.
Multiply this by itself; the product is twenty-five.
Subtract from this the twenty-one which are connected
with the square ; the remainder is four. Extract its
root ; it is two. Subtract this from the moiety of the
roots, which is five ; the remainder is three. This is the
root of the square which you required, and the square
is nine. Or you may add the root to the moiety of the
roots ; the sum is seven ; this is the root of the square
which you sought for, and the square itself is forty-
nine.
When you meet with an instance which refers you to
this case, try its solution by addition, and if that do not
serve, then subtraction certainly will. For in this case
both addition and subtraction may be employed, which
will not answer in any other of the three cases in which
* '2d case. cx^-\-a-bx
Example. a:'4-2i— lo^
5 — n/ 25 —21
5=t2
( 12 )
the number of the roots must be halved. And know,
that, when in a question belonging to this case you
have halved the number of the roots and multiplied
the moiety by itself, if the product be less than the
number of dirhems connected with the square, then the
instance is impossible;* but if the product be equal to
(8) the dirhems by themselves, then the root of the square
is equal to the moiety of the roots alone, without either
addition or subtraction.
In every instance where you have two squares, or
more or less, reduce them to one entire square, f as I
have explained under the first case.
Roots and Numbers are equal to Squares ;% for instance,
*' three roots and four of simple numbers are equal
to a square." Solution : Halve the roots ; the moiety
is one and a half. Multiply this by itself; the product
is two and a quarter. Add this to the four ; the sum is
* If in an equation, of the form x^-^-azz-bx, (|.)2 z. a,
the case supposed in the equation cannot happen. If
(1)2= a, then^=^
f cx'^'\-a=bx is to be reduced to x^ +4=—^
J 3d case cx^ =:zbx+a
Example x^ = 30; + 4
= v/(i i)2 + 4 +ii
= \/ 2j^+4 +ij
= 2j +ij=4
( 13 )
six and a quarter. Extract its root ; it is two and a
half. Add this to the moiety of the roots, which was
one and a half; the sum is four. This is the root of the
square, and the square is sixteen.
Whenever you meet with a multiple or sub-multiple
of a square, reduce it to one entire square.
These are the six cases which I mentioned in the
introduction to this book. They have now been ex-
plained. I have shown that three among them do not
require that the roots be halved, and I have taught
how they must be resolved. As for the other three, in
which halving the roots is necessary, I think it expe-
dient, more accurately, to explain them by separate
chapters, in which a figure will be given for each
case, to point out the reasons for halving.
Demonstration of the Case : " a Square and ten Roots
are equal to thirty-nine Dirhems"^
The figure to explain this a quadrate, the sides of
which are unknown. It represents the square, the
which, or the root of which, you wish to know. This is
the figure A B, each side of which may be considered
as one of its roots ; and if you multiply one of these (9)
sides by any number, then the amount of that number
may be looked upon as the number of the roots which
are added to the square. Each side of the quadrate
represents the root of the square; and, as in the instance,
* Geometrical illustration of the case, x^+ ioa: = 39
( 14 )
the roots were connected with the square, we may take
one-fourth of ten, that is to say, two and a half, and
combine it with each of the four sides of the figure.
Thus with the original quadrate A B, four new paral-
lelofrrams are combined, each having a side of the qua-
drate as its length, and the number of two and a half as
its breadth ; they are the parallelograms C, G, T, and
K. We have now a quadrate of equal, though unknown
sides ; but in each of the four corners of which a square
piece of two and a half multiplied by two and a half is
wanting. In order to compensate for this want and to
complete the quadrate, we must add (to that which we
have already) four times the square of two and a half, that
is, twenty-five. We know (by the statement) that the first
figure, namely, the quadrate representing the square,
together with the four parallelograms around it, which
represent the ten roots, is equal to thirty-nine of num-
bers. If to this we add twenty-five, which is the equivalent
of the four quadrates at the corners of the figure A B,
by which the great figure D H is completed, then we
know that this together makes sixty-four. One side
of this great quadrate is its root, tliat is, eight. If we
subtract twice a fourth of ten, that is five, from eight,
as from the two extremities of the side of the great
quadrate D H, then the remainder of such a side will
be three, and that is the root of the square, or the side
of the original figure A B. It must be observed, that
we have halved the number of the roots, and added the
product of the moiety multiplied by itself to the number
( 15 )
thirty-nine, in order to complete the great figure in its
four corners ; because the fourth of any number multi-
plied by itself, and then by four, is equal to the product
of the moiety of that number multiplied by itself.*
Accordingly, we multiplied only the moiety of the roots
by itself, instead of multiplying its fourth by itself, and
then by four. This is the figure :
(10)
u
a
c
A
R
K
T
The same may also be explained by another figure.
We proceed from the quadrate A B, which represents
the square. It is our next business to add to it the ten
roots of the same. We halve for this purpose the ten,
so that it becomes five, and construct two quadrangles
on two sides of the quadrate A B, namely, G and D,
the length of each of them being ^ve, as the moiety of
the ten roots, whilst the breadth of each is equal to a
side of the quadrate A B. Then a quadrate remains
opposite the corner of the quadrate A B. This is equal
to five multiplied by five : this five being half of the
number of the roots which we have added to each of the
two sides of the first quadrate. Thus we know that
<b \-
* M^r=(k)
( 16 )
the first quadrate, which is the square, and the two
quadrangles on its sides, which are the ten roots, make
together thirty-nine. In order to complete the great
quadrate, there wants only a square of five multiplied
(11) by five, or twenty-five. This we add to thirty-nine, in
order to complete the great square S H. The sum is
sixty-four. We extract its root, eight, which is one of
the sides of the great quadrangle. By subtracting from
this the same quantity which we have before added,
namely five, we obtain three as the remainder. This is
the side of the quadrangle A B, which represents the
square; it is the root of this square, and the square
itself is nine. This is^ the figure : —
G-
B
26
D
Demonstration of the Case : « a Square and twenty-me
Dirhems are equal to ten Boots,'* *^
We represent the square by a quadrate A D, the
length of whose side we do not know. To this we join a
parallelogram, the breadth of which is equal to one of
the sides of the quadrate A D, such as the side H N.
This paralellogram is H B. The length of the two
* Geometrical illustration of the case, a:' + 2 1 = 1
ox
( 17 )
figures together is equal to the line H C. We know
that its length is ten of numbers ; for every quadrate
has equal sides and angles, and one of its sides multi-
plied by a unit is the root of the quadrate, or multiplied
by two it is twice the root of the same. As it is stated,
therefore, that a square and twenty-one of numbers are
equal to ten roots, we may conclude that the length of
the line H C is equal to ten of numbers, since the line
C D represents the root of the square. We now divide
the line C H into two equal parts at the point G : the
line G C is then equal to H G. It is also evident that (12)
the line G T is equal to the line C D. At present we
add to the line G T, in the same direction, a piece
equal to the difference between C G and G T, in order
to complete the square. Then the line T K becomes
equal to K M, and we have a new quadrate of equal
sides and angles, namely, the quadrate M T. We
know that the line T K is five ; this is consequently the
length also of the other sides : the quadrate itself is
twenty-five, this being the product of the multiplication
of half the number of the roots by themselves, for five
times five is twenty-five. We have perceived that the
quadrangle H B represents the twenty-one of numbers
which were added to the quadrate. We have then cut
off a piece from the quadrangle H B by the line K T
(which is one of the sides of the quadrate M T), so that
only the part T A remains. At present we take from
the line K M the piece K L, which is equal to G K; it
then appears that the line T G is equal to M L ; more-
D
( 18 )
over, the line K L, which has been cut off from K M^
is equal to K G; consequently, the quadrangle MR is
equal to T A. Thus it is evident that the quadrangle
H T, augmented by the quadrangle M R, is equal to
the quadrangle H B, which represents the twenty-one.
The whole quadrate M T was found to be equal to
twenty-five. If we now subtract from this quadrate,
MT, the quadrangles HT and M R, which are equal
to twenty-one, there remains a small quadrate K R,
which represents the difference between twenty-five and
twenty-one. This is four ; and its root, represented by
the line R G, which is equal to G A, is two. If you
(13) subtract this number two from the line C G, which is
the moiety of the roots, then the remainder is the line
A C ; that is to say, three, which is the root of the ori-
ginal square. But if you add the number two to the
line C G, which is the moiety of the number of the
roots, then the sum is seven, represented by the line
C R, which is the root to a larger square. However,
if you add twenty-one to this square, then the sum will
likewise be equal to ten roots of the same square. Here
is the figure : —
^^r L K
A . C
K G
IT
T B
( 19 )
Demonstration of the Case : " three Roots and four of
Simple Numbers are equal to a Square"^
Let the square be represented by a quadrangle, the
sides of which are unknown to us, though they are equal
among themselves, as also the angles. This is the qua-
drate A D, which comprises the three roots and the four
of numbers mentioned in this instance. In every qua-
drate one of its sides, multiplied by a unit, is its root.
We now cut off the quadrangle H D from the quadrate
A D, and take one of its sides H C for three, which is
the number of the roots. The same is equal to R D.
It follows, then, that the quadrangle H B represents
the four of numbers which are added to the roots. Now
we halve the side C H, which is equal to three roots, at
the point G ; from this division we construct the square
H T, which is the product of half the roots (or one and (14)
a half) multiplied by themselves, that is to say, two and
a quarter. We add then to the line G T a piece equal
to the line A H, namely, the piece T L ; accordingly
the line G L becomes equal to A G, and the line K N
equal to T L. Thus a new quadrangle, with equal
sides and angles, arises, namely, the quadrangle G M ;
and we find that the line A G is equal to M L, and the
same line A G is equal to G L. By these means the
line C G remains equal to N R, and the line M N
equal to T L, and from the quadrangle H B a piece
equal to the quadrangle K L is cut off.
* Geometrical illustration of the 3d case, x- = 3^ + 4
( 20 )
But we know that the quadrangle A R represents the
four of numbers which are added to the three roots.
The quadrangle A N and the quadrangle K L are to-
gether equal to the quadrangle A R, which represents
the four of numbers.
We have seen, also, that the quadrangle G M com-
prises the product of the moiety of the roots, or of one
and a half, multiplied by itself; that is to say two and
a quarter, together with the four of numbers, which are
represented by the quadrangles A N and K L. There
remains now from the side of the great original quadrate
A D, which represents the whole square, only the moiety
of the roots, that is to say, one and a half, namely, the
line G C. If we add this to the line A G, which is
the root of the quadrate G M, being equal to two and
a half; then this, together with C G, or the moiety of
the three roots, namely, one and a half, makes four,
which is the line A C, or the root to a square, which
is represented by the quadrate A D. Here follows
the figure. This it was which we were desirous to
explain.
(1^) B M A
R
N
( 21 )
We have observed that every question which requires
equation or reduction for its solution, will refer you to
one of the six cases which I have proposed in this
book. I have now also explained their arguments.
Bear them, therefore, in mind.
ON MULTIPLICATION.
I SHALL now teach you how to multiply the unknown
numbers, that is to say, the roots, one by the other, if
they stand alone, or if numbers are added to them, or if
numbers are subtracted from them, or if they are sub-
tracted from numbers ; also how to add them one to the
other, or how to subtract one from the other.
Whenever one number is to be multiplied by another,
the one must be repeated as many times as the other
contains units.*
If there are greater numbers combined with units to
be added to or subtracted from them, then four multi-
plications are necessary ;f namely, the greater numbers
by the greater numbers, the greater numbers by the
* If or is to be multiplied by y, x is to be repeated as
many times as there are units in t/.
f If X zt « is to be multiplied by j/ =t b, x is to be mul-
tiplied by y, X is to be multiplied by i, a is, to be multiplied
by y, and a is to be multiplied by b.
( 22 )
units, the units by the greater numbers, and the units
by the units.
If the units, combined with the greater numbers, are
positive, then the last multiplication is positive ; if they
are both negative, then the fourth multiplication is like-
vi^ise positive. But if one of them is positive, and one
(16) negative, then the fourth multiplication is negative.*
For instance, " ten and one to be multiplied by ten
and two."f Ten times ten is a hundred ; once ten is
ten positive ; twice ten is twenty positive, and once two
is two positive; this altogether makes a hundred and
thirty-two.
But if the instance is " ten less one, to be multiplied
by ten less one,"t then ten times ten is a hundred ; the
* In multiplying y^xzha) by (ydb^)
-{■ax-\-b = -{-ab
— ax —bz=-\-ab
■\-aX'-b=—ab
— ax +b=—ab
t (io + i)x(io + 2)
= 10X10.... 100
+ 1 XIO 10
4- 2X10 20
+ 1X2 2
+ 132
X (10-1) (10-1)
= 10X 10.. +100
— IX 10.. — 10
— IX 10.. — 10
— IX -1.. + 1
+ 81
( 23 )
negative one by ten is ten negative ; the other negative
one by ten is likewise ten negative, so that it becomes
eighty : but the negative one by the negative one is
one positive, and this makes the result eighty-one.
Or if the instance be " ten and two, to be multipled
by ten less one,"* then ten times ten is a hundred, and
the negative one by ten is ten negative; the positive
two by ten is twenty positive ; this together is a hun-
dred and ten ; the positive two by the negative one
gives two negative. This makes the product a hundred
and eight.
I have explained this, that it might serve as an intro-
duction to the multiplication of unknown sums, when
numbers are added to them, or when numbers are
subtracted from them, or when they are subtracted from
numbers.
For instance : " Ten less thing (the signification of
thing being root) to be multipled by ten."f You
begin by taking ten times ten, which is a hundred ; less
thing by ten is ten roots negative; the product is there-
fore a hundred less ten things.
* (10 + -2)X(10— l) =
\
10X10.... 100
— 1 xio —lO
+ 10X 2 +20
- IX 2.. . . — 2
lOS
f (10— jc)x io=iox 10 — iox=:ioo~
-loj;.
( 24 )
If the instance be : " ten and thing to be multiplied
by ten,"* then you take ten times ten, which is a hun-
dred, and thing by ten is ten things positive ; so that the
product is a hundred plus ten things.
If the instance be : " ten and thing to be multiplied
(17) by itself,"t then ten times ten is a hundred, and ten
times thing is ten things ; and again, ten times thing is
ten things ; and thing multiplied by thing is a square
positive, so that the whole product is a hundred dir-
hems and twenty things and one positive square.
If the instance be : " ten minus thing to be multiplied
by ten minus thing, "J then ten times ten is a hundred;
and minus thing by ten is minus ten things; and
again, minus thing by ten is minus ten things. But
minus thing multiplied by minus thing is a positive
square. The product is therefore a hundred and a
square, minus twenty things.
In like manner if the following question be proposed
to you : " one dirhem minus one-sixth to be multiplied
by one dirhem minus one-sixth ;"§ that is to say, five-
sixths by themselves, the product is five and twenty
parts of a dirhem, which is divided into six and thirty
parts, or two-thirds and one-sixth of a sixth. Compu-
tation : You multiply one dirhem by one dirhem, the
*(io+x)x io=iox 10+ loor — 100+1 oj;
f (lO+x) (I0+j;)=10 X lO+lO^+lO^ + X- — 100 + 20:r + X*'^
:j:(io— a:)x(i0-:r) r::iox 10— lox— lox+x-^ioo— 20a:+a:-
*(i-*)x(i-J)--i-^ + ixi = |+Jxi;/.e.J^,.|+ixA
{ 25 )
product is one clirhem ; then one dirhem by minus one-
sixth, that is one-sixth negative ; then, again, one dir-
hem by minus one-sixth is one-sixth negative : so far,
then, the result is two-thirds of a dirhem : but there is
still minus one-sixth to be multiplied by minus one-sixth,
which is one-sixth of a sixth positive ; the product is,
therefore, two- thirds and one sixth of a sixth.
If the instance be, " ten minus thing to be multiplied
by ten and thing," then you say,* ten times ten is a
hundred ; and minus thing by ten is ten things negative;
and thing by ten is ten things positive; and minus
thing by thing is a square positive ; therefore, the
product is a hundred dirhems, minus a square.
If the instance be, " ten minus thing to be multiplied
by thing,"t then you say, ten multiplied by thing is ten
things; and minus thing by thing is a square negative ; (18)
therefore, the product is ten things minus a square.
If the instance be, " ten and thing to be multiplied
by thing less ten,"| then you say, thing multiplied by
ten is ten things positive ; and thing by thing is a square
positive ; and minus ten by ten is a hundred dirhems
negative ; and minus ten by thing is ten things nega-
tive. You say, therefore, a square minus a hundred
dirhems ; for, having made the reduction, that is to say,
having removed the ten things positive by the ten things
* (lo— x) (io + j;) = iox 10— lojr+iox— x'-' = ioo— x2
f (lo— x) xa: = ioar— .r-
X (lo + x) (x— io) = io^-|-i;'-— loo — ioa: ~a?-— 100
E
( 26 )
negative, there remains a square minus a hundred
dirhems.
If the instance be, ^' ten dirhems and half a thing to
be multiplied by half a dirhem, minus five things,"*
then you say, half a dirhem by ten is five dirhems posi-
tive ; and half a dirhem by half a thing is a quarter of
thing positive ; and minus five things by ten dirhems is
fifty roots negative. This altogether makes five dir-
hems minus forty-nine things and three quarters of
thing. After this you multiply five roots negative by
half a root positive : it is two squares and a half negative.
Therefore, the product is five dirhems, minus two
squares and a half, minus forty-nine roots and three
quarters of a root.
If the instance be, " ten and thing to be multiplied
by thing less ten,"f then this is the same as if it were
said thing and ten by thing less ten. You say, there-
fore, thing multiplied by thing is a square positive ; and
ten by thing is ten things positive ; and minus ten by
thing is ten things negative. You now remove the
positive by the negative, then there only remains a
square. Minus ten multiplied by ten is a hundred, to
be subtracted from the square. This, therefore, alto-
gether, is a square less a hundred dirhems.
(19) Whenever a positive and a negative factor concur in
t( 1 o -f j:)(a;— 1 o) := (ar-f 1 o)(a;— 1 o) - jr- 4- 1 go:— 1 ox— 1 00 = ar2 - 1 00
( 27 )
a multiplication, such as thing positive and minus thing,
the last multiplication gives always the negative pro-
duct. Keep this in memory.
ON ADDITION and SUBTRACTION.
Know that the root of two hundred minus ten, added
to twenty minus the root of two hundred, is just ten.*
The root of two hundred, minus ten, subtracted from
twenty minus the root of two hundred, is thirty minus
twice the root of two hundred; twice the root of two
hundred is equal to the root of eight hundred.'!'
A hundred and a square minus twenty roots, added
to fifty and ten roots minus two squares,^ is a hundred
and fifty, minus a square and minus ten roots.
A hundred and a square, minus twenty roots, dimi-
nished by fifty and ten roots minus two squares, is fifty
dirhems and three squares minus thirty roots.§
I shall hereafter explain to you the reason of this by
a figure, which will be annexed to this chapter.
If you require to double the root of any known or
unknown square, (the meaning of its duplication being
* 20— V^200-j-('v/200— 10)=10
.j. 20— v/200— (-v/sOO— 10)z=30— 2\/200 = 30— \/8oo
t 50+ lox— 2x'^-f (loo+a?'-^— 20a;) = i50— 10*— ^2
§ loo+o;'-- 20a;— [50— 2x'- + ioa:] =50 + 3x''^— 30*
( 28 )
that you multiply it by two) then it will suffice to
multiply two by two, and then by the square;* the
root of the product is equal to twice the root of the
original square.
If you require to take it thrice, you multiply three
by three, and then by the square ; the root of the pro-
duct is thrice the root of the original square.
Compute in this manner every multiplication of the
roots, whether the multiplication be more or less than
two.t
(20) If you require to find the moiety of the root of the
square, you need only multiply a half by a half, which
is a quarter ; and then this by the square : the root of
the product will be half the root of the first square.]:
Follow the same rule when you seek for a third, or a
quarter of a root, or any larger or smaller quota§ of it,
whatever may be the denominator or the numerator.
Examples of this : If you require to double the root
of nine, II you multiply two by two, and then by nine:
this gives thirty- six ; take the root of this, it is six,
and this is double the root of nine.
2v'9 = v/4X9 = v/36=6
( 29 )
In the same manner, if you require to triple the root of
nine,* you multiply three by three, and then by nine :
the product is eighty-one ; take its root, it is nine, which
becomes equal to thrice the root of nine.
If you require to have the moiety of the root of nine,t
you multiply a half by a half, which gives a quarter, and
then this by nine ; the result is two and a quarter : take
its root ; it is one and a half, which is the moiety of the
root of nine.
You proceed in this manner with every root, whether
positive or negative, and whether known or unknown.
ON DIVISION.
If you will divide the root of nine by the root of four4
you begin with dividing nine by four, which gives two
and a quarter : the root of this is the number which you
require — it is one and a half.
If you will divide the root of four by the root of nine,§
you divide four by nine ; it is four-ninths of the unit :
the root of this is two divided by three ; namely, two-
thirds of the unit.
* 3v/9 = v^9X9=>/8i=9
( 30 )
If you wish to divide twice the root of nine by the
root of four^ or of any other square*, you double the
(21) root of nine in the manner above shown to you in the
chapter on Multiplication, and you divide the product by
four, or by any number whatever. You perform this in
the way above pointed out.
In like manner, if you wish to divide three roots
of nine, or more, or one-half or any multiple or sub-
multiple of ihe root of nine, the rule is always the
same :t follow it, the result will be right.
If you wish to multiply the root of nine by the root of
four,+ multiply nine by four ; this gives thirty- six ; take
its root, it is six ; this is the root of nine, multiplied by
the root of four.
Thus, if you wish to multiply the root of five by the
root of ten,§ multiply five by ten : the root of the pro-
duct is what you have required.
If you wish to multiply the root of one- third by the
root of a half, II you multiply one- third by a half: it is
one- sixth : the root of one- sixth is equal to the root of
one-third, multiplied by the root of a half.
If you require to multiply twice the root of nine by
* Sv^g
= x/^ = v/9:
V4
X >/4Xv/9=\/4X9=v/36 = 6
§ v'ioxv'5= n/5xio=v'5o
II s/^y^s/h^s/W^^s/^
( 31 )
thrice the root of fom', " then take twice the root of nine,
according to the rule above given, so that you may know
the root of what square it is. You do the same with
respect to the three roots of four in order to know what
must be the square of such a root. You then multiply
these two squares, the one by the other, and the root of
the product is equal to twice the root of nine, multiplied
by thrice the root of four.
You proceed in this manner with all positive or ne-
gative roots.
Demomtratiom, (22)
The argument for the root of two hundred, minus ten,
added to twenty, minus the root of two hundred, may be
elucidated by a figure :
Let the line A B represent the root of two hundred ;
let the part from A to the point C be the ten, then the
remainder of the root of two hundred will correspond to
the remainder of the line A B, namely to the line C B.
Draw now from the point B a line to the point D, to
represent twenty ; let it, therefore, be twice as long as
the line A C, which represents ten; and mark a part of
it from the point B to the point H, to be equal to the
line A B, which represents the root of two hundred;
then the remainder of the twenty will be equal to the
part of the line, from the point H to the point D. As
3\/4 X 2 ^9 = >v/9 X4 X ^/4 X 9 - v/36 X 36=36
( .^s )
our object was to add the remainder of the root of two
hundred, after the subtraction of ten, that is to say, the
hne C B, to the line H D, or to twenty, minus the root
of two hundred, we cut off from the line B H a piece
equal to C B, namely, the line S H. We know already
that the line A B, or the root of two hundred, is equal to
the line B H, and that the line A C, which represents the
ten, is equal to the line S B, as also that the remainder
of the line A B, namely, the line C B is equal to the
remainder of the line B H, namely, to S H. Let us
add, therefore, this piece S H, to the line H D. We
have already seen that from the line B D, or twenty, a
piece equal to A C, which is ten, was cut off, namely,
the piece B S. There remains after this the line S D,
which, consequently, is equal to ten. This it was that
we intended to elucidate. Here follows the figure.
(23) AJ
S BT S S
The argument for the root of two hundred, minus ten,
to be subtracted from twenty, minus the root of two
hundred, is as follows. Let the line A B represent the
root of two hundred, and let the part thereof, from A to
the point C, signify the ten mentioned in the instance.
We draw now from the point B, a line towards the point
D, to signify twenty. Then we trace from B to the
( 33 )
point H, the same lengtli as the leiigtli of the line which
represents the root of two liundred ; that is of the line
A B. We have seen that the line C B is the remainder
from the twenty, after the root of two hundred has been
subtracted. It is our purpose, therefore, to subtract
the line C B from the line H D ; and we now draw from
the point B, a line towards the point S, equal in length
to the line A C, which represents the ten. Then the
whole line S D is equal to S B, plus B D, and we per-
ceive that all this added together amounts to thirty.
We now cut off from the line H D, a piece equal to
C B, namely, ,the line H G ; thus we find that the line
G D is the remainder from the line S D, which signifies
thirty. We see also that the line B H is the root of
two hundred and that the line S B and B C is likewise
the root of two hundred. Kow the line H G is equal
to C B ; therefore the piece subtracted from the line
S D, which represents thirty, is equal to twice the
root of two hundred, or once the root of eight hundred. (^^)
This it is that we wished to elucidate.
Here follows the figure :
1> G M B S
As for the hundred and square minus twenty roots
added to fifty, and ten roots minus two squares, this does
F
( 34 )
not admit of any figure, because there are three diffe-
rent species, viz. squares, and roots, and numbers, and
nothing corresponding to them by which they might
be represented. We had, indeed, contrived to con-
struct a figure also for this case, but it was not suffi-
ciently clear.
The elucidation by words is very easy. You know
that you have a hundred and a square, minus twenty
roots. When you add to this fifty and ten roots, it be-
comes a hundred and fifty and a square, minus ten roots.
The reason for these ten negative roots is, that from the
twenty negative roots ten positive roots were subtracted
by reduction. This being done, there remains a hun-
dred and fifty and a square, minus ten roots. With the
hundred a square is connected. If you subtract from
this hundred and square the two squares negative con-
nected with fifty, then one square disappears by reason
of the other, and the remainder is a hundred and fifty,
minus a square, and minus ten roots.
This it was that we wished to explain.
( 35 )
OF THE SIX PROBLEMS.
Before the chapters on computation and the several (25)
species thereof, I shall now introduce six problems, as
instances of the six cases treated of in the beginning of
this work. I have shown that three among these cases,
in order to be solved, do not require that the roots
be halved, and I have also mentioned that the calculat-
ing by completion and reduction must always neces-
sarily lead you to one of these cases. I now subjoin
these problems, which will serve to bring the sub-
ject nearer to the understanding, to render its com-
prehension easier, and to make the arguments more
perspicuous.
First Problem,
I have divided ten into two portions ; I have multi-
plied the one of the two portions by the other ; after
this I have multiplied the one of the two by itself,
and the product of the multiplication by itself is four
times as much as that of one of the portions by the
other.*
Computation : Suppose one of the portions to be
thing, and the other ten minus thing : you multiply
* x2--4^jlo— a;)=:40a;— 4x2
5*2=400:
X =8; (10— a;)=2
( 36 )
thing by ten minus thing ; it is ten things minus a
square. Then multiply it by four, because the in-
stance states " four times as much." The result will be
four times the product of one of the parts multiplied by
the other. This is forty things minus four squares.
After this you multiply thing by thing, that is to say,
one of the portions by itself. This is a square, which
is equal to forty things minus foursquares. Reduce it
now by the four squares, and add them to the one
square. Then the equation is : forty things are equal
to five squares ; and one square will be equal to eight
roots, that is, sixty-four ; the root of this is eight, and
this is one of the two portions, namely, that which is to
(26) be multiplied by itself. The remainder from the ten
is two, and that is the other portion. Thus the question
leads you to one of the six cases, namely, that of
" squares equal to roots." Remark this.
Second Problem,
I have divided ten into two portions : I have multi-
plied each of the parts by itself, and afterwards ten by
itself: the product often by itself is equal to one of the
two parts multiplied by itself, and afterwards by two
and seven-ninths; or equal to the other multiplied by
itself, and afterwards by six and one-fourth.*
* \o^=x^
X2^
100 =rx2
^ 9
^^xioo
=x'
36=:i:2*
6=x
( 37 )
Computation : Suppose one of the parts to be thing,
and the other ten minus thing. You multiply thing by
itself, it is a square; then by two and seven-ninths,
this makes it two squares and seven- ninths of a square.
You afterwards multiply ten by ten ; it is a hundred,
which much be equal to two squares aaid seven-ninths
of a square. Reduce it to one square, through division
by nine twenty- fifths ;^ this being its fifth and four-
fifths of its fifth, take now also the fifth and four-fifths
of the fifth of a hundred ; this is thirty-six, which is
equal to one square. Take its root, it is six. This is
one of the two portions ; and accordingly the other is
four. This question leads you, therefore, to one of the
six cases, namely, " squares equal to numbers."
Third Problem,
I have divided ten into two parts. I have afterwards
divided the one by the other, and the quotient was four.f
Computation : Suppose one of the two parts to be (27)
thing, the other ten minus thing. Then you divide ten
minus thing by thing, in order that four may be ob-
tained. You know that if you multiply the quotient
by the divisor, the sum which was divided is restored.
X
10 — XZZ.^
io=5x
2=j:
( 38 )
In the present question tlie quotient is four and the
divisor is thing. Multiply, therefore, four by thing ;
the result is four things, which are equal to the sum to
be divided, which was ten minus thing. You now
reduce it by thing, which you add to the four things.
Then we have ^\e things equal to ten ; therefore one
thing is equal to two, and this is one of the two portions.
This question refers you to one of the six cases,
namely, " roots equal to numbers."
Fourth Problem.
I have multiplied one- third of thing and one dirhem
by one-fourth of thing and one dirhem, and the product
was twenty.*
Computation : You multiply one- third of thing by
one- fourth of thing; it is one-half of a sixth of a square.
Further, you multiply one dirhem by one-third of thing,
it is one- third of thing ; and one dirhem by one-fourth
of thing, it is one-fourth of thing ; and one dirhem by
one dirhem, it is one dirhem. The result of this is : the
moiety of one-sixth of a square, and one- third of thing,
and one-fourth of thing, and one dirhem, is equal to
twenty dirhems. Subtract now the one dirhem from
* (J:c+i)(J;r+i)=20
( 39 )
these twenty dirhems, there remain nineteen dirhems,
equal to the moiety of one-sixth of a square, and one-
third of thing, and one-fourth of thing. Now make your
square a whole one : you perform this by multiplying all
that you have by twelve. Thus you have one square
and seven roots, equal to two hundred and twenty-eight
dirhems. Halve the number of the roots, and multiply
it by itself; it is twelve and one- fourth. Add this to
the numbers, that is, to two hundred and twenty-eight ; (28)
the sum is two hundred and forty and one quarter. Ex-
tract the root of this; it is fifteen and a half. Subtract
from this the moiety of the roots, that is, three and a
half, there remains twelve, which is the square required.
