The Compendious Book on Calculation by Completion and Balancing/On addition and subtraction

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On multiplication

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

On division

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Abū ʿAbdallāh Muḥammad ibn Mūsā al-Khwārizmī4187933The Algebra of Mohammed Ben Musa1831Friedrich Rosen

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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.^[1]

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.^[2]

A hundred and a square minus twenty roots, added to fifty and ten roots minus two squares,^[3] is a hundred and fifty, minus a square and minus ten roots.

A hundred and a square, minus twenty roots, diminished by fifty and ten roots minus two squares, is fifty dirhems and three squares minus thirty roots.^[4]

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 that you multiply it by two) then it will suffice to multiply two by two, and then by the square;^[5] 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 product 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.^[6]

(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.^[7]

Follow the same rule when you seek for a third, or a quarter of a root, or any larger or smaller quota^[8] of it, whatever may be the denominator or the numerator.

Examples of this: If you require to double the root of nine,^[9] 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.

In the same manner, if you require to triple the root of nine,^[10] 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^[11]. 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.

↑ $20-{\sqrt {200}}+({\sqrt {200}}-10)=10$
↑ $20-{\sqrt {200}}-({\sqrt {200}}-10)=30-2{\sqrt {200}}=30-{\sqrt {800}}$
↑ $50+10x-2x^{2}+(100+x^{2}-20x)=150-10x-x^{2}$
↑ $100+x^{2}-20x-[50-2x^{2}+10x]=50+3x^{2}-30x$
↑ $2{\sqrt {x^{2}}}={\sqrt {4x^{2}}}$
$3{\sqrt {x^{2}}}={\sqrt {9x^{2}}}$
↑ $n{\sqrt {x^{2}}}={\sqrt {n^{2}x^{2}}}$
↑ ${\tfrac {1}{2}}{\sqrt {x^{2}}}={\sqrt {\tfrac {x^{2}}{4}}}$
↑ ${\tfrac {1}{n}}{\sqrt {x^{2}}}={\sqrt {\tfrac {x^{2}}{n^{2}}}}$
↑ $2{\sqrt {9}}={\sqrt {4\times 9}}={\sqrt {36}}=6$
↑ $3{\sqrt {9}}={\sqrt {9\times 9}}={\sqrt {81}}=9$
↑ ${\tfrac {1}{2}}{\sqrt {9}}={\sqrt {\tfrac {9}{4}}}={\sqrt {2{\tfrac {1}{4}}}}=1{\tfrac {1}{2}}$

[1] $20-{\sqrt {200}}+({\sqrt {200}}-10)=10$

[2] $20-{\sqrt {200}}-({\sqrt {200}}-10)=30-2{\sqrt {200}}=30-{\sqrt {800}}$

[3] $50+10x-2x^{2}+(100+x^{2}-20x)=150-10x-x^{2}$

[4] $100+x^{2}-20x-[50-2x^{2}+10x]=50+3x^{2}-30x$

[5] $2{\sqrt {x^{2}}}={\sqrt {4x^{2}}}$
$3{\sqrt {x^{2}}}={\sqrt {9x^{2}}}$

[6] $n{\sqrt {x^{2}}}={\sqrt {n^{2}x^{2}}}$

[7] ${\tfrac {1}{2}}{\sqrt {x^{2}}}={\sqrt {\tfrac {x^{2}}{4}}}$

[8] ${\tfrac {1}{n}}{\sqrt {x^{2}}}={\sqrt {\tfrac {x^{2}}{n^{2}}}}$

[9] $2{\sqrt {9}}={\sqrt {4\times 9}}={\sqrt {36}}=6$

[10] $3{\sqrt {9}}={\sqrt {9\times 9}}={\sqrt {81}}=9$

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

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