The Compendious Book on Calculation by Completion and Balancing/Compendium on calculating by completion and reduction

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The author’s preface

The Algebra of Mohammed Ben Musa (1831)
by Abū ʿAbdallāh Muḥammad ibn Mūsā al-Khwārizmī, translated by Friedrich Rosen

On multiplication

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Abū ʿAbdallāh Muḥammad ibn Mūsā al-Khwārizmī4187067The Algebra of Mohammed Ben Musa1831Friedrich Rosen

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MOHAMMED BEN MUSA’S

COMPENDIUM

on calculating by

COMPLETION AND REDUCTION.

(3) 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 repeated at any complex number; and so forth to the utmost limit of numeration.

I 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.

A root is any quantity which is to be multiplied by itself, consisting of units, or numbers ascending, or fractions descending.^[1]

A square is the whole amount of the root multiplied by itself.

A simple number is any number which may be pronounced 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.”^[2]

(4) Of the case in which squares are equal to roots, this is an example, “A square is equal to five roots of the same;”^[3] 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;”^[4] 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;”^[5] then one square is equal to two roots; the root of the square is two, and its square is four.

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 equal to numbers; for instance, you say, “a square is equal to nine;”^[6] then this is a square, and its root is three. Or “five squares are equal to eighty;”^[7] then one square is equal to one-fifth of eighty, which is sixteen. Or “the half of the square is equal to eighteen;”^[8] 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 members; for instance, “one root equals three in number;”^[9] then the root is three, and its square nine. (5) Or “four roots are equal to twenty;”^[10] 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;”^[11] then the 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; ^[12] 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;^[13] for instance, “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 number^[14] 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 solution is the same when two squares or three, or more or less be specified;^[15] you reduce them to one single square, and in the same proportion you reduce also the roots and simple numbers which are connected therewith. (6)

For instance, “two squares and ten roots are equal to forty-eight dirhems;”^[16] 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 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 dirhems? Now halve the number of the roots; the moiety is two and a half. Multiply that by itself; the product 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 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;”^[17] that is to say, what must be the amount of a square, the moiety of which, when added to the equivalent of five of its roots, is equal to twenty-eight dirhems? 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 double 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 remainder 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.

Squares and Numbers are equal to Roots;^[18] 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 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 (8) instance is impossible;^[19] but if the product be equal to 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, ^[20] as I have explained under the first case.

Roots and Numbers are equal to Squares;^[21] 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 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 explained. 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 expedient, 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.” ^[22]

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, 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 parallelograms are combined, each having a side of the quadrate 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 numbers. 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, that 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 thirty-nine, in order to complete the great figure in its (10) four corners; because the fourth of any number multiplied by itself, and then by four, is equal to the product of the moiety of that number multiplied by itself.^[23] 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:

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 five, 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 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:—

Demonstration of the Case: “a Square and twenty-one Dirhems are equal to ten Roots.” ^[24]

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 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 multiplied 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; moreover, the line K L, which has been. cut off from K M, is equal to K G; consequently, the quadrangle M R 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, M T, the quadrangles H T 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 original 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:—

Demonstration of the Case: “three Roots and four of Simple Numbers are equal to a Square.” ^[25]

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 quadrate A D, which comprises the three roots and the four of numbers mentioned in this instance. In every quadrate 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.

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 together equal to the quadrangle A R, which represents the four of numbers.

We have seen, also, that the quadrangle G M comprises 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.

(15)

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.