This question leads you to one of the cases, namely,
" squares and roots equal to numbers."
Fifth Problem.
I have divided ten into two parts ; I have then multi-
plied each of them by itself, and when I had added the
products together, the sum was fifty-eight dirhems.*
Computation : Suppose one of the two parts to be
thing, and the other ten minus thing. Multiply ten
minus thing by itself; it is a hundred and a square
minus twenty things. Then multiply thing by thing ; it
* «2-i-{lo-a:)2=58
2 j;2 — 20*4-100 = 58
jf-^ — 100:4-50=29
jc24.2i = ioa:
a;=5d=v/25~2i=:5d:2=7 or 3
( 40 )
is a square. Add both together. The sum is a hun-
dred, phis two squares minus twenty things, which are
equal to fifty-eight dirhems. Take now the twenty
negative things from the hundred and the two squares,
and add them to fifty- eight ; then a hundred, plus two
squares, are equal to fifty-eight dirhems and twenty
things. Reduce this to one square, by taking the moiety
of all you have. It is then: fifty dirhems and a square,
which are equal to twenty-nine dirhems and ten things.
Then reduce this, by taking twenty-nine from fifty ;
there remains twenty-one and a square, equal to ten
things. Halve the number of the roots, it is five; multiply
this by itself, it is twenty-five; take from this the twenty-
one which are connected with the square, the remainder
^ ^ is four. Extract the root, it is two. Subtract this from
the moiety of the roots, namely, from five, there remains
three. This is one of the portions; the other is seven.
This question refers you to one of the six cases, namely
*' squares and numbers equal to roots."
Sixth Problem. ,
I have multiplied one-third of a root by one-fourth
of a root, and the product is equal to the root and
twenty-four dirhems.*
.3 4
( 4.1 )
Computation : Call the root thing; then one- third of
thing is multiplied by one-fourth of thing ; this is the
moiety of one-sixth of the square, and is equal to thing
and twenty -four dirhems. Multiply this moiety of one-
sixth of the square by twelve, in order to make your
square a whole one, and multiply also the thing by
twelve, which yields twelve things ; and also four-and-
twenty by twelve : the product of the whole will be two
hundred and eighty-eight dirhems and twelve roots,
which are equal to one square. The moiety of the roots
is six. Multiply this by itself, and add it to two hun-
dred and eighty-eight, it will be three hundred and
twenty-four. Extract the root from this, it is eighteen;
add this to the moiety of the roots, which was six ; the
sum is twenty-four, and this is the square sought for.
This question refers you to one of the six cases,
namely, " roots and numbers equal to squares."
VARIOUS QUESTIONS.
If a person puts such a question to you as : "I have (30)
divided ten into two parts, and multiplying one of
these by the other, the result was twenty-one;"^ then
* (lO — X)X=:21
10X-X- = 21
which is to be reduced to
a;'H2i=;io^
x=5±v/25-2l=5d=2
G
( 42 )
you know that one of the two parts is thing, and the
other ten minus thing. Multiply, therefore, thing by
ten minus thing; then you have ten things minus
a square, which is equal to twenty-one. Separate the
square from the ten things, and add it to the twenty-
one. Then you have ten things, which are equal to
twenty-one dirhems and a square. Take away the
moiety of the roots, and multiply the remaining five
by itself; it is twenty-five. Subtract from this the
twenty-one which are connected with the square ; the
remainder is four. Extract its root, it is two. Sub-
tract this from the moiety of the roots, namely, five ;
there remain three, which is one of the two parts. Or,
if you please, you may add the root of four to the
moiety of the roots; the sum is seven, which is likewise
one of the parts. This is one of the problems which
may be resolved by addition and subtraction.
If the question be : "I have divided ten into two parts,
and having multiplied each part by itself, I have sub-
tracted the smaller from the greater, and the remainder
was forty;"* then the computation is — ^you multiply ten
(31) minus thing by itself, it is a hundred plus one square
minus twenty things ; and you also multiply thing by
100 — 200; =40
100 = 200?+ 40
60 = 20X
3 = ^
( 43 )
thing, it is one square. Subtract this from a hundred
and a square minus twenty things, and you have a
hundred, minus twenty things, equal to forty dirhems.
Separate now the twenty things from a hundred, and
add them to the forty ; then you have a hundred, equal
to twenty things and forty dirhems. Subtract now forty
from a hundred ; there remains sixty dirhems, equal to
twenty things: therefore one thing is equal to three,
which is one of the two parts.
If the question be : " I have divided ten into two parts,
and having multiplied each part by itself, I have put
them together, and have added to them the difference
of the two parts previously to their multiplication, and
the amount of all this is fifty-four;"^ then the compu-
tation is this: You multiply ten minus thing by itself;
it is a hundred and a square minus twenty things.
Then multiply also the other thing of the ten by itself ;
it is one square. Add this together, it will be a hun-
dred plus two squares minus twenty things. It was
stated that the difference of the two parts before multi-
plication should be added to them. You say, therefore,
the difference between them is ten minus two things.
* (lO — a:)2-|-j;2 + (l0— x)— x~54
1 GO — 20a; + 2 x- + 1 o •— 2a: = 54
100 — 2 20;+ 2a:' =54
55-iia; + a:2=:27
X =U±^m_28=U^=7or4
( 44 )
The result is a hundred and ten and two squares minus
twenty-two things, which are equal to fifty-four dirhems.
Having reduced and equalized this, you may say, a
hundred and ten dirhems and two squares are equal to
fifty-four dirhems and twenty-two things. Reduce now
the two squares to one square, by taking the moiety of
all you have. Thus it becomes fifty-five dirhems and a
square, equal to twenty-seven dirhems and eleven things.
Subtract twenty-seven from fifty-five, there remain
(32) twenty-eight dirhems and a square, equal to eleven
things. Halve now the things, it will be five and a
half; multiply this by itself, it is thirty and a quarter.
Subtract from it the twenty-eight which are combined
with the square, the remainder is two and a fourth.
Extract its root, it is one and a half. Subtract this
from the moiety of the roots, there remain four, which
is one of the two parts.
If one say, "I have divided ten into two parts ; and
have divided the first by the second, and the second by
the first, and the sum of the quotient is two dirhemis
and one-sixth ;"* then the computation is this : If you
multiply each part by itself, and add the products
together, then their sum is equal to one of the parts
J. 10 — X X ,
* ' _L — o-L
X ^lO-a; 6
100-h 2ar^— 20a? = 4^^ -^) X 2^ = 2 ifx - 2^0;''*
^=,5~v/25— 24 = 5— 1 = 4 or 6
( 45 )
multiplied by the other, and again by the quotient
which is two and one-sixth. Multiply, therefore, ten
less thing by itself; it is a hundred and a square less
ten things. Multiply thing by thing; it is one square.
Add this together ; the sum is a hundred plus two
squares less twenty things, which is equal to thing mul-
tiplied by ten less thing ; that is, to ten things less a
square, multiplied by the sum of the quotients arising
from the division of the two parts, namely, two and
one-sixth. We have, therefore, twenty-one things and
two-thirds of thing less two squares and one-sixth, equal
to a hundred plus two squares less twenty things. Re-
duce this by adding the two squares and one-sixth to a
hundred plus two squares less twenty things, and add
the twenty negative things from the hundred plus the
two squares to the twenty- one things and two -thirds of
thing. Then you have a hundred plus four squares (33)
and one-sixth of a square, equal to forty-one things and
two- thirds of thing. Now reduce this to one square.
You know that one square is obtained from four squares
and one-sixth, by taking a fifth and one-fifth of a fifth.*
Take, therefore, the fifth and one-fifth of a fifth of all
that you have. Then it is twenty-four and a square,
equal to ten roots ; because ten is one-fifth and one-fifth
of the fifth of the forty-one things and two-thirds of a
thing. Now halve the roots; it gives five. Multiply this
4=^6^ and^\ = i+ixi
( 46 )
by itself; it is five-and-twenty. Subtract from this
the twenty-four, which are connected with the square ;
the remainder is one. Extract its root; it is one.
Subtract this from the moiety of the roots, which
is five. There remains four, which is one of the two
parts.
Observe that, in every case, where any two quantities
whatsoever are divided, the first by the second and the
second by the first, if you multiply the quotient of the
one division by that of the other, the product is always
one.^
If some one say: "You divide ten into two parts;
multiply one of the two parts by five, and divide it by
the other : then take the moiety of the quotient, and
add this to the product of the one part, multiplied by
five ; the sum is fifty dirhems ;"t then the computation
is this : Take thing, and multiply it by five. This is
now to be divided by the remainder of the ten, that is,
by ten less thing ; and of the quotient the moiety is to
be taken.
(34) You know that if you divide five things by ten less
thing, and take the moiety of the quotient, the result is
a b
-X-= 1
6 a
5x
2(10-.)'^^-^'^
2(10-x) '^-^^
( 47 )
the same as if you divide the moiety of five things by
ten less thing. Take, therefore, the moiety of five
things; it is two things and a half: and this you
require to divide by ten less thing. Now these two
things and a half, divided by ten less thing, give a
quotient which is equal to fifty less five things : for the
question states : add this (the quotient) to the one
part multiplied by five, the sum will be fifty. You
have already observed, that if the quotient, or the result
of the division, be multiplied by the divisor, the divi-
dend, or capital to be divided, is restored. Now, your
capital, in the present instance, is two things and a
half. Multiply, therefore, ten less thing by fifty less
^we things. Then you have five hundred dirhems and
five squares less a hundred things, which are equal to
two things and a half. Reduce this to one square.
Then it becomes a hundred dirhems and a square less
twenty things, equal to the moiety of thing. Separate
now the twenty things from the hundred dirhems and
square, and add them to the half thing. Then you
have a hundred dirhems and a square, equal to twenty
things and a half. Now halve the things, multiply
the moiety by itself, subtract from this the hundred,
extract the root of the remainder, and subtract this
from the moiety of the roots, which is ten and one-
fourth : the remainder is eight ; and this is one of the
portions.
If some one say : " You divide ten into two parts :
multiply the one by itself; it will be equal to the other
( 48 )
taken eighty-one times." ^ Computation : You say, ten
less thing, multiplied by itself, is a hundred plus a
(35) square less twenty things, and this is equal to eighty-
one things. Separate the twenty things from a hundred
and a square, and add them to eighty-one. It will
then be a hundred plus a square, which is equal to a
hundred and one roots. Halve the roots ; the moiety is
fifty and a half Multiply this by itself, it is two thou-
sand five hundred and fifty and a quarter. Subtract
from this one hundred ; the remainder is two thousand
four hundred and fifty and a quarter. Extract the root
from this; it is forty-nine and a half. Subtract this
from the moiety of the roots, which is fifty and a half.
There remains one, and this is one of the two parts.
If some one say : " I have purchased two measures of
wheat or barley, each of them at a certain price. I
afterwards added the expences, and the sum was equal
to the difference of the two prices, added to the diffe-
rence of the measures. "t
* (io-x)2=8ia:
100— 20a;+a;^=8i5:
a:2 + 100 = 101^7
^ = i|i_v/'if'-ioo=5ol-49i = i
f The purchaser does not make a clear enunciation of the
terms of his bargain. He intends to say, " 1 bought m
bushels of wheat, and n bushels of barley, and the wheat was
r times dearer than the barley. The sum I expended was
equal to the difference in the quantities, added to the diffe-
rence in the prices of the grain."
( 49 )
Computation : Take what numbers you please, for it
is indifferent ; for instance, four and six. Then you
say : I have bought each measure of the four for thing;
and accordingly you multiply four by thing, which gives
four things; and I have bought the six, each for the
moiety of thing, for which I have bought the four ; or,
if you please, for one-third, or one-fourth, or for any
other quota of that price, for it is indifferent. Suppose
that you have bought the six measures for the moiety of
thing, then you multiply the moiety of thing by six ;
this gives three things. Add them to the four things ;
the sum is seven things, which must be equal to the
difference of the two quantities, which is two measures,
plus the difference of the two prices, which is a moiety
of thing. You have, therefore, seven things, equal to
two and a moiety of thing. Remove, now, this moiety
of thing, by subtracting it from the seven things.
There remain six things and a half, equal to two dir- (36)
hems: consequently, one thing is equal to four-thir-
teenths of a dirhem. The six measures were bought,
each at one-half of thing; that is, at two-thirteenths of
a dirhem. Accordingly, the expenses amount to eight-
and-twenty thirteenths of a dirhem, and this sum is
equal to the difference of the two quantities; namely.
If X is the price of the barley, rx is the price of the
wheat ; whence, mrx -\- nx zz (m — n) + (rx — x) ; ,\ x =
m^n - , , , . (mr-^-n) X (m— w)
— r— 7 and the sum expended is - ^^ . ^ . ^ — T"-
( 50 )
the two measures, the arithmetical equivalent for which
is six-and-twenty thirteenths, added to the difference of
the two prices, which is two-thirteenths : both diffe-
rences together being likewise equal to twenty-eight
parts.
If he say: "There are two numbers,* the difference
of which is two dirhems. I have divided the smaller by
the larger, and the quotient was just half a dirhem."f
Suppose one of the two numbers* to be thing, and the
other to be thing plus two dirhems. By the division
of thing by thing plus two dirhems, half a dirhem
appears as quotient. You have already observed, that
by multiplying the quotient by the divisor, the capital
which you divided is restored. This capital, in the
present case, is thing. Multiply, therefore, thing and
two dirhems by half a dirhem, which is the quotient;
the product is half one thing plus one dirhem ; this is
equal to thing. Remove, now, half a thing on account
* In the original, ** squares." The word square is used
in the text to signify either, ist, a square, properly so called,
fractional or integral; 2d, a rational integer, not being a
square number ; 3d, a rational fraction, not being a square ;
4th, a quadratic surd, fractional or integral.
x-j- 2
h
x-f 2__x
2 ~"2
=z 1 and or -f 2 = 4
*=-i-=5 + >
( 51 )
of the other half thing; there remains one dirhem,
equal to half a thing. Double it, then you have one
thing, equal to two dirhems. Consequently, the other
number* is four.
If some one say: "I have divided ten into two parts;
I have multiplied the one by ten and the other by itself,
and the products were the same;"f then the computa-
tion is this : You multiply thing by ten ; it is ten things.
Then multiply ten less thing by itself; it is a hundred (37)
and a square less twenty things, which is equal to ten
things. Reduce this according to the rules, which I
have above explained to you.
In like manner, if he say: " I have divided ten into
two parts ; I have multiplied one of the two by the
other, and have then divided the product by the diffe-
rence of the two parts before their multiplication, and the
result of this division is five and one- fourth :"J the com-
putation will be this: You subtract thing from ten; there
remain ten less thing. Multiply the one by the other, it
is ten things less a square. This is the product of the
multiplication of one of the two parts by the other. At
* " Square " in the original,
f ioa:=(io— ary-^rzioo— 20a;+a;2
a:=:i5-\/225— 100=15— >/ 125
xJlO-x)
+ 10 — 2a: ^*
1 ox-^x^ = 51 J — 1 o\x
20jx=::a:--|-52j
jr=:ioJ-7i=3
( 52 )
present you divide this by the difference between the
two parts, which is ten less two things. The quotient
of this division is, according to the statement, five and
a fourth. If, therefore, you muliply five and one-fourth
by ten less two things, the product must be equal to the
above amount, obtained by multiplication, namely, ten
things less one square. Multiply now five and one-
fourth by ten less two squares. The result is fifty-two
dirhems and a half less ten roots and a half, which is
equal to ten roots less a square. Separate now the ten
roots and a half from the fifty-two dirhems, and add
them to the ten roots less a square ; at the same time
separate this square from them, and add it to the
fifty-two dirhems and a half. Thus you find twenty
roots and a half, equal to fifty- two dirhems and a half
and one square. Now continue reducing it, conform-
ably to the rules explained at the commencement of
this book.
(38) If the question be: "There is a square^ two-thirds of
one-fifth of which are equal to one-seventh of its root;"
then the square is equal to one root and half a seventh
of a root; and the root consists of fourteen-fifteenths
of the square.* The computation is this : You
xix^ = l
c^ = ^Ix± = i^j:
X =:lJ
Ti
( 53 )
multiply two- thirds of one-fifth of the square by
seven and a half, in order that the square may be com-
pleted. Multiply that which you have already, namely,
one-seventh of its root, by the same. The result will
be, that the square is equal to one root and half a
seventh of the root ; and the root of the square is one
and a half seventh ; and the square is one and twenty-
nine one hundred and ninety-sixths of a dirhem. Two-
thirds of the fifth of this are thirty parts of the hundred
and ninety-six parts. One-seventh of its root is like-
wise thirty parts of a hundred and ninety-six.
If the instance be : " Three-fourths of the fifth of a
square are equal to four-fifths of its root,"* then the
computation is this : You add one-fifth to the four-
fifths, in order to complete the root. This is then equal
to three and three-fourths of twenty parts, that is, to
fifteen eightieths of the square. Divide now eighty by
fifteen ; the quotient is five and one-third. This is the
root of the square, and the square is twenty-eight and
four-ninths.
If some one say : " What is the amount of a square-
rootjt which, when multiplied by four times itself,
* ixK=l^
f " Square " in the original.
( 54 )
amounts to twenty?*" the answer is this : If you mul-
tiply it by itself it will be five : it is therefore the root
of five.
If somebody ask you for the amount of a square-
root,t which when multiplied by its third amounts to
ten, J the solution is, that when multiplied by itself it
will amount to thirty ; and it is consequently the root
of thirty.
(39) If the question be : " To find a quantity t, which
when multiplied by four times itself, gives one- third of
the first quantity as product,"^ the solution is, that if
you multiply it by twelve times itself, the quantity
itself must re-appear : it is the moiety of one moiety of
one-third.
If the question be : "A square, which when multiplied
by its root gives three times the original square as pro-
duct," 1| then the solution is: that if you multiply the
root by one-third of the square, the original square is
* 4a:2 _ 20
x =\/5
t
" Square " in the
. t xxJ=io
a:2=30
a; =\/30
§ XX4X=1
original.
'
[| X^XX = ^X'
X -s
•
( 55 )
restored ; its root must consequently be three, and the
square itself nine.
If the instance be : " To i&nd a square, four roots of
which, multiplied by three roots, restore the square
with a surplus of forty-four dirhems,*** then the solution
is : that you multiply four roots by three roots, which
gives twelve squares, equal to a square and forty-four
dirhems. Remove now one square of the twelve on
account of the one square connected with the forty- four
dirhems. There remain eleven squares, equal to forty-
four dirhems. Make the division, the result will be
four, and this is the square.
If the instance be : "A square, four of the roots of
which multiplied by five of its roots produce twice the
square, with a surplus of thirty-six dirhems ;"f then the
solution is : that you multiply four roots by five roots,
which gives twenty squares, equal to two squares and
thirty -six dirhems. Remove two squares from the twenty
on account of the other two. The remainder is eigh-
teen squares, equal to thirty-six dirhems. Divide now
thirty-six dirhems by eighteen; the quotient is two,
and this is the square.
1 1x2 __ ^4
X- = 4
X r: 2
f 4xX5a:=2x2-j-36
i8x-=36
x2= 2
( 56 )
(40) In the same manner, if the question be : "A square,
multiply its root by four of its roots, and the product
will be three times the square, with a surplus of fifty
dirhems."t Computation : You multiply the root by four
roots, it is four squares, which are equal to three squares
and fifty dirhems. Remove three squares from the four ;
there remains one square, equal to fifty dirhems. One
root of fifty, multiplied by four roots of the same, gives
two hundred, which is equal to three times the square,
and a residue of fifty dirhems.
If the instance be: "A square, which when added to
twenty dirhems, is equal to twelve of its roots,"+ then
the solution is this : You say, one square and twenty
dirhems are equal to twelve roots. Halve the roots and
multiply them by themselves; this gives thirty-six.
Subtract from this the twenty dirhems, extract the
root from the remainder, and subtract it from the
moiety of the roots, which is six. The remainder is
the root of the square : it is two dirhems, and the square
is four.
If the instance be : " 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
* 4x2= 3^9^. ^Q
x^= 50
f ^2.^20 =12X
:r=6=t\/36~2o = 6±4= lo or 2
( 57 )
itself restores tlie square;"^ then the computation
is this: If you subtract one- third and three dirhems
from tlie square, there remain two-thirds of it less three
dirhems. This is the root. Multiply therefore two- thirds
of thing less three dirhems by itself. You say two-
thirds by two-thirds is four ninths of a square ; and less
two- thirds by three dirhems is two roots : and again,
two-thirds by three dirhems is two roots; and less three
dirhems by less three dirhems is nine dirhems. You (41)
have, therefore, four-ninths of a square and nine dirhems
less four roots, which are equal to one root. Add the
four roots to the one root, then you have five roots,
which are equal to four-nintlis of a square and nine
dirhems. Complete now your square ; that is, multiply
the four-ninths of a square by two and a fourth, which
gives one square ; multiply likewise the nine dirhems
by two and a quarter; this gives twenty and a quarter ;
finally, multiply the five roots by two and a quarter;
this gives eleven roots and a quarter. You have, there-
fore, a square and twenty dirhems and a quarter, equal
to eleven roots and a quarter. Reduce this according to
what I taught you about halving the roots.
* [*-(f+3)r=x
or [if-3]'^=»
X - 9, or 2A
( 58 )
If the instance be : " To find a number,* one-third
of which, when multiplied by one-fourth of it, restores
the *number,"f then the computation is : You multiply
one-third of thing by one-fourth of thing, this gives
one-twelfth of a square, equal to thing, and the square
is equal to twelve things, which is the root of one
hundred and forty-four.
If the instance be : "A number,* one-third of which
and one dirhem multiplied by one-fourth of it and two
dirhems restore the number,* with a surplus of thirteen
dirhems ;"J then the computation is this : You multiply
one- third of thing by one-fourth of thing, this gives
half one-sixth of a square; and you multiply two
dirhems by one-third of thing, this gives two-thirds
of a root; and one dirhem by one-fourth of thing
gives one-fourth of a root ; and one dirhem by two
dirhems gives two dirhems. This altogether is one-
twelfth of a square and two dirhems and eleven-
(42) twelfths of a thing, equal to thing and thirteen dir-
» «
Square " in the original.
X = 12
12 12
( 59 )
hems. Remove now two dirhems from thirteen, on
account of the other two dirhems, the remainder is
eleven dirhems. Remove then the eleven- twelfths of a
root from the one (root on the opposite side), there
remains one-twelfth of a root and eleven dirhems, equal
to one-twelfth of a square. Complete the square: that
is, multiply it by twelve, and do the same with alj you
have. The product is a square, which is equal to a
hundred and thirty-two dirhems and one root. Reduce
this, according to what I have taught you, it will be
right.
If the instance be: "A dirhem and a half to be di-
vided among one person and certain persons, so that the
share of the one person be twice as many dirhems as
there are other persons;''* then the Computation is
this :f You say, the one person and some persons are
one and thing : it is the same as if the question had
been one dirhem and a half to be divided by one and
thing, and the share of one person to be equal to two
things. Multiply, therefore, two things by one and
* The enunciation in the original is faulty, and I have
altered it to correspond with the computation. But in the
computation, x, the number of persons, is fractional f I am
unable to correct the passage satisfactorily.
X =1—^
( 60 )
thing ; it is two squares and two things, equal to one
dirhem and a half. Reduce them to one square : that
is, take the moiety of all you have. You say, there-
fore, one square and one thing are equal to three-
fourths of a dirhem. Reduce this, according to what
I have taught you in the beginning of this work.
If the instance be: "A number,* you remove one-
third of it, and one-fourth of it, and four dirhems : then
you multiply the remainder by itself, and the number,*
is restored, with a surplus of twelve dirhems :"t then
the computation is this : You take thing, and subtract
from it one-third and one-fourth; there remain five-
twelfths of thing. Subtract from this four dirhems:
(43) the remainder is five-twelfths of thing less four dirhems.
Multiply this by itself. Thus the five parts become
five-and-twenty parts ; and if you multiply twelve by
itself, it is a hundred and forty-four. This makes,
therefore, five and twenty hundred and forty-fourths
of a square. Multiply then the four dirhems twice by
the five-twelfths ; this gives forty parts, every twelve of
which make one root (forty-twelfths) ; finally, the four
* " Square" in the original.
tV4-^H4 = 4J^
^' + 232V=24ifx
ili-J-M:2|f- 24_-a;
( 61 )
dirhems, multiplied by four dirhems, give sixteen dir-
hems to be added. The forty-twelfths are equal to
three roots and one-third of a root, to be subtracted.
The whole product is, therefore, twenty-five-hundred-
and-forty-fourths of a square and sixteen, dirhems less
three roots and one-third of a root, equal to the original
number,* which is thing and twelve dirhems. Reduce
this, by adding the three roots and one-third to the
thing and twelve dirhems. Thus you have four roots
and one-third of a root and twelve dirhems. Go on
balancing, and subtract the twelve (dirhems) from six-
teen ; there remain four dirhems and five-and-twenty-
hundred-and-forty-fourths of a square, equal to four
roots and one-third. Now it is necessary to complete
the square. This you can accomplish by multiply-
ing all you have by five and nineteen twenty-fifths.
Multiply, therefore, the twenty-five-one-hundred-and-
forty-fourths of a square by five and nineteen twenty-
fifths. This gives a square. Then multiply the four (44)
dirhems by five and nineteen twenty-fifths ; this gives
twenty-three dirhems and one twenty-fifth. Then
multiply four roots and one- third by five and nineteen
twenty-fifths ; this gives twenty-four roots and twenty-
four twenty-fifths of a root. Now halve the number of
the roots : the moiety is twelve roots and twelve twenty-
fifths of a root. Multiply this by itself. It is one
hundred -and- fifty-five dirhems and four hundred-and-
Square " in the original.
( 62 )
sixty-nine six-hundred-and- twenty-fifths. Subtract
from this the twenty-three dirhems and the one twenty-
fifth connected with the square. The remainder is
one-hundred-and-thirty-two and four-hundred-and-
forty six-hundred- and-twenty-fifths. Take the root of
this : it is eleven dirhems and thirteen twenty-fifths.
Add this to the moiety of the roots, which was twelve
dirhems and twelve twenty-fifths. The sum is twenty-
four. It is the number* which you sought. When
you subtract its third and its fourth and four dirhems,
and multiply the remainder by itself, the number * is
restored, with a surplus of twelve dirhems.
If the question be : " To find a square-root,* which,
when multiplied by two-thirds of itself, amounts to
(45) five;"f then the computation is this : You multiply
one thing by two- thirds of thing; the product is two-
thirds of square, equal to five. Complete it by adding
its moiety to it, and add to five likewise its moiety.
Thus you have a square, equal to seven and a half.
Take its root ; it is the thing which you required, and
which, when multiplied by two-thirds of itself, is equal
to five.
If the instance be: "Two numbers, J the difference
* " Square " in the original,
t ^ ^ 1^ = 5
x2 = 7j
X " Squares " in the original.
( 63 )
of which is two dirhems ; you divide the small one by
the great one, and the quotient is equal to half a dir-
hem ;* then the computation is this : Multiply thing
and two dirhems by the quotient, that is a half. The
product is half a thing and one dirhem, equal to thing.
Remove now half a dirhem on account of the half dir-
hem on the other side. The remainder is one dir-
hem, equal to half a thing. Double it: then you have
thing, equal to two dirhems. This is one of the two
numbers,f and the other is four.
Instance : *' You divide one dirhem amongst a cer-
tain number of men, which number is thing. Now you
add one man more to them, and divide again one dir-
hem amongst them; the quota of each is then one-sixth
of a dirhem less than at the first time."t Computation:
You multiply the first number of men, which is thing,
by the difference of the share for each of the other
number. Then multiply the product by the first and
second number of men, and divide the product by the
— X
x-\-2 - ^
lx+ 1 ==x
i:. = l
a:=;2, x -{- 2 = 4.
' Squares " in the original
t X "" 2 -f 1 = 6
6
X^ + X = 6
v/[Ap+6-4 = a;=2
( 64 )
difference of these two numbers. Thus you obtain the
sum which shall be divided. Multiply, therefore, the
first number of men, which is thing, by the one-
sixth, which is the difference of the shares; this gives
one-sixth of root. Then multiply this by the original
number of the men, and that of the additional one,
that is to say, by thing plus one. The product is one-
sixth of square and one- sixth of root divided by one
(46) dirhem, and this is equal to one dirhem. Complete the
square which you have through multiplying it by six.
Then you have a square and a root equal to six dir-
hems. Halve the root and multiply the moiety by
itself, it is one-fourth. Add this to the six; take the
root of the sum and subtract from it the moiety of the
root, which you have multiplied by itself, namely, a
half. The remainder is the first number of men ; which
in this instance is two.
If the instance be : " To find a square-root,* which
when multiplied by two-thirds of itself amounts to
five :"f then the computation is this : If you multiply
it by itself, it gives seven and a half. Say, therefore.
* " Square " in the original,
t f ^'^ = 5
_^ = A/7i
( 65 )
it is the root of seven and a half multiplied by two-
thirds of the root of seven and a half. Multiply then
two-thirds by two-thirds, it is four-ninths ; and four-
ninths multiplied by seven and a half is three and a
third. The root of three and a third is two- thirds of
the root of seven and a half Multiply three and a
third by seven and a halt ; the product is twenty-five,
and its root is five.
If the instance be : "A square multiplied by three of
its roots is equal to five times the original square;"*
then this is the same as if it had been said, a square,
which when multiplied by its root, is equal to the first
square and two- thirds of it. Then the root of the
square is one and two-thirds, and the square is two
dirhems and seven-ninths.
If the instance be : " Remove one-third from a
square, then multiply the remainder by three roots of
the first square, and the first" square will be restored."f
Computation : If you multiply the first square, before (47)
removing two-thirds from it, by three roots of the
same, then it is one square and a half; for according
to the statement two-thirds of it multiplied by three
* af^ X sx = 5x^
x^ X X = i^x^
* = i|
^2 = 2j
f (:iP-ix^) X 3^=«- .*. f x2 X 3x=x^'
x^X3^='^i^'^
X—2 . . **> —4
K
( 66 )
roots give one square ; and, consequently, the whole of
it multiplied by three roots of it gives one square and a
half. This entire square, when multiplied by one
root, gives half a square ; the root of the square must
therefore be a half, the square one-fourth, two- thirds
of the square one-sixth, and three roots of the square
one and a half. If you multiply one-sixth by one and
a half, the product is one-fourth, which is the square.
Instance : " A square; you subtract four roots of the
same, then take one-third of the remainder; this is
equal to the four roots." The square is two hundred
and fifty-six.* Computation: You know that one-third
of the remainder is equal to four roots ; consequently,
the whole remainder must be twelve roots ; add to this
the four roots ; the sum is sixteen, which is the root of
the square.
