The Compendious Book on Calculation by Completion and Balancing/On division

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

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Abū ʿAbdallāh Muḥammad ibn Mūsā al-Khwārizmī4187936The Algebra of Mohammed Ben Musa1831Friedrich Rosen

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ON DIVISION.

If you will divide the root of nine by the root of four,^[1] 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,^[2] 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.

If you wish to divide twice the root of nine by the root of four, or of any other square^[3], 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 submultiple of the root of nine, the rule is always the same:^[4] follow it, the result will be right.

If you wish to multiply the root of nine by the root of four,^[5] 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,^[6] multiply five by ten: the root of the product is what you have required.

If you wish to multiply the root of one-third by the root of a half;^[7] 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 thrice the root of four,^[8] 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 negative roots.

Demonstrations.

(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 our object was to add the remainder of the root of two hundred, after the subtraction of ten, that is to say, the line 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)

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 point H, the same length as the length of the line which represents the root of two hundred; 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 perceive 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. Now 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. (24) This it is that we wished to elucidate.

Here follows the figure:

As for the hundred and square minus twenty roots added to fifty, and ten roots minus two squares, this does not admit of any figure, because there are three different species, viz. squares, and roots, and numbers, and nothing corresponding to them by which they might be represented. We had, indeed, contrived to construct a figure also for this case, but it was not sufficiently 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 becomes 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 hundred 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 connected 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.

↑ ${\tfrac {\sqrt {9}}{\sqrt {4}}}={\sqrt {\tfrac {9}{4}}}={\sqrt {2{\tfrac {1}{4}}}}=1{\tfrac {1}{2}}$
↑ ${\tfrac {\sqrt {9}}{\sqrt {4}}}={\sqrt {\tfrac {9}{4}}}={\tfrac {2}{3}}$
↑ ${\tfrac {2{\sqrt {9}}}{\sqrt {4}}}={\sqrt {\tfrac {36}{4}}}={\sqrt {9}}=3$
↑ ${\tfrac {m{\sqrt {p^{2}}}}{\sqrt {q^{2}}}}={\sqrt {\tfrac {m^{2}p^{2}}{q^{2}}}}$
↑ ${\sqrt {4}}\times {\sqrt {9}}={\sqrt {4\times 9}}={\sqrt {36}}=6$
↑ ${\sqrt {10}}\times {\sqrt {5}}={\sqrt {5\times 10}}={\sqrt {50}}$
↑ ${\sqrt {\tfrac {1}{2}}}\times {\sqrt {\tfrac {1}{3}}}={\sqrt {{\tfrac {1}{2}}\times \ {\tfrac {1}{3}}}}={\sqrt {\tfrac {1}{6}}}$
↑ $3{\sqrt {4}}\times 2{\sqrt {9}}={\sqrt {9\times 4}}\times {\sqrt {4\times 9}}={\sqrt {36\times 36}}=36$

[1] ${\tfrac {\sqrt {9}}{\sqrt {4}}}={\sqrt {\tfrac {9}{4}}}={\sqrt {2{\tfrac {1}{4}}}}=1{\tfrac {1}{2}}$

[2] ${\tfrac {\sqrt {9}}{\sqrt {4}}}={\sqrt {\tfrac {9}{4}}}={\tfrac {2}{3}}$

[3] ${\tfrac {2{\sqrt {9}}}{\sqrt {4}}}={\sqrt {\tfrac {36}{4}}}={\sqrt {9}}=3$

[4] ${\tfrac {m{\sqrt {p^{2}}}}{\sqrt {q^{2}}}}={\sqrt {\tfrac {m^{2}p^{2}}{q^{2}}}}$

[5] ${\sqrt {4}}\times {\sqrt {9}}={\sqrt {4\times 9}}={\sqrt {36}}=6$

[6] ${\sqrt {10}}\times {\sqrt {5}}={\sqrt {5\times 10}}={\sqrt {50}}$

[7] ${\sqrt {\tfrac {1}{2}}}\times {\sqrt {\tfrac {1}{3}}}={\sqrt {{\tfrac {1}{2}}\times \ {\tfrac {1}{3}}}}={\sqrt {\tfrac {1}{6}}}$

[8] $3{\sqrt {4}}\times 2{\sqrt {9}}={\sqrt {9\times 4}}\times {\sqrt {4\times 9}}={\sqrt {36\times 36}}=36$

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