↑ By the word root, is meant the simple power of the unknown quantity.
↑ $cx^{2}=bx$ $cx^{2}=a$ $bx=a$
↑ $x^{2}=5x\;\;\therefore \;\;x=5$
↑ ${\tfrac {x^{2}}{3}}=4x\;\;\therefore \;\;x^{2}=12x\;\;\therefore \;\;x=12$
↑ $5x^{2}=10x\;\;\therefore \;\;x^{2}=2x\;\;\therefore \;\;x=2$
↑ $x^{2}=9\;\;x=3$
↑ $5x^{2}=80\;\;\therefore \;\;x^{2}={\tfrac {80}{5}}=16$
↑ ${\tfrac {x^{2}}{2}}=18\;\;\therefore \;\;x^{2}=36\;\;\therefore \;\;x=6$
↑ $x=3$
↑ $4x=20\;\;\therefore \;\;x=5$
↑ ${\tfrac {x}{2}}=10\;\;\therefore \;\;x=20$
↑ The three cases considered are,
${\begin{aligned}{\text{1st.}}\;cx^{2}+bx=a\\{\text{2d.}}\;cx^{2}+a=bx\\{\text{3d.}}\;cx^{2}=bx+a\end{aligned}}$
↑ ${\begin{aligned}{\text{1st case}}:\;cx^{2}+bx=a\\{\text{Example}}\;x^{2}+10x=39\\x={\sqrt {({\tfrac {1}{2}})^{2}+39}}-{\tfrac {10}{2}}\\={\sqrt {64}}-5\\=8-5=3\end{aligned}}$
↑ i.e the coefficient.
↑ $cx^{2}+bx=a$ is to be reduced to the form $x^{2}+{\tfrac {b}{c}}x={\tfrac {a}{c}}$
↑ ${\begin{aligned}2x^{2}+10x=48\\x^{3}+5x=24\\x={\sqrt {\left({\tfrac {5}{2}}\right)^{2}+24}}-{\tfrac {5}{2}}\\={\sqrt {6{\tfrac {1}{4}}+24}}-2{\tfrac {1}{2}}\\=5{\tfrac {1}{2}}-2{\tfrac {1}{2}}=3\end{aligned}}$
↑ ${\tfrac {x^{2}}{2}}+5x=28$
$x^{2}+10x=56$
$x={\sqrt {\left({\tfrac {10}{2}}\right)^{2}+56}}-{\tfrac {10}{2}}$
$x={\sqrt {25+56}}-5$
$x={\sqrt {81}}-5$
$=9-5=4$
↑ ${\begin{aligned}{\text{2d case.}}\;&cx^{2}+a=bx\\{\text{Example.}}\;&x^{2}+21=10x\\x&={\tfrac {10}{2}}\pm {\sqrt {({\tfrac {10}{2}})^{2}-21}}\\&=5\pm {\sqrt {25-21}}\\&=5\pm {\sqrt {4}}\\&=5\pm 2\end{aligned}}$
↑ If in an equation, of the form $x^{2}+a=bx,({\tfrac {b}{2}})^{2}<a$ , the case supposed in the equation cannot happen. If $({\tfrac {b}{2}})^{2}=a$ , then $x={\tfrac {b}{2}}$
↑ $cx^{2}+a=bx$ is to be reduced to $x^{2}+{\tfrac {a}{c}}={\tfrac {b}{c}}x$
↑ ${\begin{aligned}{\text{3d case}}\;cx^{2}&=bx+a\\{\text{Example}}\;x^{2}&=3x+4\\x^{2}&={\sqrt {({\tfrac {3}{2}})^{2}+4}}+{\tfrac {3}{2}}\\&={\sqrt {1{\tfrac {1}{4}}^{2}+4}}+1{\tfrac {1}{2}}\\&={\sqrt {2{\tfrac {1}{4}}+4}}+1{\tfrac {1}{2}}\\&={\sqrt {6{\tfrac {1}{4}}}}+1{\tfrac {1}{2}}\\&={\sqrt {2{\tfrac {1}{2}}}}+1{\tfrac {1}{2}}=4\\\end{aligned}}$
↑ Geometrical illustration of the case, $x^{2}+10x=39$
↑ $4\times {({\tfrac {b}{4}})}^{2}={({\tfrac {b}{2}})}^{2}$
↑ Geometrical illustration of the case, $x^{2}+21=10x$
↑ Geometrical illustration of the 3d case, $x^{2}=3x+4$

[1] By the word root, is meant the simple power of the unknown quantity.