Instance : " A square ; you remove one root from it;
and if you add to this root a root of the remainder, the
sum is two dirhems."f Then, this is the root of a
3
.^^
a;^-4x
= I2ar
x2=
16*
X = 16 .-.
X^ =: 256
t Vx^-^
X -j- X = 2
Vx^-x
= 2 - X
r2— a?=44-a;2 — 42;
0:^+3^ =
= 4 + ^2
3^ =
= 4
X =
H
( 67 )
square, which, when added to the root of the same
square, less one root, is equal to two dirhenis. Sub-
tract from this one root of the square, and subtract also
from the two dirhems one root of the square. Then
two dirhems less one root multiplied by itself is four
dirhems and one square less four roots, and this is equal
to a square less one root. Reduce it, and you find a
square and four dirhems, equal to a square and three
roots. Remove square by square ; there remain three
roots, equal to four dirhems ; consequently, one root is
equal to one dirhem and one-third. This is the root of
the square, and the square is one dirhem and seven-
ninths of a dirhem. (48)
Instance : " Subtract three roots from a square, then
multiply the remainder by itself, and the square is
restored."* You know by this statement that the re-
mainder must be a root likewise; and that the square
consists of four such roots; consequently, it must be
sixteen.
»
(x-' -
- 3^)2 =
X-
x'^-
■3^ = ^
x'
= 4x
X
= 4
( 68 )
ON MERCANTILE TRANSACTIONS.
You know that all mercantile transactions of people,
such as buying and selling, exchange and hire, com-
prehend always two notions and four numbers, which
are stated by the enquirer ; namely, measure and price,
and quantity and sum. The number which expresses
the measure is inversely proportionate to the number
which expresses the sum, and the number of the price
inversely proportionate to that of the quantity. Three
of these four numbers are always known, one is un-
known, and this is implied when the person inquiring
says tww much ? and it is the object of the question.
The computation in such instances is this, that you try
the three given numbers ; two of them must necessarily
be inversely proportionate the one to the other. Then
you multiply these two proportionate numbers by each
other, and you divide the product by the third given
number, the proportionate of which is unknown. The
quotient of this division is the unknown number, which
the inquirer asked for ; and it is inversely proportionate
to the divisor.*
Examples. — For the first case : If you are told, " ten
(49) for six, how much for four ?" then ten is the measure ;
* If a is given for h, and A for B, then a i b :: A : B or
aB=.bA.'.a — -- and ^=—7 .
B A
( 69 )
six is the price ; the expression how much implies the
unknown number of the quantity; and four is the
number of the sum. The number of the measure,
which is ten, is inversely proportionate to the number
of the sum, namely, four. Multiply, therefore, ten
by four, that is to say, the two known proportionate
numbers by each other ; the product is forty. Divide
this by the other known number, which is that of the
price, namely, six. The quotient is six and two-
thirds; it is the unknown number, implied in the words
of the question " how much f it is the quantity, and
inversely proportionate to the six, which is the price.
For the second case : Suppose that some one ask this
question : " ten for eight, what must be the sum for
four ?" This is also sometimes expressed thus : " What
must be the price of four of them ?" Ten is the number
of the measure, and is inversely proportionate to the
unknown number of the sum, which is involved in the
expression how much of the statement. Eight is the
number of the price, and this is inversely proportionate
to the known number of the quantity, namely, four.
Multiply now the two known proportionate numbers one
by the other, that is to say, four by eight. The product
is thirty-two. Divide this by the other known number,
which is that of the measure, namely, ten. The quo-
tient is three and one- fifth; this is the number of the
sum, and inversely proportionate to the ten which was
the divisor. In this manner all computations in matters
of business may be solved.
( ^0 )
If somebody says, " a workman receives a pay of ten
(^^) dirhems per month ; how much must be his pay for six
days?" Then you know that six days are one-fifth of
the month; and that his portion of the dirhems must
be proportionate to the portion of the month. You
calculate it by observing that one month, or thirty
days, is the measure, ten dirhems the price, six days
the quantity, and his portion the sum. Multiply the
price, that is, ten, by the quantity, which is propor-
tionate to it, namely, six ; the product is sixty. Divide
this by thirty, which is the known number of the mea-
sure. The quotient is two dirhems, and this is the sum.
This is the proceeding by which all transactions con-
cerning exchange or measures or weights are settled.
MENSURATION.
Know that the meaning of the expression *' one by
one'* is mensuration : one yard (in length) by one yard
(in breadth) being understood.
Every quadrangle of equal sides and angles, which
has one yard for every side, has also one for its area.
Has such a quadrangle two yards for its side, then the
area of the quadrangle is four times the area of a qua-
drangle, the side of which is one yard. The same takes
place with three by three, and so on, ascending or
descending : for instance, a half by a half, which gives
( n )
a quarter, or other fractions, always following the same
rule. A quadrate, every side of which is half a yard, is (51)
equal to one-fourth of the figure which has one yard for
its side. In the same manner, one-third by one-third,
or one-fourth by one-fourth, or one-fifth by one-fifth,
or two-thirds by a half, or more or less than this, al-
ways according to the same rule.
One side of an equilateral quadrangular figure,
taken once, is its root ; or if the same be multiplied by
two, then it is like two of its roots, whether it be small
or great.
If you multiply the height of any equilateral triangle
by the moiety of the basis upon which the line marking
the height stands perpendicularly, the product gives
the area of that triangle.
In every equilateral quadrangle, the product of one
diameter multiplied by the moiety of the other will be
equal to the area of it.
In any circle, the product of its diameter, multiplied
by three and one-seventh, will be equal to the peri-
phery. This is the rule generally followed in practical
life, though it is not quite exact. The geometricians
have two other methods. One of them is, that you
multiply the diameter by itself; then by ten, and
hereafter take the root of the product ; the root will be
the periphery. The other method is used by the astro-
nomers among them : it is this, that you multiply the
diameter by sixty-two thousand eight hundred and
thirty- two and then divide the product by twenty
( ^2 )
thousand ; the quotient is the periphery. Both methods
come very nearly to the same effect.*
If you divide the periphery by three and one-seventh,
the quotient is the diameter.
The area of any circle will be found by multiplying
the moiety of the circumference by the moiety of the
diameter; since, in every polygon of equal sides and
(52) angles, such as triangles, quadrangles, pentagons, and
so on, the area is found by multiplying the moiety of
the circumference by the moiety of the diameter of the
middle circle that may be drawn through it.
If you multiply the diameter of any circle by itself,
and subtract from the product one-seventh and half
one-seventh of the same, then the remainder is equal
to the area of the circle. This comes very nearly to the
same result with the method given above, t
Every part of a circle may be compared to a bow.
It must be either exactly equal to half the circum-
ference, or less or greater than it. This may be ascer-
tained by the arrow of the bow. When this becomes
equal to the moiety of the chord, then the arc is
* The three formulas are,
1st, 3\d=^p i.e. 3.1428 c?
2d, \/iod'^=p i.e. 3.i622'7c?
3d, -» I.e. 3.14166/
20000 '
+ The area of a circle whose diameter is c? is tt—-.
' 4
,icV=Ci-^-rx-,K^.
( ^3 )
exactly the moiety of the circumference: is it shorter
than the moiety of the chord, then the bow is less than
half the circumference; is the arrow longer than half
the chord, then the bow comprises more than half the
circumference.
If you want to ascertain the circle to which it be-
longs, multiply the moiety of the chord by itself, divide
it by the arrow, and add the quotient to the arrow,
the sum is the diameter of the circle to which this bow
belongs.
If you want to compute the area of the bow, mul-
tiply the moiety of the diameter of the circle by the
moiety of the bow, and keep the product in mind.
Then subtract the arrow of the bow from the moiety
of the diameter of the circle, if the bow is smaller than
half the circle ; or if it is greater than half the circle,
subtract half the diameter of the circle from the arrow
of the bow. Multiply the remainder by the moiety of
the chord of the bow, and subtract the product from
that which you have kept in mind if the bow is smaller (53)
than the moiety of the circle, or add it thereto if the
bow is greater than half the circle. The sum after the
addition, or the remainder after the subtraction, is the
area of the bow.
The bulk of a quadrangular body will be found by
multiplying the length by the breadth, and then by the
height.
If it is of another shape than the quadrangular (for
instance, circular or triangular), so, however, that a
L
( ^4 )
line representing its height may stand perpendicularly
on its basis, and yet be parallel to the sides, you must
calculate it by ascertaining at first the area of its basis.
This, multiplied by the height, gives the bulk of the
body.
Cones and pyramids, such as triangular or quadran-
gular ones, are computed by multiplying one- third of
the area of the basis by the height.
Observe, that in every rectangular triangle the two
short sides, each multiplied by itself and the products
added together, equal the product of the long side mul-
tiplied by itself.
The proof of this is the follovi^ing. We draw a qua-
drangle, with equal sides and angles A B C D. We
divide the line A C into two moieties in the point H,
from which we draw a parallel to the point R. Then
we divide, also, the line A B into two moieties at the
point T^ and draw a parallel to the point G. Then the
quadrate A B C D is divided into four quadrangles of
equal sides and angles, and of equal area ; namely, the
squares AK, CK, BK, and DK. Now, we draw from
f54) ^^^ point H to the point T a line which divides the
quadrangle AK into two equal parts: thus there arise
two triangles from the quadrangle, namely, the triangles
A T H and H K T. We know that A T is the moiety
of A B, and that A H is equal to it, being the moiety of
AC; and the line TH joins them opposite the right
angle. In the same manner we draw lines from T to
R, and from R to G, and from G to H. Thus from
( ^5 )
all the squares eight equal triangles arise, four of which
must, consequently, be equal to the moiety of the great
quadrate AD. We know that the line AT multiplied
by itself is like the area of two triangles, and AK gives
the area of two triangles equal to them ; the sum of
them is therefore four triangles. But the line HT,
multiplied by itself, gives likewise the area of four such
triangles. We perceive, therefore, that the sum of AT
multiplied by itself, added to AH multiplied by itself,
is equal to TH multiplied by itself. This is the
observation which we were desirous to elucidate. Here
is the figure to it :
Quadrangles are of five kinds : firstly, with right (55)
angles and equal sides ; secondly, with right angles and
unequal sides ; thirdly, the rhombus, with equal sides
and unequal angles ; fourthly, the rhomboid, the length
of which differs from its breadth, and the angles of
which are unequal, only that the two long and the two
short sides are respectively of equal length; fifthly,
quadrangles with unequal sides and angles.
First kind. — The area of any quadrangle with equal
sides and right angles, or with unequal sides and right
( 76 )
angles, may be found by multiplying the length by the
breadth. The product is the area. For instance : a
quadrangular piece of ground, every side of which has
five yards, has an area of five-and- twenty square yards.
Here is its figure.
Second kind. — A quadrangular piece of ground, the
two long sides of which are of eight yards each, while
the breadth is six. You find the area by multiplying
six by eight, which yields forty- eight yards. Here is
(56) the figure to it :
Third kind, the Rhombus. — Its sides are equal: let
each of them be five, and let its diagonals be, the one
eight and the other six yards. You may then compute
the area, either from one of the diagonals, or from
both. As you know them both, you multiply the one
by the moiety of the other, the product is the area :
that is to say, you multiply eight by three, or six by
four ; this yields twenty-four yards, which is the area.
( 77 )
If you know only one of the diagonals, then you are
aware, that there are two triangles, two sides of each
of which have every one five yards, while the third is
the diagonal. Hereafter you can make the computa-
tion according to the rules for the triangles.* This is
the figure :
The fourth kind, or Rhomboid, is computed in the
same way as the rhombus. Here is the figure to it :
/
4
iii—
4
8
3
/3
The other quadrangles are calculated by drawing a (57)
diagonal, and computing them as triangles.
Triangles are of three kinds, acute-angular, obtuse-
angular, or rectangular. The peculiarity of the rec-
tangular triangle is, that if you multiply each of its
two short sides by itself, and then add them together,
their sum will be equal to the long side multiplied by
itself. The character of the acute-angled triangle is
^ If the two diagonals are d and d\ and the side 5, the
area of the rhombus is _ = ^ x v/*"— — •
2 V 4
( 78 )
this : if you multiply every one of its two short sides
by itself, and add the products, their sum is more
than the long side alone multiplied by itself. The
definition of the obtuse-angled triangle is this : if you
multiply its two short sides each by itself, and then add
the products, their sum is less than the product of the
long side multiplied by itself.
The rectangular triangle has two cathetes and an
hypotenuse. It may be considered as the moiety of a
quadrangle. You find its area by multiplying one of
its cathetes by the moiety of the other. The product
is the area.
Examples. — A rectangular triangle; one cathete being
(58) six yards, the other eight, and the hypotenuse ten.
You make the computation by multiplying six by four :
this gives twenty-four, which is the area. Or if you
prefer, you may also calculate it by the height, which
rises perpendicularly from the longest side of it : for
the two short sides may themselves be considered as
two heights. If you prefer this, you multiply the
height by the moiety of the basis. The product is the
area. This is the figure :
Second kind. — Kri equilateral triangle with acute
angles, every side of which is ten yards long. Its area
( T9 )
may be ascertained by the line representing its height
and the point from which it rises.* Observe, that in
every isosceles triangle, a line to represent the height
drawn to the basis rises from the latter in a right
angle, and the point from which it proceeds is always
situated in the midst of the basis ; if, on the contrary,
the two sides are not equal, then this point never lies
in the middle of the basis. In the case now before us
we perceive, that towards whatever side we may draw
the line which is to represent the height, it must
necessarily always fall in the middle of it, where the
length of the basis is five. Now the height will be
ascertained thus. You multiply five by itself; then
multiply one of the sides, that is ten, by itself, which
gives a hundred. Now you subtract from this the
product of five multiplied by itself, which is twenty-five. (59)
The remainder is seventy-five, the root of which is the
height. This is a line common to two rectangular tri-
angles. If you want to find the area, multiply the
root of seventy-five by the moiety of the basis, which is
five. This you perform by multiplying at first five by
itself; then you may say, that the root of seventy-five is
to be multiplied by the root of twenty-five. Multiply
seventy-five by twenty-five. The product is one thou-
sand eight hundred and seventy-five ; take its root, it is
* The height of the equilateral triangle whose side is lo,
is s/ 10'^ — 5^ rt v/75, and the area of the triangle is
5 v/75 = 25 \/3
( 80 )
the area : it is forty-three and a little. * Here is the
figure :
There are also acute-angled triangles, with different
sides. Their area will be found by means of the line
representing the height and the point from which it
proceeds. Take, for instance, a triangle, one side of
which is fifteen yards, another fourteen, and the third
thirteen yards. In order to find the point from which
the line marking the height does arise, you may take
for the basis any side jou choose ; e. g. that which is
fourteen yards long. The point from which the line
(60) representing the height does arise, lies in this basis at
an unknown distance from either of the two other
sides. Let us try to find its unknown distance from
the side which is thirteen yards long. Multiply this
distance by itself; it becomes an [unknown] square.
Subtract this from thirteen multiplied by itself; that is,
one hundred and sixty-nine. The remainder is one
hundred and sixty-nine less a square. The root from
this is the height. The remainder of the basis is four-
teen less thing. We multiply this by itself; it becomes
one hundred and ninety-six, and a square less twenty-
* The root is 43. 3 +
( 81 )
eight things. We subtract this from fifteen multiplied
by itself; the remainder is twenty-nine dirhems and
twenty-eight things less one square. The root of this
is the height. As, therefore, the root of this is the
height, and the root of one hundred and sixty-nine less
square is the height likewise, we know that they both are
the same.* Reduce them, by removing square against
square, since both are negatives. There remain twenty-
nine [dirhems] plus twenty-eight things, which are
equal to one hundred and sixty-nine. Subtract now
twenty-nine from one hundred and sixty-nine. The
remainder is one hundred and forty, equal to twenty-
eight things. One thing is, consequently, five. This is
the distance of the said point from the side of thirteen
yards. The complement of the basis towards the other
side is nine. Now in order to find the height, you
multiply five by itself, and subtract it from the conti-
guous side, which is thirteen, multiplied by itself. The
remainder is one hundred and forty-four. Its root is the
height. It is twelve. The height forms always two (gi)
right angles with the basis, and it is called the column^
on account of its standing perpendicularly. Multiply
the height into half the basis, which is seven. The
* \/T69 — «2 = 29 -h 28a: — a?2
1G3 = 29 + 28a;
140 =z 28x
5 =^
M
( 82 )
product is eighty-four, which is the area. Here is the
figure :
The third species is that of the obtuse-angled triangle
with one obtuse angle and sides of different length.
For instance, one side being six, another five, and the
third nine. The area of such a triangle will be found
by means of the height and of the point from which a
line representing the same arises. This point can,
within such a triangle, lie only in its longest side. Take
therefore this as the basis : for if you choose to take one
of the short sides as the basis, then this point would
fall beyond the triangle. You may find the distance
of this point, and the height, in the same manner,
which I have shown in the acute- angled triangle; the
whole computation is the same. Here is the figure :
We have above treated at length of the circles, of
their qualities and their computation. The following
(62) is an example : If a circle has seven for its diameter,
then it has twenty-two for its circumference. Its area
you find in the following manner : Multiply the moiety
( 83 )
of the diameter, which is three and a half, by the moiety
of the circumference, which is eleven. The product is
thirty-eight and a half, which is the area. Or you may
also multiply the diameter, which is seven, by itself: this
is forty-nine; subtracting herefrom one-seventh and half
one-seventh, which is ten and a half, there remain thirty-
eight and a half, which is the area. Here is the figure:
If some one inquires about the bulk of a pyramidal
pillar, its base being four yards by four yards, its
height ten yards, and the dimensions at its upper ex-
tremity two yards by two yards ; then we know already
that every pyramid is decreasing towards its top, and
that one-third of the area of its basis, multiplied by the
height, gives its bulk. The present pyramid has no top.
We must therefore seek to ascertain what is wanting
in its height to complete the top. We observe, that the
proportion of the entire height to the ten, which we
have now before us, is equal to the proportion of four
to two. Now as two is the moiety of four, ten must
likewise be the moiety of the entire height, and the
whole height of the pillar must be twenty yards. At
present we take one-third of the area of the basis,
that is, five and one-third, and multiply it by the
length, which is twenty* The product is one hundred (6^)
( 84 )
and six yards and two-thirds. Herefrom we must then
subtract the piece, which we have added in order to
complete the pyramid. This we perform by multiply-
ing one and one-third, which is one-third of the pro-
duct of two by two, by ten : this gives thirteen and a
third. This is the piece which we have added in order
to complete the pyramid. Subtracting this from one
hundred and six yards and two-thirds, there remain
ninety- three yards and one-third : and this is the bulk
of the mutilated pyramid. This is the figure :
If the pillar has a circular basis, subtract one-seventh
and half a seventh from the product of the diameter
multiplied by itself, the remainder is the basis.
If some one says : " There is a triangular piece of
land, two of its sides having ten yards each, and the
basis twelve ; what must be the length of one side of a
quadrate situated within such a triangle ?" the solution
is this. At first you ascertain the height of the trian-
gle, by multiplying the moiety of the basis^ (which is
six) by itself, and subtracting the product, which is
thirty-six, from one of the two short sides multiplied
by itself, which is one hundred ; the remainder is
( 85 )
sixty-four: take the root from this; it is eight. This (64)
is the height of the triangle. Its area is, therefore,
forty-eight yards : such being the product of the height
multiplied by the moiety of the basis, which is six.
Now we assume that one side of the quadrate inquired
for is thing. We multiply it by itself; thus it becomes
a square, which we keep in mind. We know that
there must remain two triangles on the two sides of the
quadrate, and one above it. The two triangles on
both sides of it are equal to each other : both having
the same height and being rectangular. You find their
area by multiplying thing by six less half a thing,
which gives six things less half a square. This is the
area of both the triangles on the two sides of the qua-
drate together. The area of the upper triangle will be
found by multiplying eight less thing, which is the
height, by half one thing. The product is four things
less half a square. This altogether is equal to the area
of the quadrate plus that of the three triangles: or,
ten things are equal to forty-eight, which is the area of
the great triangle. One thing from this is four yards
and four-fifths of a yard ; and this is the length of any
side of the quadrate. Here is the figure :
•s| ^^ 3i
( 86 )
ON LEGACIES.
On Capital^ and Money lent.
(65) " A MAN dies, leaving two sons behind him, and
bequeathing one-third of his capital to a stranger. He
leaves ten dirhems of property and a claim of ten dir-
hems upon one of the sons."
Computation : You call the sum which is taken out
of the debt thing. Add this to the capital, which is ten
dirhems. The sum is ten and thing. Subtract one-third
of this, since he has bequeathed one-third of his pro-
perty, that is, three dirhems and one-third of thing.
The remainder is six dirhems and two-thirds of thing.
Divide this between the two sons. The portion of
each of them is three dirhems and one-third plus one-
third of thing. This is equal to the thing which was
sought for.* Reduce it, by removing one-third from
* If a father dies, leaving n sons, one of whom owes the
father a sum exceeding an wth part of the residue of the
father's estate, after paying legacies, then such son retains
the whole sum which he owes the father : part, as a set-off
against his share of the residue, the surplus as a gift from
the father.
In the present example, let each son's share of the residue
be equal to x,
§ [io-|-^] =2.1; /, i+a: — 30? /, 10 = 2j: ,\x — ^.
The stranger receives 5 ; and the son, who is not indebted
to the father, receives 5.
( 87 )
thing, on account of the other third of thing. There
remain two-thirds of thing, equal to three dirhems and
one-third. It is then only required that you complete
the thing, by adding to it as much as one half of the
same ; accordingly, you add to three and one-third as
much as one-half of them : This gives five dirhems,
which is the thing that is taken out of the debts.
If he leaves two sons and ten dirhems of capital and
a demand of ten dirhems against one of the sons, and
bequeaths one-fifth of his property and one dirhem to
a stranger, the computation is this : Call the sum which
is taken out of the debt, thing. Add this to the pro-
perty ; the sum is thing and ten dirhems. Subtract
one-fifth of this, since he has bequeathed one-fifth of (66)
his capital, that is, two dirhems and one-fifth of thing ;
the remainder is eight dirhems and four-fifths of thing.
Subtract also the one dirhem which he has bequeathed;
there remain seven dirhems and four-fifths of thing.
Divide this between the two sons ; there will be for each
of them three dirhems and a half plus two-fifths of
thing ; and this is equal to one thing.^ Reduce it by
subtracting two-fifths of thing from thing. Then you
have three-fifths of thing, equal to three dirhems and a
half. Complete the thing by adding to it two-thirds of
the same : add as much to the three dirhems and a half,
I [lo-f-a;] — 1=20; ,'. flio+a:2—i=x
The stranger receives ;^[io + \^ ] + 1 ~4^
( 88 )
namely, two dirhems and one-third ; the sum is five and
five-sixths. This is the thing, or the amount which is
taken from the debt.
If he leaves three sons, and bequeaths one- fifth of his
property less one dirhem, leaving ten dirhems of capital
and a demand of ten dirhems against one of the sons,
the computation is this : You call the sum which is
taken from the debt thing. Add this to the capital ;
it gives ten and thing. Subtract from this one-fifth of
it for the legacy : it is two dirhems and one-fifth of
thing. There remain eight dirhems and four-fifths of
thing ; add to this one dirhem, since he stated " less
one dirhem." Thus you have nine dirhems and four-
fifths of thing. Divide this between the three sons.
There will be for each son three dirhems, and one-
fifth and one- third and one-fifth of thing. This equals
one thing. ^ Subtract one-fifth and one- third of one-
(6T) fifth of thing from thing. There remain eleven-
fifteenths of thing, equal to three dirhems. It is now
required to complete the thing. For this purpose, add
to it four-elevenths, and do the same with the three
dirhems, by adding to them one dirhem and one-
eleventh. Then you have four dirhems and one-
eleventh, which are equal to thing. This is the sum
which is taken out of the debt.
JThe stranger receives y[xo-|-^x] — i = i-3^t
( 89 )
Ow another Species of Legacy.
*' A man dies, leaving his mother, his wife, and two
brothers and two sisters by the same father and mother
with himself ; and he bequeaths to a stranger one-ninth
of his capital."
Computation:* You constitute their shaifes by taking
them out of forty-eight parts. You know that if you
take one-ninth from any capital, eight-ninths of it will
remain. Add now to the eight-ninths one-eighth of the
same, and to the forty-eight also one-eighth of them,
namely, six, in order to complete your capital. This
gives fifty-four. The person to whom one-ninth is
bequeathed receives six out of this, being one-ninth of
the whole capital. The remaining forty-eight will be
distributed among the heirs, proportionably to their
legal shares.
If the instance be: "A woman dies, leaving her
husband, a son, and three daughters, and bequeathing
* It appears in the sequel (p. 96) that a widow is enti-
tled to l^th, and a mother to ^th of the residue ; J + e^iJ?
leaving ^ of the residue to be distributed between two bro-
thers and two sisters ; that is, '^ between a brother and a
sister; but in what proportion these 17 parts are to be
divided between the brother and sister does not appear in
the course of this treatise.
Let the whole capital of the testator = 1
and let each 48th share of the residue =x
8:^48^ ... ^=Qx :, ^=x
that is, each 48th part of the residue -^^th of the whole
capital.
( 90 )
to a stranger one-eighth and one- seventh of her capi-
(68)tal;" then you constitute the shares of the heirs, by
taking them out of twenty.* Take a capital, and sub-
tract from it one -eighth and one-seventh of the same.
The remainder is, a capital less one-eighth and one-
seventh. Complete your capital by adding to that
vi^hich you have already, fifteen forty-one parts. Mul-
tiply the parts of the capital, which are twenty, by
forty-one ; the product is eight hundred and twenty.
Add to it fifteen forty-one parts of the same, which are
three hundred : the sum is one thousand one hundred
and twenty parts. The person to whom one-eighth
and one-seventh were bequeathed, receives one-eighth
and one-seventh of this. One seventh of it is one hun-
dred and sixty, and one-eighth one hundred and forty.
Subtracting this, there remain eight hundred and
twenty parts for the heirs, proportionably to their legal
shares.
* A husband is entitled to :ith of the residue, and the
sons and daughters divide the remaining |ths of the residue
in such proportion, that a son receives twice as much as a
daughter. In the present instance, as there are three daughters
and one son, each daughter receives } of |, = 2%» ^^ ^^^
residue, and the son, /^. Since the stranger takes 1-+^ =
Jf of the capital, the residue =41- of the capital, and each
^^th share of the residue=Jj^ X^=^^ of the capital.
The stranger, therefore, receives i| =t^~, =T¥o^ff of the
capital.
( 91 )
On another Species of Legacies^'^ viz.
If nothing has been imposed on some of the heirs,t
and something has been imposed on others ; the legacy
amounting to more than one-third. It must be known,
that the law for such a case is, that if more than one-
third of the legacy has been imposed on one of the
heirs, this enters into his share ; but that also those on
whom nothing has been imposed must, nevertheless,
contribute one-third.
Example: " A woman dies, leaving her husband, a
son, and her mother. She bequeaths to a person two-
fifths, and to another one- fourth of her capital. She
imposes the two legacies together on her son, and on
her mother one moiety (of the mother's share of the
residue) ; on her husband she imposes nothing but one-
third, (which he must contribute, according to the
* The problems in this chapter may be considered as
belonging rather to Law than to Algebra, as they contain
little more than enunciations of the law of inheritance in
certain complicated cases.
f If some heirs are, by a testator, charged with payment
of bequests, and other heirs are not charged with payment
of any bequests whatever : if one bequest exceeds in amount
Jd of the testator's whole property ; and if one of his heirs
is charged with payment of more than Jd of such bequest ?
then, whatever share of the residue such heir is entitled to
receive, the like share must he pay of the bequest where-
with he is charged, and those heirs whom the testator has
not charged with any payment, must each contribute towards
paying the bequests a third part of their several shares of
the residue.
( 92 )
law)."* Computation; You constitute the shares of the
(^9) heritage, by taking them out of twelve parts : the son
receives seven of them, the husband three, and the
mother two parts. You know that the husband must
give up one- third of his share; accordingly he retains
twice as much as that which is detracted from his share
for the legacy. As he has three parts in hand, one of
these falls to the legacy, and the remaining two parts
he retains for himself. The tw^o legacies together are
imposed upon the son. It is therefore necessary to
subtract from his share two-fifths and one-fourth of the
same. He thus retains seven twentieths of his entire
original share, dividing the whole of it into twenty
equal parts. The mother retains as much as she con-
tributes to the legacy ; this is one (twelfth part), the
entire amount of what she had received being two parts.
* If the bequests stated in the present example were charged
on the heirs collectively, the husband would be entitled to ^,
the mother to J of the residue : ^-f ^— x%; ^^^ remainder J_
would be the son's share of the residue ; but since the
bequests, J+i — 1-| of the capital, are charged upon the son
and mother, the law throws a portion of the charge on the
husband.
TheHusband contributes J x §■ =20X2^0, and retains i X ^ =40X^i(y
The Mother ^ x h =2ox-^lj^, ^ x i =20x^^0
The Son • • .tVxM = 91 x^, 1^X2^=49x^1^-
Total contributed = if J Total retained = i£g
I- 4- J^ — JL4._5_ — JLA
The Legatee, to whom the f are bequeathed, receives -^^ x i|i =
The Legatee, to whom J is bequeathed, receives /^ x Jf^ = •'^ ^^ — -
3 120
( 93 )
Take now a sum, one -fourth of which may be di-
vided into thirds, or of one-sixth of which the moiety
may be taken ; this being again divisible by twenty.
Such a capital is two hundred and forty. The mother
receives one-sixth of this, namely, forty ; twenty from
this fall to the legacy, and she retains twenty for her-
self. The husband receives one-fourth, namely, sixty ;
from which twenty belong to the legacy, so that he
retains forty. The remaining hundred and forty belong
to the son ; the legacy from this is two-fifths and one-
fourth, or ninety-one; so that there remain fortyr
nine. The entire sum for the legacies is, therefore,
one hundred and thirty-one, which must be divided
among the two legatees. The one to whom two-fifths
were bequeathed, receives eight-thirteenths of this;
the one to whom one-fourth was devised, receives five-
thirteenths. If you wish distinctly to express the
shares of the two legatees, you need only to multiply (70)
the parts of the heritage by thirteen, and to take them
out of a capital of three thousand one hundred and
twenty.
But if she had imposed on her son (payment of) the
two-fifths to the person to whom the two-fifths were
bequeathed, and of nothing to the other legatee ; and
upon her mother (payment of) the one-fourth to the
person to whom one-fourth was granted, and of nothing
to the other legatee; and upon her husband nothing
besides the one- third (which he must according to law
contribute) to both ; then you know that this one-third
( 94, )
comes to the advantage of the heirs collectively ; and
the legatee of the two-fifths receives eight- thirteenths,
and the legatee of the one-fourth receives five-thir-
teenths from it. Constitute the shares as I have shown
above, by taking twelve parts ; the husband receives
one-fourth of them, the mother one-sixth, and the son
that which remains.^ Computation : You know that at
all events the husband must give up one-third of his
share, which consists of three parts. The mother must
likewise give up one-third, of which each legatee par-
takes according to the proportion of his legacy. Be-
sides, she must pay to the legatee to whom one- fourth is
bequeathed, and whose legacy has been imposed on her,
as much as the difference between the one-fourth and his
5-t-4 -g— 20
The Husband, who would be entitled to j^ of the residue, is
not charged by the Testator with any bequest.
The Mother who would be entitled to J of the residue, is
charged with the payment of |^ to the Legatee A.
The Son, who would be entitled to yV of the residue, is
charged with payment off to the Legatee B.
The Husband! , , o i • , « ,
contributes I 4 X i = 780x^3^^ ; retamsix|=4|65.
The Mother . . . i [i + _8^ X ^] = 7 1 o x ^^-V^ ; retains ^^^o^
The Son t'itH + A x*] - 2884X^ ; retains ff Jf
Total contributed = 437 *; Total retained =4|:8 6
The Legatee A, to whom 1 is") 5 ^ 4374 _5 X4J74
bequeathed, receives J 1^ ^ tj^^ — ^b 4 8 (J
, The Legatee B, to whom f are 1 j, ^ 4" 4 - b x 4 , , 4
bequeathed, receives / ^j »37" ~" JsiOJS
( 95 ) '
poftion of the one-third, namely, nineteen one hundred
and fifty-sixths of her entire share, considering her share
as consisting of one hundred and fifty-six parts. His
portion of the one-third of her share is twenty parts.
But what she gives him is one-fourth of her entire share,
namely, thirty-nine parts. One third of her share is
taken for both legacies, and besides nineteen parts
which she must pay to him alone. The son gives to the
legatee to whom two- fifths are bequeathed as much as
the difference between two-fifths of his (the son's) share CTl)
and the legatee's portion of the one-third, namely,
thirty-eight one hundred and ninety-fifths of his (the
son's) entire share^ besides the one-third of it which is
taken off from both legacies. The portion which he
(the legatee) receives from this one-third, is eight-
thirteenths of it, namely, forty (one hundred and ninety-
fifths); and what the son contributes of the two-fifths
from his share is thirty-eight. These together make
seventy-eight. Consequently, sixty-five will be taken
from the son, as being one-third of his share, for both
legacies, and besides this he gives thirty-eight to the
one of them in particular. If you wish to express the
parts of the heritage distinctly, you may do so with
nine hundred and sixty-four thousand and eighty.
On another Species of Legacies,
" A man dies, leaving four sons and his wife ; and
bequeathing to a person as much as the share of one
( 96 )
of the sons less the amount of the share of the widow."
Divide the heritage into thirty-two parts. The widow
receives one-eighth,* namely, four; and each son seven.
Consequently the legatee must receive three- sevenths of
the share of a son. Add, therefore, to the heritage
three-sevenths of the share of a son, that is to say>
three parts, which is the amount of the legacy. This
gives thirty-five, from which the legatee receives three;
and the remaining thirty-two are distributed among
the heirs proportionably to their legal shares.
If he leaves two sons and a daughter, f and bequeaths
to some one as much as would be the share of a third
son, if he had one; then you must consider, what
(72) would be the share of each son, in case he had three.
Assume this to be seven, and for the entire heritage
* A widow is entitled to Jth of the residue ; therefore
Jths of the residue are to be distributed among the sons of
the testator. Let x be the stranger's legacy. The widow's
share =lnf; each son's share =ix§[i~a?]; and a son's
share, minus the widow's share = [1 — i] Lll5 :
1 X
/. a;=|.i_^ .-. x=^-^; i-^^lf A son's share ■:^^;
the widow's share = ^.
f A son is entitled to receive twice as much as a daughter.
Were there three sons and one daughter, each son would
receive f ths of the residue. Let x be the stranger's legacy.
.'. f[i— a:]=ar .*. a; = f, andi— a; = -^
Each Son's share = | {\—x\ = | x J = J^
The Daughter's share =;!■ [i — x] =^-g
The Stranger's legacy =f =i^
( 97 )
take a number, one-fifth of which may be divided into
sevenths, and one-seventh of which may be divided into
fifths. Such a number is thirty-five. Add to it two-
sevenths of the same, namely, ten. This gives forty-
five. Herefrom the legatee receives ten, each son four-
teen, and the daughter seven.
If he leaves a mother, three sons, and a daughter,
and bequeaths to some one as much as the share of one
of his sons less the amount of the share of a second
daughter, in case he had one ; then you distribute the
heritage into such a number of parts as may be divided
among the actual heirs, and also among the same, if a
second daughter were added to them.* Such a number
is three hundred and thirty-six. The share of the
second daughter, if there were one, would be thirty-
five, and that of a son eighty ; their difference is forty-
five, and this is the legacy. Add to it three hundred
and thirty-six, the sum is three hundred and eighty-
one, which is the number of parts of the entire heritage.
* Let X be the stranger's legacy ; i — x is the residue.
A widow's share of the residue is ^th ; there remains
J [i— j:], to be distributed among the children.
Since there are 3 sons, and 1 daughter, 1 2 ^ 5r-__ i
a son's share is / t ^ ^L J
Were there 3 sons and 2 daughters, ^lixsri— j:1
daughter's share would be J s ^l J
The difference = ^^ Kf[i -x]
1 _a;rj:|^|; the widow's share - ^
the daughter's share — ^^
( 98 )
If he leaves three sons, and bequeaths to some one
as much as the share of one of his sons, less the share
of a daughter, supposing he had one, plus one-third
of the remainder of the one-third; the computation
will be this :* distribute the heritage into such a number
of parts as may be divided among the actual heirs, and
also among them if a daughter were added to them.
Such a number is twenty-one. Were a daughter among
the heirs, her share would be three, and that of a son
seven. The testator has therefore bequeathed to the
(73) legatee four-sevenths of the share of a son, and one-
third of what remains from one-third. Take therefore
one-third, and remove from it four-sevenths of the
share of a son. There remains one- third of the capital
less four-sevenths of the share of a son. Subtract now
one-third of what remains of the one-third, that is to
say, one-ninth of the capital less one-seventh and one-
third of the seventh of the share of a son ; the remainder
* Since there are 3 sons, each son's share of the residue n J.
Were there 3 sons and a daughter, the daughter's share
would be \.
i-i=4
3
Let X be the stranger's legacy, and v a son's share
Then i—x = ^v
and i^a:^%-{.i-±v-i[i^^v]=^3v
.-. f + frr3AxV' orf=J^«
.*. f=V '^ •*• *' = 2TT = ^ ^°^'^ share
X zz -^^ = the stranger's legacy.
( 99 )
is two-ninths of the capital less two-sevenths and two-
thirds of a seventh of the share of a son. Add this to
the two-thirds of the capital ; the sum is eight- ninths
of the capital less two-sevenths and two thirds of a
seventh of the share of a son, or eight twenty-one parts
of that share, and this is equal to three shares. Re-
duce this, you have then eight-ninths of the capital,
equal to three shares and eight twenty-one parts of a
share. Complete the capital by adding to eight-ninths
as much as one- eighth of the same, and add in the
same proportion to the shares. Then you find the
capital equal to three shares and forty-five fifty-sixth
parts of a share. Calculating now each share equal to
fifty-six, the whole capital is two hundred and thirteen,
the first legacy thirty-two, the second thirteen, and of
the remaining one hundred and sixty-eight each son
takes fifty-six.
On another Species of Legacies.
" A woman dies, leaving her daughter, her mother,
and her husband, and bequeaths to some one as much
as the share of her mother, and to another as much as
one-ninth of her entire capital."* Computation : You
begin by dividing the heritage into thirteen parts, two
* In the former examples (p. 90) when a husband and a
mother were among the heirs, a husband was found to be
entitled to ^=^~^ and a mother to 6=y% of the residue.
Here a husband is stated to be entitled to -f.^ , and a mother
to -f^ of the residue.
( 100 )
of which the mother receives. Now you perceive that the
C^^) legacies amount to two parts plus one-ninth of the en-
tire capital. Subtracting this, there remains eight-ninths
of the capital less two parts, for distribution among
the heirs. Complete the capital, by making the eight-
ninths less two parts to be thirteen parts, and adding
two parts to it, so that you have fifteen parts, equal
to eight-ninths of capital; then add to this one-
eighth of the same, and to the fifteen parts add like-
wise one- eighth of the same, namely, one part and
seven-eighths ; then you have sixteen parts and seven-
eighths. The person to whom one-ninth is bequeathed,
receives one-ninth of this, namely, one part and seven-
eighths ; the other, to whom as much as the share of
the mother is bequeathed, receives two parts. The
remaining thirteen parts are divided among the heirs,
according to their legal shares. You best determine
the respective shares by dividing the whole heritage
into one hundred and thirty-five parts.
If she has bequeathed as much as the share of the
husband and one- eighth and one-tenth of the capital,*
Let -f^ of the residue =zv
.'. urr-jf^ of the capital
A mother's share— ^i^^^
A husband's share of the residue is -^^
.•. vrrg3j^; a husband's share =^^-^q
The stranger's legacy = ||^
( 101 )
then you begin by dividing the heritage into thirteen
parts. Add to this as much as the share of the hus-
band, namely, three; thus you have sixteen. This is
what remains of the capital after the deduction of one-
eighth and one-tenth, that is to say, of nine-fortieths.
The remainder of the capital, after the deduction of
one-eighth and one- tenth, is thirty-one fortieths of the
same, v^^hich must be equal to sixteen parts. Complete
your capital by adding to it nine thirty-one parts of the
same, and multiply sixteen by thirty-one, which gives
four hundred and ninety-six ; add to this nine thirty-
one parts of the same, which is one hundred and forty- ^75)
four. The sum is six hundred and forty. Subtract
one-eighth and one-tenth from it, which is one hun-
dred and forty-four, and as much as the share of the
husband, which is ninety- three. There remains four
hundred and three, of which the husband receives
ninety-three, the mother sixty-two, and every daughter
one hundred and twenty-four.
If the heirs are the same,* but that she bequeaths to
a person as much as the share of the husband, less
one-ninth and one-tenth of what remains of the capital,
•••W [1-3] =13^
... 1^9= [13-1-1^9] «
The husband's share =y^t
The stranger's legacy =Yf§y
( 102 )
after the subtraction of that share, the computation is
this : Divide the heritage into thirteen parts. The
legacy from the whole capital is three parts, after the
subtraction of which there remains the capital less three
parts. Now, one-ninth and one-tenth of the remain-
ing capital must be added, namely, one-ninth and one-
tenth of the whole capital less one- ninth and one-tenth
of three parts, or less nineteen- thirtieths of a part ; this
yields the capital and one-ninth and one -tenth less
three parts and nineteen-thirtieths of a part, equal to
thirteen parts. Reduce this, by removing the three
parts and nineteen-thirtieths from your capital, and
adding them to the thirteen parts. Then you have
the capital and one-ninth and one-tenth of the same,
equal to sixteen parts and nineteen-thirtieths of a part.
Reduce this to one capital, by subtracting from it
nineteen one-hundred-and-ninths. There remains a
(76) capital, equal to thirteen parts and eighty one-hundred-
and-ninths. Divide each part into one hundred and
nine parts, by multiplying thirteen by one hundred
and nine, and add eighty to it. This gives one thou-
sand four hundred and ninety-seven parts. The share
of the husband from it is three hundred and twenty-
seven parts.
If some one leaves two sisters and a wife,* and be-
queaths to another person as much as the share of a
* When the heirs are a wife, and 2 sisters, they each
inherit ^ of the residue.
Let
( 103 )
sister less one- eighth of what remains of the capital
after the deduction of the legacy, the computation is
this : You consider the heritage as consisting of twelve
parts. Each sister receives one-third of what remains
of the capital after the subtraction of the legacy ; that
is, of the capital less the legacy. You perceive that
one-eighth of the remainder plus the legacy equals the
share of a sister ; and also, one-eighth of the remainder
is as much as one-eighth of the whole capital less one-
eighth of the legacy ; and again, one-eighth of the
capital less one-eighth of the legacy added to the legacy
equals the share of a sister, namely, one-eighth of the
capital and seven-eighths of the legacy. The whole
capital is therefore equal to three- eighths of the capital
plus three and five-eighth times the legacy. Subtract
now from the capital three- eighths of the same. There
remain five-eighths of the capital, equal to three and
five-eighth times the legacy ; and the entire capital is
equal to five and four-fifth times the legacy. Conse-
quently, if you assume the capital to be twenty-nine,
the legacy is five, and each sister's share eight.
Let X be the stranger's iegacy.
3 [^~""^]= ^ sister's share
i[l-a;]-J[l-a:]=ar
... /^[i~x]=ar .•.A=fl>^
and a sister's share =^
( 104 )
On another Species of Legacies.
" A man dies, and leaves four sons, and bequeaths
tx) some person as much as the share of one of his sons;
and to another, one-fourth of what remains after the
deduction of the above share from one-third." You
perceive that this legacy belongs to the class of those
V*^) which are taken from one-third of the capital.* Compu-
tation : Take one-third of the capital, and subtract
from it the share of a son. The remainder is one-
third of the capital less the share. Then subtract from
it one-fourth of what remains of the one-third, namely,
one-fourth of one- third less one-fourth of the share.
The remainder is one-fourth of the capital less three-
fourths of the share. Add hereto two-thirds of the
capital : then you have eleven-twelfths of the capital less
three-fourths of a share, equal to four shares. Reduce
this by removing the three-fourths of the share from the
capital, and adding them to the four shares. Then you
have eleven- twelfths of the capital, equal to four shares
and three-fourths. Complete your capital, by adding
to the four shares and three-fourths one-fourth of the
same. Then you have five shares and two-elevenths,
* Let the first bequest r= v; and the second =1/
Then i—v —y — 41;
i.e.f+^-«-J[^-»]=4«
•••|+f»-''] = 4»
•••*+A=[4+J]» •••«=',?«
.*. u=:i_i; the 2d bequest =3^
( 105 )
equal to the capital. Suppose, now, every share to be
eleven ; then the whole square will be fifty-seven ; one-
third of this is nineteen ; from this one share, namely,
eleven, must be subtracted ; there remain eight. The
legatee, to whom one-fourth of this remainder was be-
queathed, receives two. The remaining six are re-
turned to the other two-thirds, which are thirty-eight.
Their sum is forty-four, which is to be divided amongst
the four sons; so that each son receives eleven.
If he leaves four sons, and bequeaths to a person as
much as the share of a son, less one-fifth of what re-
mains ffom one-third after the deduction of that share,
then this is likewise a legacy, which is taken from one-
third.^ Take one-third, and subtract from it one
share ; there remains one-third less the share. Then
return to it that which was excepted, namely, one-fifth
of the one-third less one-fifth of the share. This gives
one-third and one-fifth of one-third (or two-fifths) (78)
less one share and one-fifth of a share. Add this to
two-thirds of the capital. The sum is, the capital and
one-third of one-fifth of the capital less one share and
one-fifth of a share, equal to four shares. Reduce this
by removing one share and one-fifth from the capital,
or fx^-v-f^[i-v]=r4v
^ , and the stranger's legacy = ^
( 106 )
and add to it the four shares. Then you have the
capital and one -third of one-fifth of the capital, which
are equal to five shares and one- fifth. Reduce this to
one capital, by subtracting from what you have the
moiety of one-eighth of it, that is to say, one-sixteenth.
Then you find the capital equal to four shares and
seven-eighths of a share. Assume now thirty-nine as
capital; one- third of it will be thirteen, and one share
eight ; what remains of one-third, after the deduction
of that share, is five, and one-fifth of this is one. Sub-
tract now the one, which was excepted from the legacy ;
the remaining legacy then is seven ; subtracting this
from the one- third of the capital, there remain six.
Add this to the two-thirds of the capital, namely, to
the twenty-six parts, the sum is thirty-two; which,
when distributed among the four sons, yields eight for
each of them.
If he leaves three sons and a daughter,* and be-
queaths to some person as much as the share of a
* Since there are three sons and one daughter, the daugh-
ter receives i, and each son |^ths of the residue.
If the 1st legacy = 1), the 2d =:^, and therefore a daugh-
ter's share = v,
... QJ — ii-?i) . — 1B8
The 2d legacy = ..^ = r?g,.
( 107 )
daughter, and to another one-fifth and one-sixth of
what remains of two-sevenths of the capital after the
deduction of the first legacy ; then this legacy is to be
taken out of two-sevenths of the capital. Subtract
from two-sevenths the share of the daughter: there
remain two -sevenths of the capital less that share.
Deduct from this the second legacy, which comprises (T9)
one-fifth and one-sixth of this remainder : there remain
one-seventh and four-fifteenths of one-seventh of the
capital less nine teen-thirtieths of the share. Add to
this the other five-sevenths of the capital: then you
have six-sevenths and four-fifteenths of one-seventh of
the capital less nineteen thirtieths of the share, equal to
seven shares. Reduce this, by removing the nineteen
thirtieths, and adding them to the seven shares : then
you have six- sevenths and four-fifteenths of one-seventh
of capital, equal to seven shares and nineteen-thirtieths.
Complete your capital by adding to every thing that
you have eleven ninety-fourths of the same ; thus the
capital will be equal to eight shares and ninety-nine
one hundred and eighty-eighths. Assume now the
capital to be one thousand six hundred and three ; then
the share of the daughter is one hundred and eighty-
eight. Take two- sevenths of the capital ; that is, four
hundred and fifty-eight. Subtract from this the share,
which is one hundred and eighty-eight ; there remain
two hundred and seventy. Remove one-fifth and one-
sixth of this, namely, ninety-nine ; the remainder is
one hundred and seventy-one. Add thereto five-
( 108 )
sevenths of the capital, which is one thousand one
hundred and forty-five. The sum is one thousand three
(80) hundred and sixteen parts. This may be divided into
seven shares, each of one hundred and eighty-eight
parts ; then this is the share of the daughter, whilst
every son receives twice as much.
If the heirs are the same, and he bequeaths to some
person as much as the share of the daughter, and to
another person one-fourth and one-fifth out of what
remains from two-fifths of his capital after the deduc-
tion of the share ; this is the computation :* You must
observe that the legacy is determined by the two fifths.
Take two-fifths of the capital and subtract the shares :
the remainder is, two-fifths of the capital less the share.
Subtract from this remainder one-fourth and one-fifth
of the same, namely, nine- twentieths of two-fifths, less
as much of the share. The remainder is one-fifth
and one-tenth of one fifth of the capital less eleven-
twentieths of the share. Add thereto three-fifths of the
Let the ist legacy =v = a daughter's share
Let the 2d legacy =y
1— V — ^=7^
•••4+*-»'-A[f-«]=7«
••4+M [#-"] =7"
and the 2d legacy, y, =tA
( 109 )
capital : the sum is four- fifths and one- tenth of one-
fifth of the capital, less eleven-twentieths of the share,
equal to seven shares. Reduce this by removing the
eleven-twentieths of a share, and adding them to the
seven shares. Then you have the same four-fifths and
one-tenth of one-fifth of capital, equal to seven shares
and eleven-twentieths. Complete the capital by adding
to any thing that you have nine forty-one parts. Then
you have capital equal to nine shares and seventeen
eighty-seconds. Now assume each portion to consist
of eighty- two parts ; then you have seven hundred and
fifty-five parts. Two-fifths of these are three hundred (81)
and two. Subtract from this the share of the daughter,
which is eighty-two ; there remain two hundred and
twenty. Subtract from this one-fourth and one-fifth,
namely, ninety-nine parts. There remain one hun-
dred and twenty- one. Add to this three-fifths of the^
capita], namely, four hundred and fifty-three. Then
you have five hundred and seventy-four, to be divided
into seven shares, each of eighty- two parts. This is
the share of the daughter ; each son receives twice as
much.
If the heirs are the same, and he bequeaths to a
person as much as the share of a son, less one-fourth
and one-fifth of what remains of two-fifths (of the
capital) after the deduction of the share; then you see
that this legacy is likewise determined by two- fifths.
Subtract two shares (of a daughter) from them, since
every son receives two (such) shares; there remain
( HO )
two-fifths of the capital less two (such) shares. Add
thereto what was excepted from the legacy, namely,
one-fourth and one-fifth of the two-fifths less nine-
tenths of (a daughter's) share.* Then you have two-
fifths and nine-tenths of one-fifth of the capital less two
(daughter's) shares and nine-tenths. Add to this
three- fifths of the capital. Then you have one -capital
and nine- tenths of one-fifth of the capital less two
(daughter's) shares and nine- tenths, equal to seven (such)
shares. Reduce this by removing the two shares and
nine-tenths and adding them to the seven shares. Then
you have one capital and nine-tenths of one-fifth of the
capital, equal to nine shares of a daughter and nine-
(82) tenths. Reduce this to one entire capital, by deduct-
ing nine fifty-ninths from what you have. There re-
mains the capital equal to eight such shares and twenty-
three fifty-ninths. Assume now each share (of a
daughter) to contain fifty-nine parts. Then the whole
heritage comprizes four hundred and ninety-five parts.
Two-fifths of this are one hundred and ninety-eight
* v = i. of the residue = a daughter's share.
2v = a son's share
»e-f+f-2^+A[f-H =7V
.-. v = ^^; a son's share = i^f
and the legacy to the stranger = -^^
( 111 )
parts. Subtract therefrom the two shares (of a daugh-
ter) or one hundred and eighteen parts; there remain
eighty parts. Subtract now that which was excepted,
namely, one-fourth and one fifth of these eighty, or
thirty-six parts ; there remain for the legatee eighty-
two parts. Deduct this from the parts in the total
number of parts in the heritage, namely, four hundred
and ninety-five. There remain four hundred and thir-
teen parts to be distributed into seven shares; the
daughter receiving (one share or) fifty-nine (parts), and
each son twice as much.
If he leaves two sons and two daughters, and be-
queaths to some person as much as the share* of a
* Since there are two sons and two daughters, each son
receives J, and each daughter ^ of the residue. Let
V = a daughter's share.
Let the ist legacy =a:=v— J [3— v]
2d =3,=:i,_i[j_a;_i,]
and 3d = ^
i-e. f_^V + J-.x~t; + J[i^-ar-T;] =61,
or7 + B^L35a .'.v=:^ = k
The 1 st Legacy =x = ^j
The 2d =y = ij
A son's share =:J
( 112 )
daughter less one-fifth of what remains from one-third
after the deduction of that share; and to another
person as much as the share of the other daughter less
one-third of what remains from one-third after the de-
duction of all this ; and to another person half one- sixth
of his entire capital ; then you observe that all these
legacies are determined by the one-third. Take one-
third of the capital, and subtract from it the share of a
daughter ; there remains one-third of the capital less
one share. Add to this that which was excepted,
namely, one-fifth of the one-third less one-fifth of the
share : this gives one-third and one-fifth of one-third of
(83) the capital less one and one-fifth portion. Subtract
herefrom the portion of the second daughter ; there
remain one- third and one-fifth of one- third of the
capital less two portions and one-fifth. Add to this
that which was excepted; then you have one- third
and three-fifths of one-third, less two portions and
fourteen-fifteenths of a portion. Subtract herefrom
half one-sixth of the entire capital : there remain
twenty-seven sixtieths of the capital less the two
shares and fourteen-fifteenths, which are to be sub-
tracted. Add thereto two-thirds of the capital, and
reduce it, by removing the shares which are to be sub-
tracted, and adding them to the other shares. You
have then one and seven-sixtieths of capital, equal to
eight shares and fourteen-fifteenths. Reduce this to
one capital by subtracting from every thing that you
have seven-sixtieths. Then let a share be two hundred
( 113 )
and one;* the whole capital will be one thousand six
hundred and eight.
If the heirs are the same, and he bequeaths to a
person as much as the share of a daughter, and one-
fifth of what remains from one- third after the deduction
of that share ; and to another as much as the share of
the second daughter and one-third of what remains
from one-fourth after the deduction of that share;
then, in the computation,! you must consider that the
two legacies are determined by one-fourth and one-
third. Take one- third of the capital, and subtract from
it one share ; there remains one- third of the capital
less one share. Then subtract one-fifth of the re-
mainder, namely, one-fifth of one-third of the capital,
less one-fifth of the share ; there remain four-fifths of
one-third, less four-fifths of the share. Then take also
one-fourth of the capital, and subtract from it one (84)
share ; there remains one-fourth of the capital, less one
share. Subtract one-third of this remainder : there
The common denominator 1608 is unnecessarily great.
f Let X be the 1st legacy ; 1/ the 2d ; v a daughter's share.
1 — X — y—Qv
Theni-i--Hi-^-i[i-"]+i-^-i[i-^J-6«
. 51 — 112„ . 51 _ 153_
<r — 212 . ,/ — 214^
( 114 )
remain two-thirds of one-fourth of the capital, less two-
thirds of one share. Add this to the remainder from
the one-third of the capital ; the sum will be twenty-
six sixtieths of the capital, less one share and twenty-
eight sixtieths. Add thereto as much as remains of
the capital after the deduction of one-third and one-
fourth from it; that is to say, one-fourth and one-
sixth; the sum is seven teen-twentieths of the capital,
equal to seven shares and seven-fifteenths. Complete
the capital, by adding to the portions which you have
three-seventeenths of the same. Then you have one
capital, equal to eight shares and one-hundred-and-
twenty hundred-and-fifty-thirds. Assume now one share
to consist of one-hundred-and- fifty-three parts, then
the capital consists of one thousand three hundred and
forty-four. The legacy determined by one- third, after
the deduction of one share, is fifty-nine ; and the legacy
determined by one-fourth, after the deduction of the
share, is sixty- one.
If he leaves six sons, and bequeaths to a person as
much as the share of a son and one-fifth of what remains
of one- fourth ; and to another person as much as the
share of another son less one-fourth of what remains
of one-third, after the deduction of the two first lega-
cies and the second share; the computation is this:*
You subtract one share from one-fourth of the capital ;
* Let x be the legacy to the ist stranger
and 7/ 2d ; v- a son's share
( 115 )
there remains one-fourth less the share. Remove then (85)
one-fifth of what remains of the one-fourth, namely,
half one-tenth of the capital less one-fifth of the share.
Then return to the one- third, and deduct from it half
one-tenth of the capital, and four-fifths of a share, and
one other share besides. The remainder then is one-
third, less half one-tenth of the capital, and less one
share and four-fifths. Add hereto one-fourth of the
remainder, which was excepted, and assume the one-
third to be eighty; subtracting from it half one-tenth of
the capital, there remain of it sixty-eight less one
share and four-fifths. Add to this one-fourth of it,
namely, seventeen parts, less one-fourth of the shares
to be subtracted from the parts. Then you have
eighty-five parts less two shares and one- fourth. Add
this to the other two- thirds of the capital, namely, one
hundred and sixty parts. Then you have one and one-
eighth of one-sixth of capital, less two shares and one-
fourth, equal to six shares. Reduce this, by remov-
ing the shares which are to be subtracted, and adding
1 —x—y=^v
i.e. §-f j-— jc — u-t-J [J— x— u]=6u .
.-. x-v^^s, and 3^=©-^^
( 116 )
them to the other shares. Then you have one and one-
eighth of one- sixth of capital, equal to eight shares
and one-fourth. Reduce this to one capital, by sub-
tracting from the parts as much as one forty-ninth of
them. Then you have a capital equal to eight shares
and four forty-ninths. Assume now every share to be
forty-nine ; then the entire capital will be three hun-
dred and ninety-six : the share forty-nine ; the legacy
(86) determined by one-fourth, ten ; and the exception from
the second share will be six.
On the Legacy with a Dirhem.
" A man dies, and leaves four sons, and bequeaths
to some one a dirhem, and as much as the share of a
son, and one-fourth of what remains from one-third
after the deduction of that share." Computation :* Take
* Let the capital = i ; a dirhem =^ ;
the legacy —a:; and a son's share —v
1 — X=:^>C
•••f+i-«-i[*-v]-^=4t^
.-. H-^ = ¥^
.*. ii of the capital — Af of a dirhem —v
and Jf of the capital +ff of a dirhem = j:, the legacy.
If we assume the capital to be so many dirhems, or a
dirhem to be such a part of the capital, we shall obtain the
{ 117 )
one third of the capital and subtract from it one share;
there remains one-third, less one share. Then sub-
tract one-fourth of the remainder, namely, one-fourth
of one-third, less one-fourth of the share ; then sub-
tract also one dirhem ; there remain three-fourths of
one- third of the capital, that is, one-fourth of the
capital, less three-fourths of the share, and less one
dirhem. Add this to two-thirds of the capital. The
sum is eleven- twelfths of the capital, less three-fourths
of the share and less one dirhem, equal to four shares.
Reduce this by removing three-fourths of the share
and one dirhem ; then you have eleven-twelfths of the
capital, equal to four shares and three- fourths, plus
one dirhem. Complete your capital, by adding to the
shares and one dirhem one-eleventh of the same. Then
you have the capital equal to five shares and two-
elevenths and one dirhem and one-eleventh. If you (8*7)
wish to exhibit the dirhem distinctly, do not complete
your capital, but subtract one from the eleven on
account of the dirhem, and divide the remaining ten by
the portions, which are four and three-fourths. The
quotient is two and two-nineteenths of a dirhem.
Assuming, then, the capital to be twelve dirhems, each
value of the son's share in terms of a dirhem, or of the
capital only.
Thus, if we assume the capital to be 1 2 dirhems,
V - ;^f [ 1 1 — 1 ]5 = V"t ^ = 2y% dirhems,
x=if [13 + 4] ^=W^ = 3ii dirhems.
( 118 )
share will be two dirhems and two-nineteenths. Or, if
you wish to exhibit the share distinctly, complete your
square, and reduce it, when the dirhem will be eleven
of the capital.
If he leaves five sons, and bequeaths to some per-
son a dirhem, and as much as the share of one of the
sons, and one-third of what remains from one-third,
and again, one-fourth of what remains from the one-
third after the deduction of this, and one dirhem more ;
then the computation is this:^ You take one-third, and
subtract one share ; there remains one- third less one
share. Subtract herefrom that which is still in your
hands, namely, one-third of one-third less one- third of
the share. Then subtract also the dirhem ; there re-
main two- thirds of one-third, less two- thirds of the
share and less one dirhem. Then subtract one -fourth
of what you have, that is, one-eighteenth, less one-
sixth of a share and less one-fourth of a dirhem, and
* Let the legacy =a:; and a son's share =r
•P=v«
.*. I^f of the capital —f^^ of a dirhem —v
.*. Jl of the capital + y^/ of a dirhem =x, the legacy.