[2] $cx^{2}=bx$ $cx^{2}=a$ $bx=a$

[3] $x^{2}=5x\;\;\therefore \;\;x=5$

[4] ${\tfrac {x^{2}}{3}}=4x\;\;\therefore \;\;x^{2}=12x\;\;\therefore \;\;x=12$

[5] $5x^{2}=10x\;\;\therefore \;\;x^{2}=2x\;\;\therefore \;\;x=2$

[6] $x^{2}=9\;\;x=3$

[7] $5x^{2}=80\;\;\therefore \;\;x^{2}={\tfrac {80}{5}}=16$

[8] ${\tfrac {x^{2}}{2}}=18\;\;\therefore \;\;x^{2}=36\;\;\therefore \;\;x=6$

[9] $x=3$

[10] $4x=20\;\;\therefore \;\;x=5$

[11] ${\tfrac {x}{2}}=10\;\;\therefore \;\;x=20$

[12] The three cases considered are,
${\begin{aligned}{\text{1st.}}\;cx^{2}+bx=a\\{\text{2d.}}\;cx^{2}+a=bx\\{\text{3d.}}\;cx^{2}=bx+a\end{aligned}}$

[13] ${\begin{aligned}{\text{1st case}}:\;cx^{2}+bx=a\\{\text{Example}}\;x^{2}+10x=39\\x={\sqrt {({\tfrac {1}{2}})^{2}+39}}-{\tfrac {10}{2}}\\={\sqrt {64}}-5\\=8-5=3\end{aligned}}$

[14] i.e the coefficient.

[15] $cx^{2}+bx=a$ is to be reduced to the form $x^{2}+{\tfrac {b}{c}}x={\tfrac {a}{c}}$

[16] ${\begin{aligned}2x^{2}+10x=48\\x^{3}+5x=24\\x={\sqrt {\left({\tfrac {5}{2}}\right)^{2}+24}}-{\tfrac {5}{2}}\\={\sqrt {6{\tfrac {1}{4}}+24}}-2{\tfrac {1}{2}}\\=5{\tfrac {1}{2}}-2{\tfrac {1}{2}}=3\end{aligned}}$

[17] ${\tfrac {x^{2}}{2}}+5x=28$
$x^{2}+10x=56$
$x={\sqrt {\left({\tfrac {10}{2}}\right)^{2}+56}}-{\tfrac {10}{2}}$
$x={\sqrt {25+56}}-5$
$x={\sqrt {81}}-5$
$=9-5=4$

[18] ${\begin{aligned}{\text{2d case.}}\;&cx^{2}+a=bx\\{\text{Example.}}\;&x^{2}+21=10x\\x&={\tfrac {10}{2}}\pm {\sqrt {({\tfrac {10}{2}})^{2}-21}}\\&=5\pm {\sqrt {25-21}}\\&=5\pm {\sqrt {4}}\\&=5\pm 2\end{aligned}}$

[19] If in an equation, of the form $x^{2}+a=bx,({\tfrac {b}{2}})^{2}<a$ , the case supposed in the equation cannot happen. If $({\tfrac {b}{2}})^{2}=a$ , then $x={\tfrac {b}{2}}$

[20] $cx^{2}+a=bx$ is to be reduced to $x^{2}+{\tfrac {a}{c}}={\tfrac {b}{c}}x$

[21] ${\begin{aligned}{\text{3d case}}\;cx^{2}&=bx+a\\{\text{Example}}\;x^{2}&=3x+4\\x^{2}&={\sqrt {({\tfrac {3}{2}})^{2}+4}}+{\tfrac {3}{2}}\\&={\sqrt {1{\tfrac {1}{4}}^{2}+4}}+1{\tfrac {1}{2}}\\&={\sqrt {2{\tfrac {1}{4}}+4}}+1{\tfrac {1}{2}}\\&={\sqrt {6{\tfrac {1}{4}}}}+1{\tfrac {1}{2}}\\&={\sqrt {2{\tfrac {1}{2}}}}+1{\tfrac {1}{2}}=4\\\end{aligned}}$

[22] Geometrical illustration of the case, $x^{2}+10x=39$

[23] $4\times {({\tfrac {b}{4}})}^{2}={({\tfrac {b}{2}})}^{2}$

[24] Geometrical illustration of the case, $x^{2}+21=10x$

[25] Geometrical illustration of the 3d case, $x^{2}=3x+4$

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