If the capital = %^ dirhems, or J of the capital =7 J dirhems,
V =: f f dirhems = 3^1^ dirhems.
( 119 )
subtract also the second dirhem ; the remainder is half
one-third of the capital, less half a share and less one
dirhem and three fourths ; add thereto two-thirdsof the
capital, the sum is five-sixths of the capital, less one
half of a share, and less one dirhem and three-fourths,
equal to five shares. Reduce this, by removing the (88)
half share and the one dirhem and three-fourths,
and adding them to the (five) shares. Then you
have five-sixths of capital, equal to five shares and a
half plus one dirhem and three-fourths. Complete )
your capital, by adding to five shares and a half and
to one dirhem and three- fourths, as much as one-fifth
of the same. Then you have the capital equal to six
shares and three-fifths plus two dirhems and one-
tenth. Assume, now, each share to consist of ten
parts, and one dirhem likewise of ten ; then the ca-
pital is eighty-seven parts. Or, if you wish to exhibit
the dirhem distinctly, take the one-third, and subtract
from it the share; there remains one-third, less one
share. Assume the one-third (of the capital) to be
seven and a half (dirhems). Subtract one- third of what
you have, namely, one-third of one-third;* there
remain two- thirds of one- third, less two-thirds of the
share : that is, five dirhems, less two- thirds of the
share. Then subtract one, on account of the one
dirhem, and you retain four dirhems, less two-thirds
* There is an omission here of the words *' less one third
of a share."
( 120 ) ,
of the share. Subtract now one-fourth of what you
have, namely, one part less one-sixth of ^ share ;
and remove also one part on account of the one
dirhem; the remainder, then, is two parts less half
a share. Add this to the two-thirds of the capital,
which is fifteen (dirhems). Then you have seventeen
parts less half a share, equal to five shares. Reduce
this, by removing half a share, and adding it to the
five shares. Then it is seventeen parts, equal to
/89) five shares and a half. Divide now seventeen by five
and a half; the quotient is the value of one share,
namely, three dirhems and one-eleventh ; and one-
third (of the capital) is seven and a half (dirhems).
If he leaves four sons, and bequeaths to some person
as much as the share of one of his sons, less one-
fourth of what remains from one-third after the deduc-
tion of the share, and one dirhem; and to another
one-third of what remains from the one-third, and one
dirhem; then this legacy is determined by one- third.*
* Let the ist legacy be x, the 2d y; and a son's share rr v
1 —X — J/z:z4.V
i.e. f H-v+i {i-v]-i-i \i-v+i (l-„)-Jj- J=4„
'•e-f+f[*-«+i(i-<')-JJ-J=4»
••• l+A-f«-f ^=4»
( 121 )
Take one-third of the capital, and subtract from it one
share ; there remains one-third, less one share ; add
hereto one-fourth of what you have : then it is one-
third and one-fourth of one-third, less one share and
one-fourth. Subtract one dirhem ; there remains one-
third of one and one-fourth, less one dirhem, and less
one share and one-fourth. There remains from the
one-third as much as five-eighteenths of the capital, less
two-thirds of a dirhem, and less five-sixths of a share.
Now subtract the second dirhem, and you retain five-
eighteenths of the capital, less one dirhem and two-
thirds, and less five-sixths of a share. Add to this
two-thirds of the capital, and you have seventeen-
eighteenths of the capital, less one dirhem and two-
thirds, and less five-sixths of a share, equal to four
shares. Reduce this, by removing the quantities
which are to be subtracted, and adding them to the
shares; then you have seventeen-eighteenths of the
capital, equal to four portions and five-sixths plus one
dirhem and two-thirds. Complete your capital by (90)
adding to the four shares and five-sixths, and one
dirhem and two-thirds, as much as one- seventeenth of
the same. Assume, then, each share to be seventeen,
and also one dirhem to be seventeen.* The whole
capital will then be one hundred and seventeen.
If you wish to exhibit the dirhem distinctly, proceed
with it as I have shown you.
* Capital =;f|v + f^ J .-. ifv=i7, and5=:i7, capital=ii7
R
( 122 )
If he leaves three sons and two daughters, and
bequeaths to some person as much as the share of a
daughter plus one dirhem ; and to another one-fifth of
what remains from one-fourth after the deduction of
the first legacy, plus one dirhem ; and to a third per-
son one-fourth of what remains from one- third after
the deduction of all this, plus one dirhem ; and to a
fourth person one-eighth of the whole capital, requiring
all the legacies to be paid off by the heirs generally :
then you calculate this by exhibiting the dirhems dis-
tinctly, which is better in such a case.^ Take one-fourth
of the capital, and assume it to be six dirhems ; the
entire capital will be twenty-four dirhems. Subtract
one share from the one-fourth; there remain six
dirhems less one share. Subtract also one dirhem;
there remain five dirhems less one share. Subtract
* Let the legacies to the three first legatees be, severally,
X, ^, z; the fourth legacy = J ; and let a daughters' share
Then !i-^-^^-a:-.j^-l\l~x-,]-^ = Sv
but ^-x-i/^i^i+l^a:-^ [l~^] -^
^=-h%hnm^> y=iUT+uu^> ^=^'^+iUP
( 123 )
one-fifth of this remainder; there remain four dirhems,
less four-fifths of a share. Now deduct the second
dirhem, and you retain three dirhems, less four-fifths
of a share. You know, therefore, that the legacy
which was determined by one-fourth, is three dirhems,
less four-fifths of a share. Return now to the one-
third, which is eight, and subtract from it three dir-
hems, less four-fifths of a share. There remain five ^ ^
dirhems, less four-fifths of a share. Subtract also one-
fourth of this and one dirhem, for the legacy ; you then
retain two dirhems and three-fourths, less three -fifths of
a share. Take now one-eighth of the capital, namely,
three ; after the deduction of one-third, you retain one-
fourth of a dirhem, less three -fifths of a share. Return
now to the two- thirds, namely, sixteen, and subtract
from them one-fourth of a dirhem less three- fifths of a
share ; there remain of the capital fifteen dirhems and
three-fourths, less three-fifths of a share, which are
equal to eight shares. Reduce this, by removing three-
fifths of a share, and adding them to the shares, which
are eight. Then you have fifteen dirhems and three-
fourths, equal to eight shares and three-fifths. Make
the division: the quotient is one share of the whole
capital, which is twenty-four (dirhems). Every daugh-
ter receives one dirhem and one-hundred- and- forty-
three one-hundred-and-seventy-second parts of a dir-
hem.*
v=JQ^gL of the capital— ^/^\ of a dirhem. If we assume
( 124 )
If you prefer to produce the shares distinctly, take
one-fourth of the capital, and subtract from it one
share; there remains one-fourth of the capital less
one share. Then subtract from this one dirhem:
then subtract one-fifth of the remainder of one-fourth,
which is one-fifth of one-fourth of the capital, less one-
fifth of the share and less one-fifth of a dirhem ; and
subtract also the second dirhem. There remain four-
fifths of the one-fourth less four-fifths of a share, and
less one dirhem and four- fifths. The legacies paid out
of one fourth amount to twelve two-hundred-and-
(92) fortieths of the capital and four- fifths of a share, and
one dirhem and four-fifths. Take one-third, which is
eighty, and subtract from it twelve, and four- fifths of a
share, and one dirhem and four-fifths, and remove
one-fourth of what remains, and one dirhem. You
retain, then, of the one-third, only fifty-one, less three-
fifths of a share, less two dirhems and seven-twentieths.
Subtract herefrom one-eighth of the capital, which is
thirty, and you retain twenty- one, less three- fifths of
a share, and less two dirhems and seven-twentieths,
and two-thirds of the capital, being equal to eight
shares. Reduce this, by removing that which is to
be subtracted, and adding it to the eight shares. Then
you have one hundred and eighty-one parts of the
the capital to be equal to 24 dirhems
=45-80 5=: ijll dirhems
V = 18^X 24-564 dirhems =lAi ^rifAl ^
2064 2064
( 123 )
capital, equal to eight shares and three-fifths, plus
two dirhems and seven twentieths. Complete your
capital, by adding to that which you have fifty-nine one-
hundred-and-eighty-one parts. Let, then, a share be
three hundred and sixty two, and a dirhem likewise
three hundred and sixty-two.* The whole capital is
then five thousand two hundred and fifty-six, and the
legacy out of one-fourtht is one thousand two hundred
and four, and that out of one-third is four hundred and
ninety-nine, and the one- eighth is six hundred and
fifty-seven.
On Completement.
" A woman dies and leaves eight daughters, a mo- (93)
ther, and her husband, and bequeaths to some per-
son as much as must be added to the share of a
daughter to make it equal to one-fifth of the capital ;
and to another person as much as must be added to the
share of the mother to make it equal to one-fourth of
* The capital = 2^4u +411 S
If we assume v -362, and J = 362, the capital =5256
Then 07=724; ^ = 480; 2 = 499; J^h of capital =657.
f The text ought to stand " the two first legacies are
instead of " the legacy out of one-fourth is."
The first legacy is 724
, The second 480
the first + second legacy =r 1204
( 126 )
the capital."* Computation: Determine the parts of
the residue, which in the present instance are thir-
teen. Take the capital, and subtract from it one-fifth
of the same, less one part, as the share of a daugh-
ter : this being the first legacy. Then subtract also
one- fourth, less two parts, as the share of the mother :
this being the second legacy. There remain eleven -
twentieths of the capital, which, when increased by
three parts, are equal to thirteen parts. Remove now
from thirteen parts the three parts on account of the
three parts (on the other side), and you retain eleven-
twentieths of the capital, equal to ten parts. Complete
the capital, by adding to the ten parts as much as nine-
elevenths of the same ; then you find the capital equal
to eighteen parts and two- elevenths. Assume now
each part to be eleven ; then the whole capital is two
hundred, each part is eleven ; the first legacy will be
twenty-nine, and the second twenty-eight.
If the case is the same, and she bequeaths to
some person as much as must be added to the share
(94) of the husband to make it equal to one-third, and to
another person as much as must be added to the share
of the mother to make it equal to one-fourth ; and to a
* In this case, the modier has -fj ; and each daughter has
J^ of the residue.
i.e. i_i.-fv— i+2u = i3u
( 127 )
third as much as must be added to the share of a
daughter to make it equal to one-fifth ; all these lega-
cies being imposed on the heirs generally : then you
divide the residue into thirteen parts.* Take the
capital, and subtract from it one-third, less three parts,
being the share of the husband ; and one-fourth, less
two parts, being the share of the mother ; and lastly,
one-fifth less one part, being the share of a daughter.
The remainder is thirteen-sixtieths of the capital, which,
when increased by six parts, is equal to thirteen parts.
Subtract the six from the thirteen parts: there re-
main thirteen-sixtieths of the capital, equal to seven
parts. Complete your capital by multiplying the seven
parts by four and eight-thirteenths, and you have a
capital equal to thirty-two parts and four- thirteenths.
Assuming then each part to be thirteen, the whole
capital is four hundred and twenty.
If the case is the same, and she bequeaths to some
person as much as must be added to the share of the
mother to make it one-fourth of the capital; and to
another as much as must be added to the portion of a
daughter, to make it one-fifth of what remains of the
capital, after the deduction of the first legacy; then
* i-lJ-3^]-[i-2u]-[i-^]=i3v
i.e. i-A_i-i = 7v
( 1S8 )
you constitute the parts of the residue by taking them
out of thirteen.* Take the capital, and subtract from
it one-fourth less two parts; and again, subtract one-
fifth of what you retain of the capital, less one part;
then look how much remains of the capital after the
deduction of the parts. This remainder, namely, three-
fifths of the capital, when increased by two parts and
three-fifths, will be equal to thirteen parts. Subtract
two parts and three-fifths from thirteen parts, there
remain ten parts and two- fifths, equal to three-fifths of
capital. Complete the capital, by adding to the parts
which you have, as much as two-thirds of the same.
Then you have a capital equal to seventeen parts and
one-third. Assume a part to be three, then the capital
is fifty-two, each part three ; the first legacy will be
seven, and the second six.
If the case is the same, and she bequeaths to some
person as much as must be added to the share of the
mother to make it one-fifth of the capital, and to ano-
ther one-sixth of the remainder of the capital ; then
* 1— ^— ^=i3u
( 129 )
the parts are thirteen.* Take the capital, and subtract
from it one-fifth less two parts; and again, subtract
one-sixth of the remainder. You retain two-thirds of
the capital, which, when increased by one part and
two-thirds, are equal to thirteen parts. Subtract the
one part and two-thirds from the thirteen parts : there
remain two thirds of the capital, equal to eleven parts
and one -third. Complete your capital, by adding to
the parts as much as their moiety ; thus you find the
capital equal to seventeen parts. Assume now the
capital to be eighty-five, and each part five ; then the
first legacy is seven, and the second thirteen, and the
remaining sixty-five are for the heirs.
If the case is the same, and she bequeaths to some
person as much as must be added to the share of the
mother, to make it one-third of the capital, less that
sum which must be added to make the share of a
daughter equal to one-fourth of what remains of the
capital after the deduction of the above complement ;
then the parts are thirteen.f Take the capital, and (96)
* \—x—y= 13U
07 = ^-217; y = i\\-x\
.-. J[# + 2U] =13U
... | = ^t, ... ^ = ^; x = ^; y^y^
t i-a:-f?^=i3v; and^ = J— 2u; y=\\\—x\-v
.-. 1— a:+i[i— ar]-u=i3u
.-. I f 1 — a:] = 14U .-. f [J+ 2uJ = Hu
... 5^2^3„ ... ^^^5^. x^y = ^^
s
( 130 )
subtract from it one-third less two parts, and add to
the remainder one-fourth (of such remainder) less one
part ; then you have five-sixths of the capital and one
part and a half, equal to thirteen parts. Subtract
one part and a half from thirteen parts. There re-
main eleven parts and a half, equal to five-sixths of
the capital. Complete the capital, by adding to the
parts as much as one-fifth of them. Thus you find
the capital equal to thirteen parts and four-fifths.
Assume, now, a part to be five, then the capital is
sixty-nine, and the legacy four.
" A man dies, and leaves a son and five daughters,
and bequeaths to some person as much as must be
added to the share of the son to complete one-fifth
and one-sixth, less one-fourth of what remains of one-
third after the subtraction of the complement."* Take
one-third of the capital, and subtract from it one-fifth
and one-sixth of the capital, less two (seventh) parts ;
so that you retain two parts less four one hundred and
twentieths of the capital. Then add it to the excep-
tion, which is half a part less one one hundred and
* Since there are five daughters and one son, each
daughter receives i, and the son f of the residue.
••• l-HffeH-H-^7^
( 131 )
twentieth, and you have two parts and a half less five
one hundred and twentieths of capital. Add hereto
two-thirds of the capital, and you have seventy-five
one hundred and twentieths of the capital and two
parts and a half, equal to seven parts. Subtract, now,
two parts and a half from seven, and you retain seventy-
five one hundred and twentieths, or five-eighths, equal
to four parts and a half. Complete your capital, by (9T)
adding to the parts as much as three-fifths of the same,
and you find the capital equal to seven parts and one-
fifth part. Let each part be five ; the capital is then
thirty-six, each portion five, and the legacy one.
If he leaves his mother, his wife, and four sisters,
and bequeaths to a person as much as must be added to
the shares of the wife and a sister, in order to make them
equal to the moiety of the capital, less two-sevenths of
the sum which remains from one-third after the deduc-
tion of that complement; the Computation is this :* If
* From the context it appears, that when the heirs of the
residue are a mother, a wife, and 4. sisters, the residue is to
be divided into 13 parts, of which the wife and one sister,
together, take 5 : therefore the mother and 3 sisters, toge-
ther, take 8 parts. Each sister, therefore, must take not
less than -^^, nor more than -f^. In the case stated at page
102, a sister was made to inherit as much as a wife ; in the
present case that is not possible ; but the widow must take
not less than ^-^ ; and each sister not more than -fj. Proba-
bly, in this case, the mother is supposed to inherit ^-j ; the
wife y\ ; each sister ^■^,
( 13a )
you take the moiety from one-third, there remains one-
sixth. This is the sum excepted. It is the share of the
wife and the sister. Let it be five (thirteenth) parts.
What remains of the one-third is five parts less one-
sixth of the capital. The two-sevenths which he has
excepted are two-sevenths of five parts less two-
sevenths of one-sixth of the capital. Then you have
six parts and three-sevenths, less one-sixth and two-
sevenths of one- sixth of the capital. Add hereto
two-thirds of the capital; then you have nineteen
forty-seconds of the capital and six parts and three-
sevenths, equal to thirteen parts. Subtract herefrom
the six parts and three-sevenths. There remain nine-
teen forty-seconds of the capital, equal to six parts
and four- sevenths. Complete your capital by adding
to it its double and four-nineteenths of it. Then you
find the capital equal to fourteen parts, and seventy
(98) one hundred and thirty-thirds of a part. Assume one
part to be one hundred and thirty-three; then the
whole capital is one thousand nine hundred and thirty-
.-. ^=^V and the residue -^^
The author unnecessarily takes 7x276=1932 for the
-common denominator.
( 133 )
two; each part is one hundred and thirty-three, the
completion of it is three hundred and one, and the
exception of one-third is ninety-eight, so that the re-
maining legacy is two hundred and three. For the
heirs remain one thousand seven hundred and twenty-
nine.
COMPUTATION OF RETURNS.*
On Marriage in Illness,
" A man, in his last illness, marries a wife, paying
(^ marriage settlement of) one hundred dirhems,
besides which he has no property, her dowry being
* The solutions which the author has given of the remain-
ing problems of this treatise, are, mathematically consider-
ed, for the most part incorrect. It is not that the problems,
when once reduced into equations, are incorrectly worked
out ; but that in reducing them to equations, arbitrary as-
sumptions are made, which are foreign or contradictory to
the data first enounced, for the purpose, it should seem, of
forcing the solutions to accord with the established rules of
inheritance, as expounded by Arabian lawyers.
The object of the lawyers in their interpretations, and of
the author in his solutions, seems to have been, to favour
heirs and next of kin ; by limiting the power of a testator,
during illness, to bequeath property, or to emancipate slaves;
and by requiring payment of heavy ransom for slaves whom
a testator might, during illness, have directed to be eman-
cipated.
( 134 )
ten dirhems. Then the wife dies, bequeathing one-
third of her property. After this the husband dies."*
Computation : You take from the one hundred that
which belongs entirely to her, on account of the
dowry, namely, ten dirhems ; there remain ninety dir-
hems, out of which she has bequeathed a legacy. Call
the sum given to her (by her husband, exclusive of her
dowry) thing; subtracting it, there remain ninety
dirhems less thing. Ten dirhems and thing are al-
ready in her hands; she has disposed of one -third of
her property, which is three dirhems and one-third,
and one-third of thing ; there remain six dirhems and
* Let .9 be the sum, including the dowry, paid by the
man, as a marriage settlement ; d the dowry ; x the gift to
the wife, which she is empowered to bequeath if she pleases.
She may bequeath, if she pleases, d-{-x\ she actually
does bequath ^ [^+d;] ; the residue is f [o^+o;], of which
one half, viz. i [^+j^] goes to her heirs, and the other half
reverts to the husband
.*. the husband's heirs have s — [^ + :r] + i [c? -f x] or
«— f [ff+x] ; and since what the wife has disposed of, exclu-
sive of the dowry, is x, twice which sum the husband is to
receiwe, S'-^[d-\-x\='2x .'. ^['^s -<2d\=x. But 5=100;
flf=io .•.07=35; </+a; = 45; ^[g?+o;]=15. Therefore
the legacy which she bequeaths is 15, her husband receives
15, and her other heirs, 15. The husband's heirs receive
2X = 70.
But had the husband also bequeathed a legacy, then, as
we shall see presently, the law would have defeated, in part,
the woman's intentions.
( 135 )
two-thirds plus two-thirds of thing, the moiety of
which, namely, three dirhems and one-third plus one-
third of thing, returns as his portion to the husband.*
Thus the heirs of the husband obtain (as his share)
ninety-three dirhems and one-third, less two- thirds of
thing ; and this is twice as much as the sum given to (99)
the woman, which was thing, since the woman had
power to bequeath one-third of all which the husband
left;t and twice as much as the gift to her is two
things. Remove now the ninety-three and one-third,
from two-thirds of thing, and add these to the two
things. Then you have ninety-three dirhems and one-
third equal to two things and two- thirds. One thing is
three-eighths of it, namely, as much as three-eighths
of the ninety-three and one- third, that is, thirty- five
dirhems.
If the question is the same, with this exception only,
that the wife has ten dirhems of debts, and that she
bequeaths one-third of her capital ; then the Computar
* In other cases, as appears from pages 92 and 93, a
husband inherits one-fourth of the residue of his wife's es-
tate, after deducting the legacies which she may have
bequeathed. But in this instance he inherits half the re-
sidue. If she die in debt, the debt is first to be deducted
from her property, at least to the extent of her dowry (see
the next problem.)
f When the husband makes a bequest to a stranger, the
third is reduced to one-sixth. Vide p. 137.
( 136 )
tion is as follows :^ Give to the wife the ten dirhems of
her dowry, so that there remain ninety dirhems, out of
which she bequeaths a legacy. Call the gift to her
thing ; there remain ninety less thing. At the disposal
of the woman is therefore ten plus thing. From this
her debts must be subtracted, which are ten dirhems.
She retains then only thing. Of this she bequeaths
one-third, namely, one-third of thing : there remains
two-thirds of thing. Of this the husband receives by
inheritance the moiety, namely, one-third of thing.
The heirs of the husband obtain, therefore, ninety
dirhems, less two-thirds of thing ; and this is twice as
much as the gift to her, which was thing ; that is, two
things. Reduce this, by removing the two-thirds of
thing from ninety, and adding them to two things.
Then you have ninety dirhems, equal to two things
and two- thirds. One thing is three-eighths of this;
that is to say, thirty-three dirhems and three-fourths,
which is the gift (to the wife).
If he has married her, paying (a marriage settle-
* The same things being assumed as in the last example,
s - [fZ-fx] remains with the husband ; d goes to pay the
debts of the wife ; and | reverts from the wife to the hus-
band.
.'. s — d-^x-2x .'. ^[s-'d]—x
.-. if 5= 100, and d= lo, x='33| ; she bequeaths iij; iij
reverts to her husband; and her other heirs receive iii.
The husband's heirs receive 2x = 67^.
( 137 )
ment of one hundred dirhems, her dowry being ten (100)
dirhems, and he bequeaths to some person one-third ot
his property; then the computation is this:^ Pay to
the woman her dowry, that is, ten dirhems ; there re-
main ninety dirhems. Herefrom pay the gift to her,
thing; then pay likewise to the legatee who is to
receive one-third, thing : for the one-third is divided
* This case is distinguished from that in page 133 by
two circumstances ; first, that the woman does not make
any bequest ; second, that the husband bequeaths one-third
of his property.
Suppose the husband not to make any bequest. Then,
since the woman had at her disposal d-^-x, but did not make
any bequest, ^ [c?+x] reverts to her husband ; and the like
amount goes to her other heirs.
.'. s -[d-\-x]-\-l[d-^a:]=2x .*. x = -^[2s — d]
and since 5= 100, and rf= 10 ; a;=38; d-\-x = 4.S;
^ [d^x] = 24 reverts to the husband, and the like sum goes
to her other heirs ; and 2x^'j6, belongs to the husband's
heirs.
Now suppose the husband to bequeath one-third of his
property. The law here interferes with the testator's right
of bequeathing ; and provides that whatever sum is at the
disposal of the wife, the same sum shall be at the disposal
of the husband ; and that the sum to be retained by the
husband's heirs shall be twice the sum which the husband
and wife together may dispose of.
.*. s — ^[d+x] — x=z4x
... ^[2s-d]=x; if 5 = 100, andfl?=lo ; a: = iy\o = i7^;
d'\-x=2'j^^ ; I [d'\-x]z=i^-^j reverts to the husband, and
the like sum goes to the other heirs of the woman ; 1 7^^ is
what the husband bequeaths ; and 69-3I3- = 4x goes to the
husband's heirs.
T
( 138 )
into two moieties between them^ since the wife cannot
take any thing, unless the husband takes the same.
Therefore give, likewise, to the legatee who is to have
one-third, thing. Then return to the heirs of the hus-
band. His inheritance from the woman is five dirhems
and half a thing. There remains for the heirs of the
husband ninety-five less one thing and a half, which
is equal to four things. Reduce this, by removing one
thing and a half, and adding it to the four things.
There remain ninety-five, equal to five things and a
half. Make them all moieties ; there will be eleven
moieties; and one thing will be equal to seventeen
dirhems and three-elevenths, and this will be the
legacy.
" A man has married a wife paying (a marriage set-
tlement of) one hundred dirhems, her dowry being ten
dirhems; and she dies before him, leaving ten dirhems,
and bequeathing one-third of her capital; afterwards
the husband dies, leaving one hundred and twenty dir-
hems, and bequeathing to some person one-third of his
capital/' Computation r* Give to the wife her dowry,
* Let c be the property which the wife leaves, besides d
the dowry, and x the gift from the husband. She bequeaths
J [c + G?+a;]; J [c-|-rf-{-^] goes to her husband; and ^
[c-f fl?+a:] to her other heirs. The husband leaves property
5, out of which must be paid the dowry, d', the gift to the
wife, X ; and the bequest he makes to the stranger, x ; and
his heirs receive from the wife's heirs \ [c-\'d-\-x]
( 139 )
namely, ten dirhems; then one hundred and ten dirhems
remain for the heirs of the husband. From these the (101)
gift to the wife is thing, so that there remain one
hundred and ten dirhems less thing; and the heirs
of the woman obtain twenty dirhems plus thing. She
bequeaths one- third of this, namely, six dirhems and
two-thirds, and one-third of thing. The moiety of the
residue, namely, six dirhems and two-thirds plus one-
third of thing, returns to the heirs of the husband : so
that one hundred and sixteen and two-thirds, less two-
thirds of thing, come into their hands. He has be-
queathed one-third of this, which is thing. There
remain, therefore, one hundred and sixteen dirhems
and two-thirds less one thing and two-thirds, and this
is twice as much as the husband's gift to the wife
added to his legacy to the stranger, namely, four
things. Reduce this, and you find one hundred and
sixteen dirhems and two-thirds, equal to five things
and two-thirds. Consequently one thing is equal to
s —d—2x+^[c-{-d+x]= 4x, according to the law of inhe-
ritance.
.% ss + c-2d=i^x, and 3: = 31+^-2^
' 1 7
If 5=120, ci=:io, and 0?= 10, x = ^^z=20\^
c+rf-f a; = 40i:?; J [c-^d+x] = i3t^t
The wife bequeaths 1 3^7^ ; 1 3-ff go to her husband, and
13^9^ to her other heirs.
The husband bequeaths to the stranger 2oif ; he gives the
same sum to the wife ; and 4x= 82j\ go to his heirs.
( 140 )
twenty dirhems and ten-seventeenths ; and this is the
legacy.
On Emancipation in Illness.
" Suppose that a man on his death-bed were to eman-
cipate two slaves ; the master himself leaving a son and
a daughter. Then one of the two slaves dies, leaving
a daughter and property to a greater amount than his
price.^" You take two-thirds of his price, and what the
other slave has to return (in order to complete his
(102) ransom). If the slave die before the master, then the
son and the daughter of the latter partake of the heri-
tage, in such proportion, that the son receives as much
as the two daughters together. But if the slave die
after the master, then you take two-thirds of his value
and what is returned by the other slave, and distribute
* From the property of the slave, who dies, is to be de-
ducted and paid to the master's heirs, first, two-thirds of
the original cost of that slave, and secondly what is wanting
to complete the ransom of the other slave. Call the amount
of these two sums p ; and the property which the slave
leaves «.
Next, as to the residue of the slaves' property :
First. If the slave dies before the master, the master's
son takes J [^^-rf; the master's daughter :J- [<»-7?], and
the slave's daughter J [««—/?] •
Second. If the slave dies after the master ; the master's
son is to receive f p, and the master's daughter ^p ; and then
the master's son takes J [«—/>]» and the slave's daughter
( 141 )
it between the son and the daughter (of the master), in
such a manner, that the son receives twice as much as
the daughter ; and what then remains (from the heri-
tage of the slave) is for the son alone, exclusive of the
daughter; for the moiety of the heritage of the slave
descends to the daughter of the slave, and the other
moiety, according to the law of succession, to the son
of the master, and there is nothing for the daughter (of
the master).
It is the same, if a man on his death-bed emancipates
a slave, besides whom he has no capital, and then the
slave dies before his master.
If a man in his illness emancipates a slave, besides
whom he possesses nothing, then that slave must ran-
som himself by two-thirds of his price. If the master has
anticipated these two-thirds of his price and has spent
them, then, upon the death of the master, the slave
must pay two- thirds of what he retains.^ But if the
master has anticipated from him his whole price and
spent it, then there is no claim against the slave, since
he has already paid his entire price.
" Suppose that a man on his death-bed emancipates
a slave, whose price is three hundred dirhems, not
having any property besides ; then the slave dies, leav-
ing three hundred dirhems and a daughter." The
* The slave retains one-third of his price ; and this 4ie
must redeem at two-thirds of its value ; namely at f x ^ = |
of his original price.
( 142 )
computation is this :* Call the legacy to the slave thing.
He has to return the remainder of his price, after the
deduction of the legacy, or three hundred less thing.
This ransom, of three hundred less thing, belongs to
the master. Now the slave dies, and leaves thing and a
(103) daughter. She must receive the moiety of this, namely,
one half of thing ; and the master receives as much.
Therefore the heirs of the master receive three hundred
less half a thing, and this is twice as much as the le-
gacy, which is thing, namely, two things. Reduce this
by removing half a thing from the three hundred, and
adding it to the two things. Then you have three
hundred, equal to two things and a half. One thing
is, therefore, as much as two-fifths of three hundred,
* Let the slave's original cost be a ; the property which
he dies possessed of, cc ; what the master bequeaths to the
slave, in emancipating him, x. Then the net property
which the slave dies possessed of is cc+x — a, \[ot-\-x — a\
belongs, by law, to the master; and ^[a+a:-a] to the
slave's daughter. The master's heirs, therefore, receive the
ransom, a — a;, and the inheritance, J [ct+or— a]; that is,
J [i«-|-fl— a;]; and on the same principle as the slave, when
emancipated, is allowed to ransom himself at two-thirds of
his cost, the law of the case is that 2 are to be taken,
where 1 is given.
.*. ^Icc-^-a—x^^^ix .*. 0; = ;^ [eft-i. a]
The daughter's share of the inheritance = J[3it--2a]
The master's heirs receive |. [^-j- a\
If, as in the example, a, = a, j; = fa; the daughter's
share = \a ; the heirs of the master receive fa.
( 143 )
namely, one hundred and twenty. This is the legacy
(to the slave,) and the ransom is one hundred and eighty.
" Some person on his sick-bed has emancipated a
slave, whose price is three hundred dirhems; the slave
then dies, leaving four hundred dirhems and ten dir-
hems of debt, and two daughters, and bequeathing to
a person one -third of his capital ; the master has twenty
dirhems debts." The computation of this case is the
following:* Call the legacy to the slave thing; his ran-
som is the remainder of his price, namely, three hun-
dred less thing. But the slave, when dying, left four
hundred dirhems; and out of this sum, his ransom,
namely, three hundred less thing, is paid to the
* Let the slave's original cost=a; the property he dies
possessed of— « ; the debt he owes^g
He leaves two daughters, and bequeaths to a stranger one-
third of his capital.
The master owes debts to the amount ^it; where a -300;
c6 = 400; g = io; f^=20.
Let what the master gives to the slave, in emancipating
him =x.
Slave's ransom = a— a:; slave's property— slave's ransom =
eC'{-X — a
Slave's property — ransom— debt =fl6-|-a;— a— g
Legacy to stranger z=^[c6-\-x — a — g]
Residue =f [x+x — a — g]
The master, and each daughter, are, by law, severally
entitled to^x^{x-\-x-a — i]
The master's heirs receive altogether a— a; -\-^[x-\-x-a^i]
or ^g\a—x]+% [oi—i], which, on the principle that 2
( 144 )
master, so that one hundred dirhems and thing re-
main in the hands of the slave's heirs. Herefrom are
(first) subtracted the debts, namely, ten dirhems;
there remain then ninety dirhems and thing. Of
this he has bequeathed one- third, that is, thirty dir-
hems and one- third of thing; so that there remain for
the heirs sixty dirhems and two-thirds of thing. Of
this the two daughters receive two- thirds, namely,
forty dirhems and four-ninths of thing, and the master
(104) receives twenty dirhems and two-ninths of thing, so
that the heirs of the master obtain three hundred and
twenty dirhems Iqss seven-ninths of thing. Of this the
debts of the master must be deducted, namely, twenty
dirhems; there remain then three hundred dirhems less
are to be taken for i given, ought to be made equal
to 2X,
But the author directs that the equation for determining x be
.-. .r = J3- [7a+2 [cc—i]-gu] =108
Hence the slave receives, the debts which he owes, g =10
+ the legacy to the stranger —-^-g[Q[u—i]--6a—Sf^]= 66
+ the inheritance of 1 st daughter = ^Vl^[^~ ^1 — 4 a — 2^] = 44
+ theinheritanceof 2ddaughter= J3.[6[«;— e] — 4a — 2^]= 44
Total r; 2V[2 1 ^ + 4s— 1 4«— 7^14] = 1 64
And the master takes ^-\-2x=-^[4.cc—4.i+i4.a—jfA'j = 2^6
Had the slave died possessed of no property whatever, his
ransom would have been 200.
His ransom, here stated, exclusive of the sum which the
master inherits from him, or «— x, — 192.
( 11^ )
seven-ninths of thing ; and this sum is twice as much as
the legacy of the slave, which was thing; or, it is equal
to two things. Reduce this, by removing the seven-
ninths of thing, and adding them to two things ; there
remain three hundred, equal to two things and seven-
ninths. One thing is as much as nine twenty-fifths of
eight hundred, which is one hundred and eight ; and
so much is the legacy to the slave.
If, on his sick-bed, he emancipates two slaves, besides
whom he has no property, the price of each of them
being three hundred dirhems ; the master having anti-
cipated and spent two-thirds of the price of one of
them before he dies;^ then only one-third of the price
* Were there the first slave only, who has paid off two-
thirds of his original cost, the master having spent the
money, that slave would have to complete his ransom by
paying two-ninths of his original cost, that is 66^ (see page
141).
Were there the second slave only, who has paid off none
of his original cost, he would have to ransom himself at two-
thirds of his cost; that is by paying 200 (see also page 141).
The master's heirs, in the case described in the text, are
entitled to receive the same amount from the two slaves
jointly, viz. o.SG^, as they would be entitled to receive,
according to the rule of page 141, from the two slaves, sepa-
rately ; but the payment of the sum is differently distributed;
the slave who has paid two-thirds of his ransom being required
to pay one- ninth only of his original cost; and the slave
who has paid no ransom, being required to pay two-thirds of
his own cost, and one-ninth of the cost of the first slave.
( 146 )
of this slave, who has already paid off a part of his
ransom, belongs to the master ; and thus the master's
capital is the entire price of the one who has paid off
nothing of his ransom, and one- third of the price of
the other who has paid part of it ; the latter is one
hundred dirhems ; the other three hundred dir-
hems: one-third of the amount, namely, one hun-
dred and thirty-three dirhems and one third, is divided
into two moieties among them ; so that each of them
receives sixty-six dirhems and two-thirds. The first
slave, who has already paid two-thirds of his ran-
som, pays thirty-three dirhems and one-third; for
(\0^) sixty-six dirhems and two- thirds out of the hundred
belong to himself as a legacy, and what remains of
the hundred he must return. The second slave has
to return two hundred and thirty-three dirhems and
one-third.
" Suppose that a man, in his illness, emancipates two
slaves, the price of one of them being three hundred
dirhems, and that of the other five hundred dirhems ;
the one for three hundred dirhems dies, leaving a
daughter; then the master dies, leaving a daughter
likewise ; and the slave leaves property to the amount
of four hundred dirhems. With how much must every
one ransom himself?"* The computation is this: Call
* Let A. be the first slave ; his original cost a ; the pro-
perty he dies possessed of u ; and let B. be the second slave ;
and his cost b.
( 147 )
the legacy to the first slave, whose price is three hun-
dred dirhems, thing. His ransom is three hundred
dirhems less thing. The legacy to the second slave of
a price of five hundred dirhems is one thing and two-
thirds, and his ransom five hundred dirhems less one
thing and two- thirds {viz. his price being one and two-
thirds times the price of the first slave, whose ransom
was thing, he must pay one thing and two-thirds for
Let X be that which the master gives to A. in emanci-
pating him.
A.'s ransom is a—x; and his property, minus his ransom,
is u— a-{-x,
A.'s daughter receives J [x—a+x], and the master's heirs
receive I [x—a-{-x]
Hence the master receives altogether from A.,
a—xi-^ [*— «+^] = 5 [x + a^x.]
B.'s ransom is b > — x
a
The master's heirs receive from A. and B. together
^[cf{-a + 2b] — — [a + 2h]x; and this is to be made equal
to twice the amount of the legacies to A. and B., that is,
i[x^a + 2b]—^/^a + 2b]x=2—-x
The master's heirs receive from A., — ^^-^q^' — =293^
A.'s daughter receives [a+ij ^—i" =8oox 4^3%^o = io6§
The legacy to B. is b ^ Z,, =i88§; his ransom is
, 4a 4-46 - a
The master's heirs receive from A. and B. together
5a+66 ^^
^[« + *]"S-'-604-
I
( 14.8 )
his ransom). Now the slave for three hundred dirhems
dies, and leaves four hundred dirhems. Out of this
his ransom is paid, namely, three hundred dirhems
less thing ; and in the hands of his heirs remain one
hundred dirhems plus thing: his daughter receives the
moiety of this, namely, fifty dirhems and half a thing;
and what remains belongs to the heirs of the master,
namely, fifty dirhems and half a thing. This is
added to the three hundred less thing ; the sum is
three hundred and fifty less half a thing. Add
thereto the ransom of the other, which is five hundred
dirhems less one thing and two-thirds ; thus, the heirs
(106) of the master have obtained eight hundred and fifty
dirhems less two things and one-sixth ; and this is
twice as much as the two legacies together, which were
two things and two- thirds. Reduce this, and you have
eight hundred and fifty dirhems, equal to seven things
and a half Make the equation ; one thing will be
equal to one hundred and thirteen dirhems and one-
third. This is the legacy to the slave, whose price is
three hundred dirhems. The legacy to the other slave
is one and two- thirds times as much, namely, one
hundred and eighty-eight dirhems and eight-ninths,
and his ransom three hundred and eleven dirhems and
one-ninth.
" Suppose that a man in his illness emancipates two
slaves, the price of each of whom is three hundred dir-
hems ; then one of them dies, leaving five hundred
dirhems and a daughter; the master having left a son."
( 149 )
Computation :* Call the legacy to each of them thing;
the ransom of each will be three hundred less thing;
then take the inheritance of the deceased slave, which is
^ve hundred dirhems, and subtract his ransom, which is
three hundred less thing ; the remainder of his inhe-
ritance will be two hundred plus thing. Of this, one
hundred dirhems and half a thing return to the master
by the law of succession, so that now altogether four
hundred dirhems less a half thing are in the hands of
the master's heirs. Take also the ransom of the other
slave, namely, three hundred dirhems less thing;
then the heirs of the master obtain seven hundred dir-
* The first slave is A. ; his cost a ; his property » ; he
leaves a daughter.
The second slave is R. ; his cost b.
Then (as in page 147) J [««-a+a:] goes to the daughter;
«_|_„_|_26
and x=a
sceives [a+b]
la [a-\-a.-\-h' ]-\-Z "-h
The daughter receives \a-\-b\ , ^-^
The master receives from A. ^a4-Qb
and the master receives from A. and B. together
2 [« + *] i^
But if 6 =0 x=^ [«ft+3a]=i27T\
The daughter receives ^ [3*^— 2a] = 163^
The master receives from A ^ [5^^ + 4a] =; 336^
The master receives from B ^ [8a— «] = 172^
The master receives from A and B. . . ^-^ [*+ 3a] — 509t4:
If 5=0,
The daughter receives ^ [3*— 2fl]
The master | [«+«J, as in page 142.
( 150 )
hems less one thing and a half, and this is twice as
much as the sum of the two legacies of both, namely
(107) two things, consequently as much as four things. Re-
move from this the one thing and a half: you find
seven hundred dirhems, equal to five things and a half.
Make the equation. One thing will be one hundred
and twenty-seven dirhems and three-elevenths.
" Suppose that a man in his illness emancipate a
slave, whose price is three hundred dirhems, but who
has already paid off to his master two hundred dirhems,
which the latter has spent ; then the slave dies before
the death of the master, leaving a daughter and three
hundred dirhems."* Computation : Take the property
left by the slave, namely, the three hundred, and add
thereto the two hundred, which the master has spent ;
this together makes five hundred dirhems. Subtract
from this the ransom, which is three hundred less thing
* The slave A. dies before his master, and leaves a
daughter. His cost is a, of which he has redeemed d, which
the master has spent ; and he leaves property oc.
Then the daughter receives . . \ [oc-\-d—a-\-x]
The master receives altogether J [cc-\-a+a—x]
The master's heirs receive. ... J [ci^d-\-a—x]
And ^[u'-a + a—a:] = 2x .*. a: = ^[ci—a-{-a]
Hence the daughter receives ^ [^ei-^-ii a— 2a]=z 140
The master's heirs J [201— id + 2a] = 160
The master receives, in toto, \ [2a + 3« + 2a] = 360
If the slave had not advanced, or the master had not spent «,
the daughter would have received ^[sct-\-^d-2a] = i8o
and the master would have received J [2c6-f2^-f 2a] = 320.
( 151 )
(since his legacy is thing) ; there remain two hundred
dirhems plus thing. The daughter receives the moiety
of this, namely, one hundred dirhems plus half a thing;
the other moiety, according to the laws of inheritance,
returns to the heirs of the master, being likewise one
hundred dirhems and half a thing. Of the three hun-
dred dirhems less thing there remain only one hundred
dirhems less thing for the heirs of the master, since
two hundred are spent already. After the deduction
of these two hundred which are spent, there remain
with the heirs two hundred dirhems less half thing, and
this is equal to the legacy of the slave taken twice; or
the moiety of it, one hundred less one-fourth of thing,
is equal to the legacy of the slave, which is thing. Re-
move from this the one-fourth of thing ; then you have
one hundred dirhems, equal to one thing and one-
fourth. One thing is four-fifths of it, namely, eighty
dirhems. This is the legacy ; and the ransom is two
hundred and twenty dirhems. Add the inheritance of the
slave, which is three hundred, to two hundred, which (108)
are spent by the master. The sum is five hundred
dirhems. The master has received the ransom of two
hundred and twenty dirhems ; and the moiety of the
remaining two hundred and eighty, namely, one hun-
dred and forty, is for the daughter. Take these from
the inheritance of the slave, which is three hundred ;
there remain for the heirs one hundred and sixty dir-
hems, and this is twice as much as the legacy of the
slave, which was thing.
( 152 )
" Suppose that a man in his illness emancipates a
slave, whose price is three hundred dirhems, but who
has already advanced to the master five hundred dir-
hems; then the slave dies before the death of his mas-
ter, and leaves one thousand dirhems and a daughter.
The master has two hundred dirhems debts."* Com-
putation : Take the inheritance of the slave, which is
one thousand dirhems, and the five hundred, which the
master has spent. The ransom from this is three hun-
dred less thing. There remain therefore twelve hundred
phis thing. The moiety of this belongs to the daughter :
it is six hundred dirhems plus half a thing. Subtract
it from the property left by the slave, which was one
* A.'s price is a ; he has advanced to his master a ; he
leaves property u. He dies before his master, and leaves a
daughter.
The master's debts are ^ ; a: is vi'hat A. receives, in being
emancipated ; a — x is the ransom ; J ^ci-]-d — a-\-a:] is what
the daughter receives.
Then et—^[u-^d—a-\-x] is what remains to the master ;
and «— i [cc-l-d—a-^x]^^ is what remains to him, after
paying his debts ; and this is to be made equal to 2x,
Whence x=^[u'i-a—d—2fA]
Hence the daughter receives ^[3«:d— 2a-|-2a— ^] = 640
The mother receives, 1 i r . ^ . i /-
inclusive of the debt) i[2«+2a-2a+f.] = 36o
The master receives, 1 i r . r ^ t /:•
'exclusive of the debt I i[2u + [2a-2a-4^] = ibo
If the mode given in page 142 had been followed, it
would have given x — l[ci-{-a + a— 2ft,]
and the daughter's portion-i [3^— 2a + 3rt- it4]-740.
( 153 )
thousand dirhems: there remain four hundred dirhenis
less half thing. Subtract herefrom the debts of the
master, namely, two hundred dirhems ; there remain
two hundred dirhems less half thing, which are equal to
the legacy taken twice, which is thing ; or equal to two
things. Reduce this, by means of the half thing. Then
you have two hundred dirhems, equal to two things
and a half. Make the equation. You find one thing,
equal to eighty dirhems ; this is the legacy. Add now
the property left by the slave to the sum which he has (109)
advanced to the master : this is fifteen hundred dir-
hems. Subtract the ransom, which is two hundred
and twenty dirhems ; there remain twelve hundred and
eighty dirhems, of which the daughter receives the
moiety, namely, six hundred and forty dirhems. Sub-
tract this from the inheritance of the slave, which is
one thousand dirhems : there remain three hundred
and sixty dirhems. Subtract from this the debts of the
master, namely, two hundred dirhems ; there remain
then one hundred and sixty dirhems for the heirs of the
master, and this is twice as much as the legacy of the
slave, which was thing.
" Suppose that a man on his sick-bed emancipates a
slave, whose price is five hundred dirhems, but who has
already paid off to him six hundred dirhems. The mas-
ter has spent this sum, and has moreover three hun-
dred dirhems of debts. Now the slave dies, leaving his
mother and his master, and property to the amount of
seventeen hundred and fifty dirhems, with two hundred
X
( 154 )
dirhems debts." Computation:* Take the property left
by the slave, namely, seventeen hundred and fifty dir-
hems, and add to it what he has advanced to the mas-
ter, namely, six hundred dirhems; the sum is two
thousand three hundred and fifty dirhems. Subtract
from this the debts, which are two hundred dirhems,
and the ransom, which is five hundred dirhems less
thing, since the legacy is thing; there remain then
sixteen hundred and fifty dirhems plus thing. The
mother receives herefrom one-third, namely, five hun-
dred and fifty plus one-third of thing. Subtract now
this and the debts, which are two hundred dirhems,
from the actual inheritance of the slave, which is
seventeen hundred and fifty ; there remain one thou-
(110) sand dirhems less one -third of thing. Subtract from
this the debts of the master, namely, three hundred
* A. dies before his master, and leaves a mother. His
price was a; he has redeemed ^, which the master has
spent. The property he leaves is «. He owes debts g.
The master owes debts f^,
^ [u-^a—a-\-x — 6] is the mother's,
fit— ^[«d-f o — a-fj—g]-— 8 is the master's.
a — ^ [ec -\- a ^ a + a: '- 1] — i — fA — 2x =:the master's, after
paying his debts.
Hence xz=.\ [2u + a — ^— 2g — 3jtt]=30O
Mother's =zi\ [^u — 2a-\-2a—^i—^] — 650
Master's, without ^t , ... —^ [4«-|-2fl— 2^— 4g— 6jei]z26oo
Mother's, with ^ r=-iJ-[4<«+2a— 2^ — 4€+|i4] = 900
A. receives, inclusive of i =z\ [3*— 2a-|-2^-|-4g— /t*] — 850.
( 155 )
dirhems ; there remain seven hundred dirhems less
one-third of thing. This is twice as much as the legacy
of the slave, which is thing. Take the moiety : then
three hundred and fifty less one-sixth of thing are
equal to one thing. Reduce this, by means of the
one-sixth of thing ; then you have three hundred and
fifty, equal to one thing and one-sixth. One thing
will then be equal to six-sevenths of the three hundred
and fifty, namely, three hundred dirhems ; this is the
legacy. Add now the property left by the slave to
what the master has spent already; the sum is two
thousand three hundred and fifty dirhems. Subtract
herefrom the debts, namely, two hundred dirhems, and
subtract also the ratisom, which is as much as the price
of the slave less the legacy, that is, two hundred dir-
hems; there remain nineteen hundred and fifty dirhems.
The mother receives one- third of this, namely, six
hundred and fifty dirhems. Subtract this and the
debts, which are two hundred dirhems, from the pro-
perty actually left by the slave, which was seventeen
hundred and fifty dirhems; there remain nine hun-
dred dirhems. Subtract from this the debts of the
master, which are three hundred dirhems; there re-
main six hundred dirhems, which is twice as much as
the legacy.
" Suppose that some one in his illness emancipates a
slave, whose price is three hundred dirhems: then the
slave dies, leaving a daughter and three hundred dir-
hems ; then the daughter dies, leaving her husband and
( 156 )
three hundred dirhems; then the master ^lies." Com-
putation:^ Take the property left by the slave, which is
three hundred dirhems, and subtract the ransom, which
(111) is three hundred less thing ; there remains thing, one
half of which belongs to the daughter, while the other
half returns to the master. Add the portion of the
daughter, which is half one thing, to her inheritance,
which is three hundred ; the sum is three hundred dir-
hems plus half a thing. The husband receives the moiety
of this ; the other moiety returns to the master, namely
one hundred and fifty dirhems plus one-fourth of thing.
All that the master has received is therefore four hun-
dred and fifty less one-fourth of thing ; and this is twice
as much as the legacy ; or the moiet^ of it is as much as
* A. is emancipated by his master, and then dies, leaving
a daughter, who dies, leaving a husband. Then the master
dies.
A.'s prices a; his property a. What he receives from
the master =^.
The daughter's property = ^
A.'s ransom = a -a:. The daughter inherits J [cj— a + or],
and ^ [« — a + o;] goes to the master.
J P+2 ['*'~^+^]] goes to the daughter's husband
and -J [^+i [^-«+a;]] to the master.
Hence, according to the author, we are to make
a— a; + i[«— a+a?]+iP+J[«~a+^]] = 2x
.% a: — -^ [3ot-|-a -j- 2^] - 200
Daughter's share =^[6e^— .40-}-^] zz 100
Husband's =i [S'* - 2a + 5^] = 200
Master's =:^ [2^ + 6a + 4^] - 400.
( 15'r )
the legacy itself, namely, two hundred and twenty -five
dirhems less one- eighth thing are equal to thing. Re-
duce this by means of one-eighth of thing, which you
add to thing; then you have two hundred and twenty-
five dirhems, equal to one thing and one-eighth. Make
the equation: one thing is as much as eight-ninths
of two hundred and twenty-five, namely, two hundred
dirhems.
" Suppose that some one in his illness emancipates a
slave, of the price of three hundred dirhems ; the slave
dies, leaving five hundred dirhems and a daughter, and
bequeathing one-third of his property ; then the daugh-
ter dies, leaving her mother, and bequeathing one-
third of her property, and leaving three hundred dir-
hems." Computation:^ Subtract from the property left
* A. is emancipated, and dies, leaving a daughter, and
bequeathing one-third of his property to a stranger.
The daughter dies, leaving a mother, and bequeathing
one-third of her property to a stranger.
A.*s price is a ; his property is a
The daughter's property is ^.
A.'s ransom is a—x\ u — a-^x is his property, clear of
ransom.
^ [tfs— a + a] goes to the stranger; and the like amount to
A.'s daughter, and to the master.
3 [3^+«— fl+^] is the property left by the daughter.
^[35-j-<* — a+a:] is the bequest of the daughter to a
stranger.
f [2i^-{-u—a-\-x] is the residue, of which ^d,
viz, ^\ [35-l-<«— a+j:] is the mother's,
and 2T [3^ + * — « -}-^] is the master's ;
( 158 )
by the slave his ransom, which is three hundred dir-
hems less thing; there remain two hundred dirhems
plus thing. He has bequeathed one-third of his pro-
perty, that is, sixty-six dirhems and two-thirds plus
one-third of thing. According to the law of succession,
(112) sixty-six dirhems and two-thirds and one-third of thing
belong to the master, and as much to the daughter.
Add this to the property left by her, which is three
hundred dirhems : the sum is three hundred and sixty-
six dirhems and two-thirds and one-third of thing.
She has bequeathed one-third of her property, that is,
one hundred and twenty-two dirhems and two-ninths
and one-ninth of thing ; and there remain two hundred
and forty-four dirhems and four-ninths and two-ninths
of thing. The mother receives one-third of this,
namely, eighty- one dirhems and four-ninths and one-
third of one-ninth of a dirhem plus two-thirds of one-
ninth of thing. The remainder returns to the master ;
it is a hundred and sixty- two dirhems and eight-ninths
and two-thirds of one-ninth of a dirhem plus one-ninth
and one-third of one-ninth of thing, as his share of the
heritage.
Hence, according to the author, we are to make
a— a:+J [at— a-f j;]-{-^[3^4-<*-a-f-x] = 2a;
Therefore ^=-h [i 3<* + 14« + 1 2^J = 2 I0y5_
The daughter's share. . --^^ [2701 — 18a 4-4^] = 136^4-
The daughter's bequest = ^^ [gM — 6a-\- 24^] = 145^
The mother's share .... =/^ [3<«— 2a-f-8^] — 97^^
The master's =^2^ [i3ct-t-i4rt-hi25] = 420|^.
( 159 )
Thus the master's heirs have obtained five hundred
and twenty-nine dirhems and seventeen twenty-sevenths
of a dirhem less four-ninths and one-third of one-ninth
of thing ; and this is twice as much as the legacy, which
is thing. Halve it : You have two hundred and sixty-
four dirhems and twenty-two twenty-sevenths of a dir-
hem, less seven twenty-sevenths of thing. Reduce it by (113)
means of the seven twenty- sevenths which you add to
the one thing. This gives one hundred and sixty-four
dirhems and twenty-two twenty-sevenths, equal to one
thing and seven twenty-sevenths of thing. Make the
equation, and adjust it to one single thing, by sub-
tracting from it as much as seven thirty-fourths of the
same. Then one thing is equal to two hundred and
ten dirhems and five-seventeenths; and this is the
legacy.
*' Suppose that a man in his illness emancipates a
slave, whose price is one hundred dirhems, and makes
to some one a present* of a slave-girl, whose price is
five hundred dirhems, her dowry being one hundred
dirhems, and the receiver cohabits with her." Abu
Hanifah says : The emancipation is the more impor-
tant act, and must first be attended to.
Computation :* Take the price of the girl, which is
* The price of the slave-girl being a ; and what she
receives on being emancipated x, her ransom is a—x.
If her dowry is ec, he that receives her, takes u, + x.
( 160 )
five hundred dirhems ; and remember that the price of
the slave is one hundred dirhems. Call the legacy of
the donee thing. The emancipation of the slave, whose
price is one hundred dirhems, has already taken place.
He has bequeathed one thing to the donee. Add the
dowry, which is one hundred dirhems less one-fifth
thing. Then in the hands of the heirs are six hundred
dirhems less one thing and one-fifth of thing. This is
twice as much as one hundred dirhems and thing ; the
moiety of it is equal to the legacy of the two, namely,
three hundred less three-fifths of thing. Reduce this
by removing the three-fifths of thing from three hun-
dred, and add the same to one thing. This gives three
hundred dirhems, equal to one thing and three-fifths and
one hundred dirhems. Subtract now from three hun-
Hence, according to the author, we arc to make
a—x—Q. [_cc-\-x~\ ; whence x— ~
3
And her ransom is §[« + «]
But if a male slave be at the same time emancipated by
the master, the donee must pay the ransom of that slave.
If his price was b, b — x is his ransom.
Hence, according to the author, we are to make the sum
of the two ransoms, viz. a—X'{-b — j;=^2[«+x]
.-. a+6-2^=[3+^] X ... x = a "-±i=|-^=i25
The donee pays ransom, in respect of the slave-girl {a -x) - 375
and he pays ransom for the male slave b—~ x — 75.
( 161 )
dred the one hundred, on account of the other one
hundred. There remain two hundred dirhems, equal
to one thing and three-fifths. Make the equation with
this. One thing will be five-eighths of what you have ; (114)
take therefore five-eighths of two hundred. It is one
hundred and twenty-five. This is thing; it is the
legacy to the person to whom he had presented the
girl.
" Suppose that a man emancipates a slave of a price
of one hundi-ed dirhems, and makes to some person a
present of a slave girl of the price of five hundred dir-
hems, her dowry being one hundred dirhems; the donee
cohabits with her, and the donor bequeaths to some
other person one-third of his property." According to
the decision of Abu Hanifah, no more than one-third
can be taken from the first owner of the slave-girl ; and
this one-third is to be divided into two equal parts be-
tween the legatee and the donee. Computation:^ Take
the price of the girl, which is five hundred dirhems.
The legacy out of this is thing; so that the heirs obtain
five hundred dirhems less thing ; and the dbwTy is one
hundred less one-fifth of thing; consequently they
♦ The same notation being used as in the last example,
the equation for determining or, according to the author, is
to be
7 * r T
a—x-^0—- X — X— 2 [<:« + 2j:|
( 162 )
obtain six hundred dirhems less one thing and one-fifth
of thing. He bequeaths to some person one third of
his capital, which is as much as the legacy of the person
■who has received the girl, namely, thing. Conse-
quently there remain for the heirs six hundred less two
things and one-fifth, and this is twice as much as both
their legacies taken together, namely, the price of the
slave pluB the two things bequeathed as legacies.
Halve it, and it will by itself be equal to these lega-
cies : it is then three hundred less one and one-tenth
of thing. Reduce this by means of the one and one-
tenth of thing. Then you have three hundred, equal
to three things and one-tenth, plus one hundred dir-
hems. Remove one hundred on occount of (the
opposite) one hundred; there remain two hundred,
equal to three things and one-tenth. Make now the
reduction. One thing will be as much as thirty-one
(1 15) parts of the sum of dirhems which you have; and just
so much will be the legacy out of the two hundred ; it
is sixty-four dirhems and sixteen thirty-one parts.
" Suppose that some one emancipates a slave girl of
the price of one hundred dirhems, and makes to some
person a present of a slave girl, which is five hundred
dirhems worth ; the receiver cohabits with her, and her
dowry is one hundred dirhems ; the donor bequeaths
to some other person as much as one-fourth of his
capital.'* Abu Hanifah says : The master of the girl
cannot be required to give up more than one-third, and
the legatee, who is to receive one-fourth, must give up
( 163 )
one-fourth. Computation :* The price of the girl is
five hundred dirhems. The legacy out of this is thing;
there remain five hundred dirhems less thing. The
dowry is one hundred dirhems less one-fifth of thing;
thus the heirs obtain six hundred dirhems less one and
one-fifth of thing. Subtract now the legacy of the
person to whom one-fourth has been bequeathed,
namely, three-fourths of thing; for if one-third is thing
then one-fourth is as much as three-fourths of the same.
There remain then six hundred dirhems less one
thing and thirty-eight fortieths. This is equal to the
legacy taken twice. The moiety of it is equal to the
legacies by themselves, namely, three hundred dirhems
less thirty-nine fortieths of thing. Reduce this by
means of the latter fraction. Then you have three hun- (116)
dred dirhems, equal to one hundred dirhems and two
things and twenty-nine fortieths. Remove one hundred
on account of the other one hundred. There remain
two hundred dirhems, equal to two things and twenty-
nine-fortieths. Make the equation. You will then
find one thing to be equal to seventy-three dirhems
and forty-three one-hundred-and-ninths dirhems.
* The same notation being used as in the two former
examples, the equation for determining x, according to the
author, is
a—x + b— x—^x =2{cc+i^x]
Whence a: = ^-y^ [a-{.b^2cc]=r3jW
( 164 )
On return of the Dowry,
" A MAN, in the illness before his death, makes to
some one a present of a slave girl, besides whom he
has no property. Then he dies. The slave girl is
worth three hundred dirhems, and her dowry is one
hundred dirhems. The man to whom she has been
presented, cohabits with her." Computation:* Call
the legacy of the person to whom the girl is pre-
sented, thing. Subtract this from the donation : there
remain three hundred less thing. One-third of this
difference returns to the donor on account of dowry
(since the dowry is one-third of the price) : this is
one hundred dirhems less one-third of thing. The
donor's heirs obtain, therefore, four hundred less
one and one- third of thing, which is equal to twice
the legacy, which is thing, or to two things. Trans-
pose the one and one-third thing from the four hun-
dred, and add it to the two things ; then you have four
hundred, equal to three things and one-third. One
thing is, therefore, equal to three- tenths of it, or to one
hundred and twenty dirhems, and this is the legacy.
* Let a be the slave-girl's price — u, her dowry.
Then, according to the author, we are to make
a
Therefore x-=- — — - [a -f- u] =-f^ x 400 r^ 1 20
The donee is to receive the girl's dowry, worth 400, for 280.
( 165 )
" Or, suppose that he, in his illness, has made a pre-
sent of the slave girl, her price being three hundred,
her dowry one hundred dirhems ; and the donor dies,
after having cohabited with her." Computation :* Call
the legacy thing : the remainder is three hundred less ^
thing. The donor having cohabited with her, the
dowry remains with him, which is one- third of the
legacy, since the dowry is one-third of the price, or one-
third of thing. Thus the donor's heirs obtain three (117)
hundred less one and one-third of thing, and this is
twice as much as the legacy, which is thing, or equal
to two things. Remove the one and one-third of
thing, and add the same to the two things. Then you
have three hundred, equal to three things and one-
third. One thing is, therefore, three- tenths of it,
namely ninety dirhems. This is the legacy.
If the case be the same, and both the donor and
donee have cohabited with her ; then the Computation
* If the donor has cohabited with the slave-girl, the
donor's heirs are to retain the dowry, but must allow the
donee, in addition to the legacy a:, the further sum of - x ;
The ransom is then a—x -- x, which according to the
author is to be made equal to 2x.
Whence x=: — — =qo
The donee is to receive the girl, worth 300, for 210.
( 166 )
is this:* Call the legacy thing; the deduction is three
hundred dirhems less thing. The donor has ceded the
dowry to the donee by (the donee's) having cohabited
with her : this amounts to one-third of thing : and
the donee cedes one-third of the deduction, which is
one hundred less one-third of thing. Thus, the donor's
heirs obtain four hundred less one and two-thirds of
thing, which is twice as much as the legacy. Reduce
this, by separating the one and two-thirds of thing
from four hundred, and add them to the two things.
Then you have four hundred things, equal to three
things and two-thirds. One thing of these is three-
elevenths of four hundred ; namely, one hundred and
* If the donor has previously cohabited with the slave-
girl, it appears from the last example, that the donee is
entitled to ransom her for a—x — x.
a
If the donee cohabits with the slave-girl, it appears from
the last example but one, that he is entitled to redeem the
dowrv, <«, for a — - a?
The redemption of the girl and dowry is
a — X X-\-X X,
which, according to the author, is to be made equal to 2x.
rr>i • 0-1- 2a
Ihat is a-{-c(, x=2x
a
Whence x= — ^ — xFa-f ^1 = 1094^
The donee is to receive the girl and dowry, worth 400,
for 29915 .
( 167 )
nine dirhems and one-eleventh. This is the legacy;
The deduction is one hundred and ninety dirhems and
ten-elevenths. According to Abu Hanifah, you call
the thing a legacy, and what is obtained on account of
the dowry is likewise a legacy.
If the case be the same, but that the donor, having
cohabited with her, has bequeathed one-third of his (118)
capital, then Abu Hanifah says, that the one-third is
halved between the donee and the legatee. Computa-
tion : * Call the legacy of the person to whom the slave-
girl has been given, thing. After the deduction of it,
there remain three hundred, less thing. Then take the
dowry, which is one- third of thing; so that the donor
retains three hundred less one and one-third of thing :
the donee's legacy being, according to Abu Hanifah,
one and one- third of thing; according to other
lawyers, only thing. The legatee, to whom one-third
is bequeathed, receives as much as the legacy of the
donee, namely, one and one-third of thing. The
donor thus retains three hundred, less two things and
* The second case is here solved in a different way.
a ^ a ^
.*. X=i'
This being halved between the legatee and donee becomes
The donee receives the girl, worth 300, for 262|.
( 168 )
two- thirds — equal to twice the two legacies, which
are two things and two-thirds. The moiety of this,
namely, one hundred and fifty less one and one- third
of thing, must, therefore, be equal to the two legacies.
Reduce it, by removing one and one- third of thing,
and adding the same to the two legacies (things).
Then you find one hundred and fifty, equal to four
things. One thing is one-fourth of this, namely,
thirty-seven and a half.
If the case be, that both the receiver and the donor
have cohabited with her, and the latter has disposed of
one-third of his capital by way of legacy; then the
computation,* according to Abu Hanifah, is, that you
call the legacy thing. After the deduction of it, there
remain three hundred less thing. Then the dowry
is taken, which is one hundred less one-third of thing ;
so that there are four hundred dirhems less one and
one- third of thing. The sum returned from the dowry
is one- third of thing; and the legatee, who is to receive
one-third, obtains as much as the legacy of the first,
namely, thing and one -third of thing. Thus there
* According to the author's rule, which is purely arbi-
trary,
n "- a -•
Whence x=za -—-, — —48
The donee will have to redeem the girl and dowry,
worth 400, for 352.
{ lfi9 )
remain four hundred dirhems less three things, equal
to twice the legacy, namely, two things and two-thirds. (119)
Reduce this, by means of the three things, and you find
four hundred, equal to eight things and one-third.
Make the equation with this ; one thing will be forty-
eight dirhems.
" Suppose that a man on his sick-bed makes to ano-
ther a present of a slave-girl, worth three hundred dir-
hems, her dowry being one hundred dirhems; the
donee cohabits with her, and afterwards, being also on
his sick-bed, makes a present of her to the donor,
and the latter cohabits with her. How much does he
acquire by her, and how much is deducted?"* Com-
* We have here the only instance in the treatise of a
simple equation, involving two unknown quantities. For
what the donee receives is one unknown quantity ; and what
the donor receives back again from the donee, called by the
author *' part of thing," is the other unknown quantity.
Let what the donee receives = a;, and what the donor
receives =:^.
Then, retaining the same notation as before, according to
the author, the donee receives, on the whole
and the donor receives, on the whole
Whence x=^\ ^^^^aa-^^ [3a'H3««-2«^] ^ 102
z
But
( 170 )
putation : Take the price, which is three hundred dir-
hems ; the legacy from this is thing ; there remain witli
the donor's heirs three hundred less thing; and the
donee obtains thing. Now the donee gives to the
donor part of thing : consequently, there remains only
thing less part of thing for the donee. He returns to the
donor one hundred less one- third of thing ; but takes the
dowry, which is one-third of thing, less one -third of
part of thing. Thus he obtains one and two-thirds
thing less one hundred dirhems and less one and one-
third of part of thing. This is twice as much as part
of thing ; and the moiety of it is as much as part of
thing, namely, five-sixths of thing less fifty dirhems
and less two- thirds of part of thing. Reduce this by
removing two-thirds of part of thing and fifty dirhems.
Then you have five-sixths of thing, equal to one and
two-thirds of part of thing plus fifty dirhems. Reduce
this to one single part of thing, in order to know^ what
the amount of it is. You effect this by taking three-fifths
(120) of what you have. Then one part of thing plus thirty
dirhems is equal to half a thing ; and one-half thing
less thirty dirhems is equal to part of thing, which is
the legacy returning from the donee to the donor.
Keep this in memory.
Then return to what has remained with the donor ;
But the reasons for reducing the question to these two
equations are not given by the author, and seem to depend
on the dicta of the sages of the Arabian law.
( 171 )
this was three hundred less thing : hereto is now added
the part of thing, or one-half thing less thirty dir-
hems. Thus he obtains two hundred and seventy less
half one thing. He further takes the dowry, which is
one hundred dirhems less one-third thing, but has to
return a dowry, which is one-third of what remains of
thing after the subtraction of part of thing, namely,
one-sixth of thing and ten dirhems. Thus he retains
three hundred and sixty less thing, which is twice as
much as thing and the dowry, which he has returned.
Halve it : then one hundred and eighty less one-half
thing are equal to thing and that dowry. Reduce this,
by removing one-half thing and adding it to the thing
and the dowry : you find one hundred and eighty dir-
hems, equal to one thing and a half plus the dowry
which he has returned, and which is one-sixth thing
and ten dirhems. Remove these ten dirhems; there
remain one hundred and seventy dirhems, equal to
one and two- thirds things. Reduce this, in order
to ascertain what the amount of one thing is, by taking
three-fifths of what you have; you find that one hun-
dred and two are equal to thing, which is the legacy
from the donor to the donee: and the legacy from
the donee to the donor is the moiety of this, less
thirty dirhems, namely, twenty-one.
( 172 )
On Surrender in Illness.
(121) " Suppose that a man, on his sick-bed, deliver to
some one thirty dirhems in a measure of victuals, worth
ten dirhems; he afterwards dies in his illness ; then the
receiver returns the measure and returns besides ten
dirhems to the heirs of the deceased." Computation :
He returns the measure, the value of which is ten dir-
hems, and places to the account of the deceased twenty
dirhems ; and the legacy out of the sum so placed is
thing; thus the heirs obtain twenty less thing, and the
measure. All this together is thirty dirhems less thing,
equal to two things, or equal to twice the legacy.
Reduce it by separating the thing from the thirty, and
adding it to the two things. Then, thirty are equal to
three things. Consequently, one thing must be one-
third of it, namely, ten, and this is the sum which he
obtains out of what he places to the account of the
deceased.
" Suppose that some one on his sick-bed delivere
to a person twenty dirhems in a measure worth fifty
dirhems ; he then repeals it while still on his sick bed,
and dies after this. The receiver must, in this case,
return four-ninths of the measure, and eleven dirhems
and one-ninth."* Computation : You know that the
* Let a be the gift of money ; and the value of the mea-
sure m xa.
It appears from the context that the donee is to pay the
heirs f mff.
( 173 )
price of the measure is two and a half times as much as
the sum which the donor has given the donee in money;
and whenever the donee returns anything from the
money capital, he returns from the measure as much
as two and a half times that amount. Take now from
the measure as much as corresponds to one thing, that
is, two things and a half, and add this to what remains
from the twenty, namely, twenty less thing. Thus the
heirs of the deceased obtain twenty dirhems and one (122)
thing and a half. The moiety of this is the legacy,
namely, ten dirhems and three-fourths of thing; and
this is one-third of the capital, namely, sixteen dirhems
and two-thirds. Remove now ten dirhems on account
of the opposite ten ; there remain six dirhems and two-
thirds, equal to three-fourths of thing. Complete the
thing, by adding to it as much as one-third of the
same: and add to the six dirhems and two-thirds
It is arbitrary how he shall apportion this sum between the
money capital and the measure.
If he pays on the money capital p. a
and on the measure . • q,ma
we have the equation p. a-i-q. ma=^ ma
or p -\-q m =f m
The author assumes jp=—. q
Whence y=-|, andjt?=f, and therefore the donee pays
on the money capital. ... ^ a=ii^
and on the measure ^ ma =22^
Total 33*.
( 1^4 )
likewise one-third of the same, namely, two dirhems
and two-ninths; this yields eight dirhems and eight-
ninths, equal to thing. Observe now how much the
eight dirhems and eight-ninths are of the money
capital, which is twenty dirhems. You will find them
to be four-ninths of the same. Take now four- ninths
of the measure and also five-ninths of twenty. The
value of four-ninths of the measure is twenty-two dir-
hems and two-ninths ; and the five-ninths of the twenty
are eleven dirhems and one-ninth. Thus the heirs
obtain thirty-three dirhems and one- third, which is as
much as two-thirds of the fifty dirhems. — God is the
Most Wise !
N O T 'E S,
Page 1, line 2-5.
The neglected state of the manuscript, in which most
diacritical points are wanting, makes me very doubtful
whether I have correctly understood the author's meaning
in several passages of his preface.
In the introductory lines, I have considered the words
amplification of what might briefly have been expressed
by l^lfcib j^iJt ' through the performance of which.'* I
conceive the author to mean, that God has prescribed to
man certain duties, ^ l^^ ^^lill ^^^ u^J^^ '^ *^' f^^
iX^Vs^l, and that by performing these (&c. ijoJ^\ to-tljb)
we express our thankfulness (jx1j\ *«jI ?-^) &c.
Since my translation was made, I have had the ad-
vantage of consulting Mr. Shakespear about this pas-
sage. He prefers to read «Ju , t.j,-c>-^:uJ , and ^^y in-
stead of «Ji3 , K^^.^y^ , and ^y , and proposes to tran-
slate as follows ; Praise to God for his favours in that
which is proper for him from among his laudable deeds,
which in the performance of what he has rendered indis-
( 176 )
pensible from (or by reason of) thein on (the part of)
whoever of his creatures worships him, gives the name of
thanksgiving, and secures the increase, and preserves from
deterioration."
The construction here assumed is evidently easier than
that adopted by myself, in as far as the relative pronoun ^-iJl
representing irJw«l.sr* , is made the subject of the three sub-
sequent verbs *_aj , &c., whilst my translation presumes a
transition from the third person (as in wJbi yb u , and in ^^
irJux>) to the first (as in *JiJi &c.).
A marginal note in the manuscript explains the words
jt^^ ^J^ ^^y>3 by^l ^ ^Us^U ^^j3j ^^Aiu Jxl " The
meaning may be : we preserve from change him who en-
joys it," (viz, the divine bounty, taking <U>.U? for c-^%^-L>
<dJl f^. The change here spoken of is the forfeiture
of the divine mercy by bad actions ; for God does not
change the mercy which he bestows on men, as long as they
do not change that which is within themselves." J <dl\ ^^b
j»g..,.,c»b U U;-*> 15*^" |*y L5^ V^^ ^^^^ ^;?*^ ^--^« {Cor any
Sur, VIII. V. 55, ed. Hinck.).
Page 1, line 7.
J--j^l ^ ^^ ^j^ L5^ ^®® Coraw, Sur. v. v. 22.
ed. Hinck.
Page 1, line 14, 15.
I am particularly doubtful whether I have correctly read
and translated the words of the text from IjL.:;^-!^ to ^J>^y
Instead oi j>^ \i\j:ij>-\ I should have preferred ljL.^1
( ITT )
yi^^ benefitting others," if the verb ,^;*-.>"\ could be con-
strued with the preposition J .
Page 2 , line 1 .
To the words J^f-j iJ^J * marginal note is given in the
manuscript, which is too much mutilated to be here tran-
scribed, but which mentions the names of several authors
who first wrote on certain branches of science, and con-
cludes with asserting, that the author of the present treatise
was the first that ever composed a book on Algebra.
Page 2, line 4.
An interlinear note in the manuscript explains <^ix-j JJ
by i3j:Ji^ f*^*
Page 2, line 10.
Mohammed gives no definition of the science which he
intends to treat of, nor does he explain the words j^ Jebr^
and ^liU mokabalah^ by which he designates certain ope-
rations peculiar to the solution of equations, and which,
combined, he repeatedly employs as an expression for this
entire branch of mathematics. As the former of these words
has, under various shapes, been introduced into the several
languages of Europe, and is now universally used as the
designation of an important division of mathematical science,
I shall here subjoin a few remarks on its original sense, and
on its use in Arabic mathematical works.
The \erh jf^jabar of which the substantive J->.^'e6r is
derived, properly signifies to restore something broken,
2 A
i
( ITS )
especially to cure a fractured bone. It is thus used in the
following passage from Motanabbi (p. 143, 144, ed.
Calculi,)
iijd\ >^ \ 4>^ Jb i^\ ^j K Lj^\ \ 4«i aL—J J>jl! ^ I)
^— jW l::^! Ulic jjj*4t€. ^^ V^ '^^^ ^^**^ C^^^^T^ ^
O thou on whoml rely in whatever I hope, with whom
I seeic refuge from all that 1 dread ; whose bounteous hand
seems to me like the sea, as thy gifts are like its pearls : pity
the youthfulness of one, whose prime has been wasted by
the hand of adversity, and whose bloom has been stifled in
the prison. Men will not heal a bone which thou hast
broken, nor will they break one which thou hast healed."
Tfence the Spanish and Portuguese expression algebrista
for a person who heals fractures, or sets right a dislocated
limb.
In mathematical language, the verb^^-^ means, to make
perfect, or to complete any quantity that is incomplete or
liable to a diminution ; i. e, when applied to equations, to
transpose negative quantities to the opposite side by chang-
ing their signs. The negative quantity thus removed is
construed with the particle C-^ : thus, if a:^— 6=^23 shall be
changed into:i:2=z29, the direction is IfcJJj d:uJu (jy^j^^
j^^yL^ltj d^\ ^ i. e. literally " Restore the square from
(the deficiency occasioned to it by) the six, and add these
to the twenty-three."
( 179 )
The \evbj^ is not likewise used, when in an equation
an integer is substituted for a fractional power of the un-
known quantity : the proper expression for this is either
the second or fourth conjugation of J-*j , or the second
of "J .
The word ^\ax mokabalah is a noun of action of the
verb J-J to be in front of a thing, which in the third
conjugation is used in a reciprocal sense of two objects
being opposite one another or standing face to face; and
in the transitive sense of putting two things face to face, of
confronting or comparing two things with one another.
In mathematical language it is employed to express
the comparison between positive and negative terms in a
compound quantity, and the reduction subsequent to such
comparison. Thus loo+ioo:— ioa;4-ar2 is reduced to 100+ a:^
<U liblJi ^ Jju after we have made a comparison."
When applied to equations, it signifies, to take away
such quantities as are the same and equal on both sides.
Thus the direction for reducing x^-\-x=x'^-\-4. to x = 4. will
be expressed by JjvJJ .
In either application the verb requires the preposition
c--> before a pronoun implying the entire equation or com-
pound quantity, within which the comparison and subse-
quent reduction is to take place.
The verb Jjli is not likewise used, when the reduction of
an equation is to be performed by means of a division : the
proper term for this operation being 3j ,
( 180 )
The mathematical application of the substantives
and <UjliU will appear from the following extracts.
1. A marginal note on one of the first leaves of the
Oxford manuscript lays down the following distinction :
[Uj] j^Ul^j-^:*- A^kc J:>>- ^-^L-^M Ij^ jltf Uli l^p-^^
la)! ^lill! CI-JjU Jjj ^^^ J] l^^ ^;-,lir»-iIl
'' Jeftr is the restoration of anything defective by means
of what is complete of another kind. Mokabalah^ a noun
of action of the third conjugation, is the facing a thing :
whence it is applied to one praying, who turns his face to-
wards the kihlah. In this branch of calculation, the method
commonly employed is the restoring of something defective
in its deficiency, and the adding of an amount equal to this
restoration to the other side, so as to make the completion
(on the one side) and this addition (on the other side) to face
(or to balance) one another. As this method is frequently
resorted to, it has been named^e^r and mokabalah (or Res-
toring and Balancing), since here every thing is made com-
plete if it is deficient, and the opposite sides are made io
balance one another Mathematicians also take
( 181 )
the word mokdbalah in the sense of the removal of equal
quantities (from both sides of an equation)."
According to the first part of this gloss, in reducing
^— 5«=:iOrt to .a:r=i5«, the substitution of x in place of
x—^a would afford an instance of Jebr or restoration, and
the corresponding addition of 5a to 10«, would be an
example of /woArafta^aA or balancing. From the following
extracts it will be seen, that mokdbalah is more generally
taken in the sense stated last by the gloss.
2. IIaji Khalfa, in his bibliographical work (MS. of the
British Museum, fol. 167, recto^.) gives the following ex-
planation: ^j\x\} (HLksA ^ ^JaA} to jji 'ij\)Jj^\ ^^-Jt^^
^\)\ \AU\ Zj}^\ ^j^^ :^jIcJ ^j^^i] ^^\ ^ s.\::^:^\j
J jU::]J ^^^K y ^ kJ^^=>^ ^J^ Jebr is the adding to one
side what is negative on the other side of an equation,
owing to a subtraction, so as to equalize them. Mokdbalah
is the removal of what is positive from either sum, so as
to make them equal."
A little farther on H aji Khalfa gives further illustration
of this by an example : ^j\ J J^. li*-i» 'i\ 'ijts. UjJi ^ U^
JU ^\j}^^ ^laJUj^ ajl^ '^l^ ^^^i Jj^ ^^ ,^;>l^m
bjLj^\j^ Sft,} Jl*4l ^ J^^\ is,^\jj ^Ac ^ ^.^^uJLl
^^^oMJ i^\ iLbUlli X»-/4o- j^ ^S"^ ■^\i^^ ^J^ tj^
* This manuscript is apparently only an abridgement of Haji Khalfa's
work.
( 1H2 )
<Ui l^y^ ^^ <LljUi,[j ^^1 Jx " For instance if
we say: Ten less one thing equal to four things;' then
jebr is the removal of the subtraction, which is performed
by adding to the minuend an amount equal to the sub-
trahend: hereby the ten are made complete, that which
was defective in them being restored. An amount equal
to the subtrahend is then added to the other side of the
equation : as in the above instance, after the ten have been
made complete, one thing must be added to the four things,
which thus become five things. Mokabalah consists in
withdrawing the same amount from quantities of the same
kind on both sides of the equation ; or as others say, it is
the balancing of certain things against others, so as to
equalize them. Thus, in the above example, the ten are
balanced against the five with a view to equalize them.
This science has therefore been called by the name of
these two rules, namely, the rule o? jebr or restoration,
and of mokabalah or reduction, on account of the fre-
quent use that is made of them."
3. The following is an extract from a treatise by Abu
Abdallah al-Hosain ben Ahmed,* entided, A^JJill
• I have not been able to find any information about this writer. The
copy of the work to which I refer is comprized in the same volume with
Mohammed ben Musa's work in the Bodleian library. It boars no date.
( 183 )
^\5\} jj^\ Jya\ ^ Li\^\ or " A complete introduction
to the elements of alsrebra."
On the original meaning of the words Jebr and
mokabalah. This species of calculation is called jebr
(or completion) because the question is first brought to
an equation ...... And as, after the equation has been
formed, the practice leads in most instances to equalize
something defective with what is not defective, that
defective quantity must be completed where it is defec-
tive ; and an addition of the same amount must be made
to what is equalized to it. As this operation is frequently
employed (in this kind of calculation), it has been called
jebr : such is the original meaning of this word, and
such the reason why it has been applied to this kind of
calculation. Mokabalah is the removal of equal magni-
tudes on both sides (of the equation)."
4. In the Kholaset al Hisdb, a compendium of arith-
metic and geometry by Baha-eddin Mohammed ben al
HosAiN (died a.h. 1031, i.e. 1375 a.d.) the Arabic
( 184 )
text of which, together with a Persian commentary by
RosHAN Ali, was printed at Calcutta* (1812. 8vo.) the
following explanation is given (pp. 334. 335.) (^l^\j
^j^\ ytj y^'j] ^:: ui3 J Ji.« ^\j)^^ J^^ ^\:J!^}\ jj
iLLUU^^ Ufw< LiLJ ^t-i^yi ti ^,f^\ «LjU:i11 (jJ[:^'^\j
The side (of the equation) on which something is to be
subtracted, is made complete, and as much is added to
the other side : this is Jebr ; again those cognate quan-
tities which are equal on both sides are removed, and this
is mokabalahJ'^ The examples which soon follow, and
the solution of which Baha-eddin shows at full length,
afford ample illustration of these definitions. In page 338,
1500— i-»=^ is reduced to 1500 = 1^0;; this he says is
effected by jebr. In page 341, *^x=^x^- + ^x is reduced
to i2,x=x^^ and this he states to be the result of both
jebr and mokabalah.
The Persians have borrowed the words jebr and mokd-
balahy together with the greater part of their mathema-
tical terminology, from the Arabs. The following extract
from a short treatise on Algebra in Persian verse, by
Mohammed Nadjm-eddin Kuan, appended to the Cal-
cutta edition of the Kholdset al Hisab, will serve as an
illustration of this remark.
* A full account of this work by Mr. Strachey will be found in the
twelfth volume of the Asiatic Researches, and in Hhtton's Tracts on
mathematical and philosophical subjects, vol. ir. pp, 179-193. See also
Mutton's Mathematical Dictionary, art. Algebra,
I
( 185 )
Complete the side in which the expression ilia (less,
minus) occurs, and add as much to the other side, O
learned man : this is in correct language called jehr.
In making the equation mark this : it may happen that
some terms are cognate and equal on each side, without
distinction ; these you must on both sides remove, and this
you call moJcabalah.^^
With the knowledge of Algebra, its Arabic name was
introduced into Europe. Leonardo Bonacci of Pisa,
when beginning to treat of it in the third part of his
treatise of arithmetic, says : Incipit pars tertia de solutione
quarundam qucestionum secundum modum Algebrce et Al'
mucabalce^ scilicet oppositionis et restaurationis. That
the sense of the Arabic terms is here given in the inverted
order, has been remarked by Cossali. The definitions
of Jebr and mokabalah given by another early Italian
2b
( )«6 )
writer, Lucas Paciolus, or Lucas de Burgo, are thus
reported by Cossali : // cojnmune oggetto deW operar
loro ^ recare la equazione alia sua maggior unita- Gli
uffizj loro per questo commune intento sono contrarj :
quello delV Algebra e di restorare li extremi del diminuti ;
€ quello di Almucabala di levare da li extremi i superjlui,
Intende Fra Luca per extremi i membri delV equazione.
Since the commencement of the sixteenth century, the
word mokabalah does no longer appear in the title of
Algebraic works. Hieronymus Cardan's Latin treatise,
first published in 1545, is inscribed : Artis magncB sive de
regulis algebraicis liber unus, A work by John Scheu-
BELius, printed at Paris in 1552, is entitled : Algebrm com-
pendiosa facilisque description qua depromuntur magna
Arithmetices miracula, (See Hutton's Tracts, &c. ii.pp.
241-243.) Pelletier's Algebra appeared at Paris in
1558, under the title: De occulta parte numerorum quam
Algebram vacant, libri duo, (Hutton, 1. c. p. 245. Mon-
TUCLA, hist, des math. i. p. 613.) A Portuguese treatise,
by Pedro Nunez or Nonius, printed at Amberez in
1567, is entitled : Libra de Algebra y Arithmetica y Geo-
metria. (Montucla, 1. c. p. 615.)
In Feizi*s Persian translation of the Lilavati (written
in 1587, printed for the first time at Calcutta in 1827, 8vo.)
I do not recollect ever to have met with the word^^-^ ; but
^JjviU is several times used in the same sense as in the above
Persian extract.
( IST )
Page 3, line 3, seqq.
In the formation of the numerals, the thousand is not,
like the ten and the hundred, multiplied by the units only,
but likewise by any number of a higher order, such as
tens and hundreds : there being no special words in Arabic
(as is the case in Sanscrit) for ten-thousand, hundred-
thousand, &c.
From this passage, and another on page 10, it would
appear that our author uses the word jJLc, plur, j^jAi^,
knot or tie, as a general expression for all numerals of a
higher order than that of the units. Baron S. de Sacy,
in his Arabic Grammar, (vol. i. § 741) when explaining
the terms of Arabic grammar relative to numerals, trans-
lates J^ifi by noeuds^ and remarks : Ce sont les noms des
dixaines^ depute vingt jusqu'o, quatre-vingt-dix.
Page 3, line 9-11.
The forms of algebraic expression employed by Leo-
nardo are thus reported by Cossali (Origine, &c. deW
Algebra, i. p. 1.) • Tre consider azioni distingue Leonardo
nel numero : una assoluta, o semplice, ed e quella del
numero in se stesso ; le altre due relative, e sono quelle
di radice e di quadrato, Nominando il quadrato sog'
giugne QUI videlicet census dicitur, ed il nome di
censo ^ quello di cui in seguito si serve. That Leonardo
seems to have chosen the expression census on account of
its acceptation, which is correspondent to that of the
( 188 )
Arabic JU, has already been remarked by Mr. Cole-
BROOKE (Algebra, (fee. Dissertation, p. liv.)
Paciolo, who wrote in Italian, used the words numero,
cosa, and censo ; and this notation was retained by Tar-
TAGLIA. From the term cosa for the unknown number,
exactly corresponding in its acceptation to the Arabic ^^^
thing, are derived the expressions Ars cossica and the
German die Coss, both ancient names of the science of
Algebra. Cardan's Latin terminology is numerusy qua-
dratum^ and re*, for the latter also positio or quantitas
ignota.
Page 3, line 17.
1 have added from conjecture the words IjJtf Jjk^j'jjjcvj
which are not in the manuscript. There occur several
instances of such omissions in the work.
The order in which our author treats of the simple
equations is, 1st. x'^=px; 2d. x'^=n', 3d. px = n. Leo-
nardo had them in the same order. (See Cossali, 1. c.
p. 2.) In the Kholaset al Hisub the arrangement is, 1st.
n=zpx', 2d, px=x- ; 3d. n=x^ .
Page 5, line 9.
In the Lilavati, the rule for the solution of the case
cx'^-{-bx = a is expressed in the following stanza.
( 189 )
i. e. rendered literally into Latin :
Per mult ip lie at am radicem diminutce \yel\ auctce quantitatis
Manifestce^ addiice ad dimidiatimultiplicatoris quadratum
Radix, dimidiato multiplicatore addito \_vef\ subtracto,
In quadratum ducta — est interrogantis desiderata
quantitas.
The same is afterwards explained in prose : ^TT
Tjfti: f^"^gwr %?rt%ri; gfoi^ ^ ^
^"^^rrff f^^ ^5? rT?3f q-jff TTI^:
^^ln>, Cl i. e. A quantity, increased or diminished
by its square-root multiplied by some number, is given.
Then add the square of half the multiplier of the root to
the given number: and extract the square-root of the
sum. Add half the multiplier, if the difference were
given; or subtract it, if the sum were so. The square of
the result will be the quantity sought." (Mr. Colebrooke's
translation.)
Feizi's Persian translation of this passage runs thus:
( 190 )
With the above Sanskrit stanza from the Lilavati some
readers will perhaps be interested to compare the following
Latin verses, which Montucla (i. p. 590) quotes from
Lucas Paciolus :
Si res et census numero cocequantur, a rebus
Dimidio sumpto, censum producere debes^
Adder eque numero, cujus a radice totiens
Tolle semis rerum^ census latusque redibit.
Page 6, line 16.
<L-*4o- U^ j^ jcj-il! u. o > d: . ^ J Such instances of the
common instead of the apocopate future, after the impe-
rative, are too frequent in this work, than that they could
be ascribed to a mere mistake of the copyist : I have
accordingly given them as I found them in the manuscript.
( 191 )
Page 7, line 1.
JjeU CJ^^j ] The same structure occurs page 21,
line 15.
Page 8 J line 11.
i^jj*A\ i::^] aj^j Hadji Khalfa, in his article on
Algebra, quotes the following observation from Ibn Khal-
DUN. ^ (*:t^^^ ^^^ L/^. ij^ ^^ ^^ (j!^*^^ fj^} J^
Ibn Khaldun remarks : A report has reached us, that
some great scholars of the east have increased the number
of cases beyond six, and have brought them to upwards of
twenty, producing their accurate solutions together with
geometrical demonstrations."
Page 8, lime 17.
See Leonardo's geometrical illustration of the three
cases involving an affected square, as reported by Cossali
(i. p. 2.), and hence by Hutton (Tracts, &c., ii. p. 198.)
Cardan, in the introduction of his Ars magna^ distinctly
refers to the demonstrations of the three cases given by our
author, and distinguishes them from others which are his
own. At etiam demonstrationes, prceter tres Mahometis
et duas Lodovici (Lewis Ferrari, Cardan's pupil),
omnes nostrce sunt. — In another passage (page 20) he
blames our author for having given the demonstration of
only one solution of the case cx'^-{-az=.bx. Nee admireris,
( 192 )
says he, hanc secundam demonstraiionem aliter quam a
Mahumete explicatam^ nam ille immutata Jigura magis ex
re ostendit, sed tamem obscurius^ nee nisi unam partem
eamque pluribus.
Page 17, line 11-13.
The words from (j LjtX-s l^lj to -^JuJl ^^-Jw.•^ are writ-
ten twice over in the manuscript.
Page 19, line 12.
f^\ jl 1*^5^^*^ LT^J'^^ -' ^^^Q root of a rational or ir-
rational number." In the Kholaset al Hisdb, p. 128. 137.
369, the expression (jjai^ (lit. audible) is used instead of
aA*k< , which stands in a more distinct opposition to ^
(lit. inaudible, surd). Baha-eddin applies the same ex-
pressions also to fractions, calling ^jt^^ those for which
there are peculiar expressions in Arabic, e. g. ui-^ one-
third, and j^^\ those which must be expressed periphras-
tically by means of the word ^Jp^ a part, e. g. ^y^' *^^
^ JLc^ L.^A^ ^jA3 three twenty-fifths. See Kholaset al
Hisab, p. 150.
Page 19, line 15.
The manuscript has JW tli3wi ^^J^ . The context
requires the insertion of j^ after ^^ 5 which I have
added from conjecture.
Page 20, line 15. 17.
i^oA^\ <-,>w3.» U ] " What is proportionate to the unit,"
( 193 )
/. e. the qtiotient. This expression will be explained by
Baiia-eddin's definition of division (Kholaset al IJisaby
p. 105). j4.j-Jil,\ L^ Jc^yi ^\ <kx^ JS£. c-Jis U^\
<l1c /•^•Jili ^\ Division is the finding a number which
bears the same proportion to the unit, as the dividend bears
to the divisor."
Pflg^e21, line 17.
^j^ ] The MS. hasjj^ .
Page 24, line 6.
f^t-^xscr "j 'ijya l^ IjJkC^ J An attempt at constructing a figure
to illustrate the case of [loo-j-j;'^— 20:c] +[504-1 oo;- sx^]
has been made on the margin of the manuscript.
Page 30, line 10.
v:>w-£i to j^ J A marginal note in the manuscript
defines this in the following manner. l^jt^\ a-j1 15-^.
lie means to say : divide the ten in any manner you like,
taking four of wheat and six of barky, or four of barley
and six of wheat, or three of wheat and seven of barley,
or vice versa, or in any other way : for the solution will
hold good in all these cases. (Note from Al MozaihaJVs
Commentary).''^
Page 42, line 8.
The manuscript has a marginal note to this passage,
2 c
( 194 )
from which it appears that the inconvenience attending the
solution of this problem has already been felt by Arabic
readers of the work.
Page 45, line 16.
This instance from Mohammed's work is quoted by
Cardan (Ars Magna, p. 22, edit. Basil.) As the passage
is of some interest in ascertaining the identity of the present
work with that considered as Mohammed's production by
the early propagators of Algebra in Europe, I will here
insert part of it. Nunc autem, says Cardan, suhjungemus
aliquas qucestiones, duas ex Mauumete, reliquas nostras.
Then follows Qucestio I, Est numerus a cujus quadrato
si abjeceris ^ et 4 ipsius quadratic atque insuper 4, rc-
siduum autem in se duxeris^Jiet productum cequale qua^
drato illius numeri et etiam 12. Pones itaque quadratum
numeri incogniti quern qiiceris esse 1 rem, abjice J e^ 4
ejus, es insuper A, Jiet t2 ^ei m: 4, duo in se, fit x^j-
quadrati p: 16 m: 33 rebus, et hoc est cequali uni rei
et 12; abjice similia, Jiet 1 res cequalis -f^ quadrati
p : 4 m : 3j rebus, &c.
The problem of the Qucestio II. is in the following terms,
Fuerunt duo duces quorum unusquisque divisit militibus
suis aureos 48. Porro unus ex his habuit milites duos
plus altera, el illi qui milites habuit duos minus contigit
ut aureos quatuor plus singulis militibus daret ; quceritur
quot unicuique milites fuerint. In the present copy of
Mohammed's algebra, no such instance occurs. Yet Car-
( 195 )
DAN distinctly intimates that he derived it from our author,
by introducing the problem which immediately follows
it, with the words : Nunc autem proponamus qucestiones
nostras.
Page 46, line 18.
The manuscript has the following marginal note to this
passage : jo-lj' ^ ^„j^ j c-^^*^b J-«ju ^LJ^. ^JJ^
jSa ^Js> \Ji^j^ ^^j^ ^^ l;^*^ ^J1^ u/^ J^^^
jWj jw 4-i^ j*^!^ ^^j'^ L?^, ij^j^ u^- tJ^^
^J Ui Jl^l J^ This instance may also be solved
by means of a cube. The computation then is, that you take
the square, and remove one-third from it; there remain
two-thirds of a square. Multiply this by three roots ; you
find two cubes equal to one square. Extracting twice
the square-root of this, it will be two roots equal to a
dirhem. Accordingly one root is one-half, and the square
one-fourth.* If you remove one-third of this, there re-
mains one-sixth, and if you multiply this by three roots,
that is by one dirhem and a half, it amounts to one-fourth
of a dirhem, which is the square as he had stated."
[x- - ^ r-] X 3a; = a^^
2x= 1
( 196 )
Page 50, line 2.
I am uncertain whether my translation of the definition
which Mohammed gives of mensuration be correct. Though
the diacritical points are partly wanting in the manuscript,
there can, I believe, be no doubt as to the reading of tTie
passage.
Page 51, line 12.
I have simply translated the words <t-s4X:^i ^jitA by
"geometricians," though from the manner in which Mo-
hammed here uses that expression it would appear that he
took it in a more specific sense.
FiRUZABADi (Kamus, p. 814, ed. Calcutt.) says that
the word handasah ((LsJC^l) is originally Persian, and
that it signifies the deternrnning by measurement where
canals for water shall be dug."
The Persians themselves assign yet another meaning to
the word <!UjJcJ& hindisah, as they pronounce it : they use it
in the sense of decimal notation of numerals.*
It is a fact well known, and admitted by the Arabs
^j^^ss>- J ^ Jusr^ \ yf"^ «J^^.y c:^UK
"Hindisah is used in the sense of measurement and size ; the same word
is also applied to the signs which are written instead of the words (for
numbers) as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10." Burhani Kati.
( 197 )
themselves, that the decimal notation is a discovery for
which they are indebted to the Hindus.* At what time the
communication took place, has, I believe, never yet been
ascertained. But it seems natural to suppose that it was at
the same period, when, after the accession of the Abbaside
dynasty to the caliphat, a most lively interest for mathe-
matical and astronomical science first arose among the
Arabs. Not only the most important foreign works on
these sciences were then translated into Arabic, but learned
foreigners even lived at the court of Bagdad, and held
conspicuous situations in those scientific establishments
which the noble ardour of the caliphs had called forth.
History has transmitted to us the names of several dis-
tinguished scholars, neither Arabs by birth nor Moham-
medans by their profession, who were thus attached to the
court of Almansur and Almamun ; and we know from
* It is almost unncessary to adduce further evidence in support of this
remark. Baha-eddin, after a few preliminary remarks on numbers, says
'ij^^L\,\ hi^\ /♦l^^^ *y^\ -tU^ y *-w?j *XJj " Learned Hindus
have invented the well known nine figures for them." {Kholdset aUHisdh^
p. 16.) In a treatise on arithmetic, entitled Jlc ^1 ^I^^ .^^
/_ »\...gj \ which forms part of Sir W. Ouseley's most valuable col-
lection of Oriental manuscripts, the nine figures are simply called
A'lVj.^11 /Jl^^U See, on the subject generally. Professor von Boh-
len's work, Das alte Indien, (Kbnigsberg, 1830. 1831. 8.) vol. ii.
p. 224, and Alexander von Humboldt's most interesting disserta-
tion : Ueber die bei verschiedenen Viilkern iiblichen Si/steme von ZaM-
zeichen, &c. (Berlin, 1829. 4>.) page 24.
( 19S )
good authority, that Hindu mathematicians and astronomers
were among their number.
If we presume that the Arabic word handasah might,
as the Persian hindisah^ be taken in the sense of decimal
notation, the passage now before us will appear in an entirely
new light. The iUaJc^t J*^^, to whom our author ascribes
two particular formulas for finding the circumference of
a circle from its diameter, will then appear to be the
Hindu Mathematicians who had brought the decimal nota-
tion with them ; — and the ^^ (^H^^ «-J^^> ^^ whom the
second and most accurate of these methods is attributed, will
be the Astronomers among these Hindu Mathematicians.
This conjecture is singularly supported by the curious
fact, that the two methods here ascribed by Mohammed
to the ^-oJc^^l Jjbl actually do occur in ancient Sanskrit
mathematical works. The first formula, ^ = v^iOfl?2j occurs
in the Vijaganita (Colebrooke's translation, p. 308, 309.);
the second, ;?= ^^^^^ ^ is reducible to -7^^ , the pro-
portion given in the following stanza of Bhaskara's Lila-
vati :
" When the diameter of a circle is multiplied by three
( 199 )
thousand nine hundred and twenty-seven, and divided by
twelve hundred and fifty, the quotient is the near circum-
ference : or multiplied by twenty-two and divided by seven,
it is the gross circumference adapted to practice."* (Cole-
brooke's translation, page 87. See Feizi's Persian trans-
lation, p. 126, 127.)
1 he comcidence of ^^^^^ with -7^^ is so striking,
and the formula is at the same time so accurate, that it
seems extremely improbable that the Arabs should by
mere accident have discovered the same proportion as the
Hindus : particularly if we bear in mind, that the Arabs
themselves do not seem to have troubled themselves much
about finding an exact method. f
* The Sanskrit original of this passage affords an instance of the
figurative method of *the Hindus of expressing numbers by the names of
objects of which a certain number is known : the expressions for the
units and the lower ranks of numbers always preceding those for the
higher ones. ^ (lunar mansion) stands for 27 ; H^^ (treasure of
Kuvera) for 9 ; and 3f t5"(sacred fire) for 3 : therefore ^^T^^?^.
1^=3927. Again, ^(cypher) is 0; ^ |0| (arrow of Kamadeva)
stands for 5 ; M^ (the sun in the several months of the year), for
12
: therefore <G|C| |U|H^ = 1250. For further examples, see As,
Res. vol. XII. p. 281, ed. Calc, and the title-pages or conclusions of
several of the Sanskrit works printed at Calcutta ; — e. g. the Sutras of
Panini and the Siddhantakaumudi.
t This would appear from the very manner in which our author
introduces the several methods; but still more from the following
marginal note of the manuscript to tlje present passage : \,^^JsCi ys
( 200 )
Page 57, line 5-8.
The words between brackets are not in the manuscript;
I have supplied the apparent hiatus from conjecture.
Page 61, line 4.
A triangle of the same proportion is used to illustrate
this case in the Lilavaii (Feizi's Persian transl, p. 121.
Colebrooke's transl. of the Lilavati, p. 71. and of the
Vijaganita, p. 203.)
Page 65, line 12-14.
The words between brackets are in the manuscript writ-
ten on the marg-in. I think that the context warrants me
sufficiently for having received them into the text.
Page 66, line 5.
The words between brackets are not in the text, I give
them merely from my own conjecture.
^1 Wj^t^ Axj ^j tljOi &Ji>. ^Ic Ss>-\ tJib ^j ijit^^ ^
^-ir^' jj^ J\jJ^l 2^j^ ui to ^^;-^\j <dll ^1 A^lxJ ^ ^;jc>- ^,^
approximation, not the exact truth itself: nobody can ascertain the
exact truth of this, and find the real circumference, except the Omnis-
cient : for the line is not straight so that its exact length might be found.
This is called an approximation, in the same manner as it is said of the
square-roots of irrational numbers that they are an approximation, and
not the exact truth : for God alone knows what the exact root is. The
best method here given is, that you multiply the diameter by three and
one-seventh : for it is the easiest and quickest. God knows best !"
( 201 )
Page 71, line 8, 9.
Tlie author says, that the capital must be divided into
219320 parts: this I considered faulty, and altered it in
my translation into 964080, to make it agree with the com-
putation furnished in the note. But having recently had
an opportunity of re-examining the Oxford manuscript,
I perceive from the copious marginal notes appended to
this passage, that even among the Arabian readers con-
siderable variety of opinion must have existed as to the
common denominator, by means of which the several
shares of the capital in this case may be expressed.
One says : Uj ei-Jj ^J^j ^j ^-^ c>^ J^ J^^
^r-^. iU<j C>y^^ ^'^^^^3 ^^^ ^J^ <--^^ CT* f^ cT*^
Find a number, one-sixth of which may be divided into
fourths, and one-fourth of which may be divided into
thirds; and what thus comes forth let be divisible by hun-
dred and ninety-five. This you caimot accomplish with
any numberless than twenty-four. Multiply twenty-four by
one hundred and ninety-five : you obtain four thousand six
hundred and eighty, and this will answer the purpose."
Another :* ^j.«..>^ fs . ^ ^ j LU J^ssT ^^jl^l <^p-j ^J,^
* The numbers in this and in part of the following scholium are in
the MS. expressed by figures, which are never used in the text of
the work.
2d
( 202 )
ii l^^li y (^,wk>- ^ J CPI ^'Jc^-j ^^Ij eJ^l ybj ^ill
iT7 ^j\ ^.^^u\j vnr ^^^^\ ^^^u j " ac-
cording to another method, you may take one hundred
and fifty-six for the one-sixth of the capital. Multiply
this by six; you find nine hundred and thirty-six. Taking
from this the share of the son, which is one-third and one-
fourth, you find it five hundred and forty-six. This is not
divisible by five : therefore multiply the whole number of
parts by five ; it will then be four thousand six hundred
and eighty. Of this the mother receives four hundred and
twenty-five, the husband seven hundred and eighty, the
son two hundred and eighty-eight (twelve hundred and
eighty-eight ?), the legatee, who is to receive the two-
fifths, fourteen hundred and ninety-two, and the legatee
to whom the one-fourth is bequeathed, six hundred and
ninety-five."
Another : X>UiJj j cJ^l ^*-J ^ ^, j^] [^j] Jj
fj^j^ l^Ju*s Ujci-kj fjti^j^j tjt^y^ U^^ tt/!^ ( ^ -N**^*^ '
J-tf^ L^j^'^ C^ ^J^.J^ (jr?!? W^ j^^ c-i-JiJij j*^
( 203 )
c^j ^ ^iii ij^i li [MS. ] * ^r] Wv ^,/.. Tr
As-) iUjlj {^j^^ J <bU2JJ J L-2II ' According to another
method, the number of parts is nine thousand three hun-
dred and sixty. The computation then is, that you divide
the property left into twelve shares; of these the mother
receives two, the husband three, and the sjon seven. This
(number of parts) you multiply by twenty, since two-
fifths and one-fourth are required by the statement. Thus
you find two hundred and forty. Take the sixth of this,
namely forty, for the mother. One-third out of this she
must give up. Now, forty is not divisible by three. You
accordingly multiply the whole number of parts by three,
which makes them seven hundred and twenty. The one-
sixth of this for the mother is one hundred and twenty.
One-third of this, namely forty, goes to the legatees, and
should be divided by thirteen ; but as this is impossible,
you multiply the whole number of parts by thirteen, which
makes them nine thousand three hundred and sixty, as we
said above. Of this the mother receives eight hundred
and fifty? the son two thousand five hundred and seventy-
six, the husband one thousand five hundred and sixty, the
legatee to whom the two-fifths are bequeathed, two thou-
sand nine hundred and eighty- four, and the legatee who
is to receive one-fourth, one thousand three hundred and
ninety.'*
( 206 )
there remains nine of it, and this is the deduction from the
completement. , Subtracting it from the completement,
which is thirteen, there remains four, and this is the legacy.
as the author has said,"
Page 98, line 8.
The word l^ll^ which I have omitted in my translation
of this and of two following passages, is in the manuscript
explained by the following scholium : ^ l^ <U jL-::.^ l^*X«
^ILJ! J j^\ J jJJ\j JW J ^.^^^\j ^\ J jj-^s!^
Adequate, «. e, corresponding to her beauty, her age, her
family, her fortune, her country, the state of the times, ....
and her virginity." (Part of the gloss is to me illegible.)
The dowry varies according to any difference in all the
circumstances referred to by the scholium. See Hamil-
ton's Hedaya, vol. i. page 148.
Page 113, line 7.
The manuscript has the following marginal note (?).
^\A\ 4 'i;^\ ^lcji\ 4 l^ " The OA:r ofa slave girl
corresponds to the adequate dowry of a free-born woman ;
it is a sum of money on payment of which one of dis-
tinguished qualities corresponding to her would be mar-
ried." See Hamilton's Hedaya, vol. ii. page 71.
I am very doubtful whether I have well understood the
words in which our author quotes Abu Hanifah's opinion.
Abu Hanifah al No'man ben Thabet is well known
1
( 207 )
as an old Mohammedan lawyer of high authority. lie was
bom at Kufa, A.H. 80 (A.D. 690), and died A.H. 150
(A.D. 767). Ebn Khallikan has given a full account
of his life, and relates some interesting anecdotes of him
which bear testimony to the integrity and independence
of his character.
Page 113, line 16.
The marginal notes on this chapter of the manuscript
give an account of what the computation of the cases here
related would be according to the precepts of different
Arabian lawyers, e. g. Shafei, Abu Yussuf, &c. The
following extract of a note on the second case will be
sufficient as a specimen : i<*)J^^^ ^^ t/*^^ ^-'*l^^ '
^^ L5^ ^^^^ (JJ^ ^} ^^3 ^'^3 ^ \"".* t-j^^yi /^
Ji4>.sr< Ai£ J ^^^llJl [^^X< jtfU ^^^^ '^^^ J^^ ^ J ^^
<Upj ill J <U^ '-r^ ^ t-^yi ^l?j J«^ (*) u-:^?^^ ^i;^
j^^ L::-Jxa lH J.»«jw Ju-i:>- ^^1 Jji j^Lcj l^ls*- j^^ ^i]/Jt
i!i LUiJJ ^jji^ '-^'V.V J^ liii(*) yjj '— fl-'j:^ (^^ «--^«X«
^jl c^JJl ^:a Jt^ ^^ ^-^ Jj^. ^^ JUj j l*-^
X--^«ci. ^,<^t Jjui ci-^lij ^-^j^ \^^ ^} ^j >^V
* These names are very indistinctly written in the manuscript.
( 208 )
l^bjl AiJj ij, The solution of this questlan given by
the Khowarezmian is according to the school of Abu
YussuF Wazfar, and one of the methods of Shafei's
followers. Abu Hanifah calls the sum which the donor has
to pay on account of having cohabited with the slave-girl
likewise a legacy ; thus, according to him, the legacy is
one and one-third of thing : this is another method of
Shafei's school. According to Mohammed ben al
J.MSH, the donor has nothing to pay on account of having
cohabited with the slave grirl :* and this is aarain a method
adopted by the school of Shafei. After this method,
one-third of the donation is really paid, whilst two-thirds
become extinct : and there is no return, as the heritage has
remained unchanged. According to Abu Hanifah, you
proceed in the same manner as after the precepts of Abu
YussuF Wazfar. Thus the heirs obtain three hundred
less one and one-third of thing, which is equal to two things
and two-thirds : for what he (the donor) has to pay on
behalf of the dowry, is likewise a legacy. Completing and
reducing this, one thing is equal to seventy-five dirhems :
this is one-fourth for the slave-girl ; one-fourth of the
donation is actually paid, and three-fourths become ex-
tinct.*'
* I doubt whether this is the meaning of the original, the words from
JkA^S^ till it^lL being very indistinctly written in the MS.
.LcUUi
^--
>
.k-J ^JS.^
^^^, JWj
c;J^'.
\^ rp
i-jiasTj
J=^j
1 rc
A-^^4
>^'.
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^■j^i
i>cj
11 n
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^
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t^ -*v'^~
A n
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■ji^
^ Pr
4«J_y i(L-*i-
io^_,^^U.^
ic ic
tuijj
J^.i
ic vr
eJJ,
'^^
1 vr
rf' ^'
r^"
1^ —
J^i^cr* '!)=r-cP^iir*
j.jjij^ecr'
Ic vc
cs~^
u-^
ir Ai
*«j^l
SmjS Lw^iil
f- AV
^j
■ >.
r V
j*j
J*
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V 11
JU ^ U«- j^jlj j^U j^
J^^
n ir
Jiscli
AS.'
|v 11^
S^
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M 11
W»j
C^,
p 1 ♦♦
IL^Sl
^^^\
1 -
liaij
ti ~
IJuc
Jufi
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LL
n iM
n tn
ciJ.'^
hL',
tr Mr
.^li
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IP tn
irr
y&^ i^^\ ^ \^suu JtAJ *L<-i> u» a .A* J ^(^_5**j Uji^J
h^ i^cflfs^ ^^Iaj ^^ ^Jii AJfe^ J bJJj Ujb^J ^^ ^
ybj [^ ^jf^\j hJ^\ ^ jjj i^ iuL jjjj ^^1
cL-Ji <LJ\^^ f^\ji^ ^^ (JJ^ (^^'^ U— j'j i^Ujb^J
^J\ L!J3<i J^ Uibjj ^^y^ ^j JH^ c^lj ^j^
cLj1 <u-«c*. j^'j <)lcLj"1 Axjjl ^1 ^ j^ l^LJi
ciJjj Ujb;u> j^yijj Hj <^^1 *-^*V3 s^^r^^ rt^*^ t*^\?
ir
* u^^ 4 (^^ ^^
(*\^ ^ ^ d Ujbjj ^^ au^> 4 -k; (^^ ^^^
^. 3 J^^ ^in'. *^^ Hr* 4 ^^^ (*^ C^b"^ i^ *^-5^'
''^ J >^' 'ir:'. J^ ^M (*>^b^ ir-^ '•^^ ^J3 ^
^ L^jSU UjbjJ ^,j^ -^V^^^ '^ e;/^ (^b^ Ir^
yb^ ^^^i> J^^^ s-^ j^ UjbjJ ^/ij uiCli lKj ^j
^^ ^^^1 ^ ^cJJ^ ^^b ^^ j^^ L^^\ ^
* i^bU^U ^^J^ Uybj >jbl^J
^ 4 ^^jo ybj Uibj^ ^^^ «>:; J^ {^--^ l;^
^ Jj '^\ lL-1 JW ijw«|^ ^^ J^^ iJ ^ \SUI) J ^^j^
^jjjb ^1 ^ i3^. c5J31 Jxs^i ^d*^ Jiw^j <Lk^^!
ybj ^.^\ ti;-* l/^ ^ L5^ ^^ ^^-^ tirAr^
R
-^^^ cJ^ Jjc^, Ufc^j ^jjj ^ ^^^\ (^^ ^^Li
C^jJl ^^1 (^ JjOO ^^ ^ ^^^ c-o^ ^^/J
^/-J UjbjJ ^^ ill >^| j_ft^ ^;^^ *^^J1 ^^ ^(
jAj ^1 jci^^j *^-i, (_a^ ^ e;.jV-j uW.^^ ^^, 4
u^Jj ^(^-^ ^r^^ c:;>>^J ^*;>Uii5 2rJo 4 J-^^^ Ai>l^J
^.^ L)^^ J^^^'J ^{^S^^ L^ ^*^3 ^LS*^ ^"^^j
hUxij\ ^f^ s.\.^\ ^i^ ^^ j^^ -"lS^ iJ^^ U^
^jLs C^s> JjUii ^^-i Li^j -\^^ ^W* J^^>.
* u&;t> -r^-'Vj^j ^^ J*^. '^^V^ "^^r*^^
l^:;^ ^ 4 ^jV t>r;^ ^-H^-? ^-^ ^^ ^^
I>UiJj l^::^ JU^ ^\ ^uUi j^k:a:^^ j^^ V-* J^- (*^
^JJ ^^-^3 4 L5*^ ^^^ *-^^^ ^ ^^^ ^ f^^^
ji^3 "-J^ u^. ji^3 ^J^ ^> J^ ^lT ^^ ^ST*
j^ ,^ lL<33 jt^"^ "^ u^. J^ ji^3 ^J^
^^U^^ as Jc;.U ^^ ^j ^yd '^5^ u^ Jl ^^'^
^\sux> U^IwJ «J^1 ^V^^ L5^^ Jy 1^^ ^"^^ cJij
<U^ ,^^iL-i /^^ <jjj y&j ^1 J, J -c^ ^ LUiSj
J>X*J ^^ ,^_^ J ^i;!*^ ^ «^,Uii5 ifjj J ^A.>J
L5^ ^^Jj "^(^^ *-^ J ''l5'^ '— ^^ ^r^^ ^^
^^^\j -^U-i»l ^^1 Jj^. ^^^---^♦^j iiuUjlaJ ^^;-:u^j!l
He iLaJ J (j^^ J <^»**' y^j <^j u^ J j^
Jr^\ JcS-lj -i,^ ^^ LUilj ^^jiif-i lL-i» ^-^jll Jjtsr
-^ *Jb;J LUjOjI i^tX> ^j jLa3 ^^c-^ lIJlj ^*i. ijto
j^^\ iL^lj -^^^ LiJj ^1 J,j -t^ <jJj J ^^
,^jiLJ J^^ (Jjjj llJi Jjill i^j J2^ cUbJlj «^
ul^i J ^j^\ J^« J'^^. -^^1 ^ ji^ (*^J^ ^.^^J^
^^^\ Jx 5Jjj ^^ cU5 J ^-^^ Ll^i ^li
^ ^^Ij ^^ lUjj ^\^\ ^ Jj^t^. ^.Uii5 ^^
^^ ^^U ^j jjfilftljill lIJj iiJ L— !U>^1 /*jij J ^(^_5i^
^ ^U*J^1 C-^yi Xjjj t^Jol <i J^ -^(^ *-^
k^J^ U^ Ui^^ J^ ^^J3 ^^ J^-^ -"lt^
dx»-Jj ^U ^j LUx^j^ ^ 1-^^ jLs- »\>-l jjA3 -'Jjs*"'
^ 1-t^ yLc Jo-1 j^ "^i^f"^ i^^^j 1^^^^ J k}"^
^j x^^jLW JxrsT ^su:.s>^ ^^\ JjJ lij ^ j^"^
n
^^ cr?^:*^J '^ J^. ^J^^ u/^ ^, iSi^ ^^ er*
<d JU ^ <C?j^ (^^ ^S ^J^ Jsfj^ S^^J J^J
jtoj i|^^ ciJj ^^ ^J £>lo (J^ij i/^Jill «J-Jj
^,^li ^^ll-i, iiOi J ^^ ^ j^l l^^\ h,< tl<3 J^
^/-i ^^^1 Jj^ ^Jjj ^^ cj^j ^^ ^.Utlj^l
H5 L!J3i ^r^ ^^j ^(^ tjjjj ^Li.1 HS Jjou ^U«j\
!U
U&;J ^>^j ^j^ ^j iil^^ j^ ^J^ ijiri!^^ cT*
Jjj^ *-r^j J rt^^ ^,^ W^^^ ^jW" (j^ u^
bybj ^ Mj^ji' ^^y f^*^ iJU-— 4^ W^*^ ^i;V
jcA--i -^^c-2i uL^li ^ <^*^J^^J {»^^ <ijU-w«C>- *^jV^
^^J^ jJ:. *Jb^J LU:;<«j £j;jll ^^kX>3 <4j^ ^Lf^ U-^*>^
JtXjo uj3i t^{lav5 iLj-^jH ib.0 uJ33j -^Lj-^ (^ ^-^
j^^ il^ill i^ji^ L!i3J j^^ ^^-i» ^^ l^jcf t^j^-
IIP
ajUj! L.A^ CJ^j ^ ^^^ J^ CJisi JjU-
l;aA^j <u./K£^ j ^U yb^ 4ir?^U ^\^\ L.^ jc^l^i
kjW (J^J s^ J (^'^ ^.^ ^"^^^ ^liJ ^'V^ j^^ ^^
c^\ ^ j^\i hj\A\ ^^^^ i^j^^ "i &j\ &ju^
«U-J JxfisT ^ ^^^ * t:;:f«-^ W^ CJ^\ ^if^
ti jUy ^^^ L!j3i jj^ L^^t AJb;J aJUwid- hj^\
Ji^ ^j «tJU viJdJ J^ ls*^j1? -^lT^ U--*>^ J -'^--
^^-r>-U ^j^ ^r^^ ■'jc^ j*^ LUsIj ybj (*^k^? J*^.
^Ll\ Hj Jj^, LUiJj ^/J ^^ ^j -^^5^ cKlJ
^^^ ^yuJ X,.U: ^> ^>^ (^j^ ^>j -^Lr* -^-^
uj3i ^^ ^^li <U-JjUj ^^^^ jL^j ^Ul H5 J^.
itr
*j&^j ,^.^ J^. *^^^y^ s.^^\ ^^^ <u^ \-^
^ Uj>. jLa Lt^ ^ s\js^\ X-.^d- J ^5;J ^/^ J
^ J LU ljb;Ai: ^ *J^;t^ iJU..-^«o- If^ ^.jV <-^^
j^^o-Up Lap* J*^j (^^ ^.^ *V^' *^r^j 4?^ S^
^jd LU ^1 ^Jj ^J^. ^ ^y^y^ ^^^3 fD^
^j,^ LUc^ L)jj^\ i^dj\ 4 J^ -^ls^ U"^ j^
^jj U-^ JJ^. j^j-^ ^.UiiS lL^J ^/J -^^1 ^^
ibU LU^\ ^j;^ ^JbU j^jj X»Ujj ^^ ^U=5.1
j:^ ^^U^l iiJjj U-^ J^. (^j^ W> L5*^^ ^W:
Q
Cl^-tfjl iXJJj ^^^ et^j ^ib;J ^J^j U^J U^J ^J
U--J J Uji;J L;j.y-^^ u^ J (^"^ ^.^ y^j W^ ci-J^
^*{;^J t^i^j^^ ^j^j J^y^ L5^.^ •'lT' t^-? C^J^
ijjjj Ajb^t) «^^ tj^j pl-ji '^.j^3 ^'♦^^ i^y^ J
-*^-i» J--J J (»^i^ «--J' bJjj cLjI ^U*j UajJ (i)^-J
t/^J ti J-^^ <^i-^i»- <)j^ aj Ij|^ ^^ ^^ ciJj^
j^Ulj Ujb^J U^J ^J^J U^\^ *--^^ t-jL=:i -^(^^
^ ^jj ^ Vj:>. ^^j ^ ^ Vy=T ^3r^ 3
^^ lL^j ^^ -^^1 W^ ^>:'j -^1/r^^ '^^k
^ ^ ^-trr i::^,j^3 (j^^ 3 ^^ e>^-? ^-'^-^
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