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

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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-Khwārizmī4187931The Algebra of Mohammed Ben Musa1831Friedrich Rosen

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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 subtracted 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.^[1]

If there are greater numbers combined with units to be added to or subtracted from them, then four multiplications are necessary;^[2] namely, the greater numbers by the greater numbers, the greater numbers by the 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 likewise positive. But,if one of them is positive, and one (16) negative, then the fourth multiplication is negative. ^[3]

For instance, “ten and one to be multiplied by ten and two.”^[4] 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,”^[5] then ten times ten is a hundred; the 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,”^[6] 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 hundred 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 introduction 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.”^[7] You begin by taking ten times ten, which is a hundred; less thing by ten is ten roots negative; the product is therefore a hundred less ten things. If the instance be: “ten and thing to be multiplied by ten,”^[8] then you take ten times ten, which is a hundred, 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,”^[9] 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 dirhems and twenty things and one positive square.

If the instance be: “ten minus thing to be multiplied by ten minus thing,”^[10] 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;”^[11] 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. Computation: You multiply one dirhem by one dirhem, the product is one dirhem; then one dirhem by minus one-sixth, that is one-sixth negative; then, again, one dirhem 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,^[12] 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,”^[13] 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,”^[14] 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 negative. 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 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,”^[15] then you say, half a dirhem by ten is five dirhems positive; 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 dirhems 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,”^[16] then this is the same as if it were said thing and ten by thing less ten. You say, therefore, 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, altogether, is a square less a hundred dirhems.

(19) Whenever a positive and a negative factor concur in a multiplication, such as thing positive and minus thing, the last multiplication gives always the negative product. Keep this in memory.

↑ If $x$ is to be multiplied by $y,x$ is to be repeated as many times as there are units in $y$ .
↑ If $x\pm a$ is to be multiplied by $y\pm b$ , $x$ is to be multiplied by $y$ , $x$ is to be multiplied by $b$ , $a$ is to be multiplied by $y$ , and $a$ is to be multiplied by $b$ .
↑ ${\begin{aligned}{\text{In multiplying}}\;(x\pm a)\;{\text{by}}\;(y\pm b)\\+a\times +b=+ab\\-a\times -b=+ab\\+a\times -b=-ab\\-a\times +b=-ab\end{aligned}}$
↑ ${\begin{aligned}&(10+1)\times (10+2)\\&=10\times 10\ldots 100\\&+1\times 10\ldots 10\\&+2\times 10\ldots 20\\&+1\times 2\ldots 2\\&+132\end{aligned}}$
↑ ${\begin{aligned}&(10-1)(10-1)\\&=10\times 10\cdot \cdot +100\\&+1\times 10\cdot \cdot -10\\&+1\times 10\cdot \cdot -10\\&+1\times -1\cdot \cdot +1\\&+81\end{aligned}}$
↑ ${\begin{aligned}(10+2)&\times (10-1)=\\10\times 10&\ldots 100\\-1\times 10&\ldots -10\\+10\times 2&\ldots +20\\-1\times 2&\ldots -2\\&108\end{aligned}}$
↑ $(10-x)\times 10=10\times 10-10x=100-10x.$
↑ $(10+x)\times 10=10\times 10+10x=100+10x$
↑ $(10+x)(10+2)=10\times 10+10x+10x+x^{2}=100+20x+x^{2}$
↑ $(10-x)\times (10-2)=10\times 10-10x-10x+x^{2}=100-20x+x^{2}$
↑ $(1-{\tfrac {1}{6}})\times (1-{\tfrac {1}{6}})=1-{\tfrac {1}{3}}+{\tfrac {1}{6}}\times {\tfrac {1}{6}}={\tfrac {2}{3}}+{\tfrac {1}{6}}\times {\tfrac {1}{6}}{\text{; i.e}}\;{\tfrac {25}{36}}={\tfrac {2}{3}}+{\tfrac {1}{6}}\times {\tfrac {1}{6}}$
↑ $(10-x)(10+x)=10\times 10-10x+10x-x^{2}=100-x^{2}$
↑ $(10-x)\times x=10x-x^{2}$
↑ $(10+x)(x-10)=10x+x^{2}-100-10x=x^{2}-100$
↑ $(10+{\tfrac {x}{z}})({\tfrac {1}{2}}-5x)={\tfrac {10}{2}}+{\tfrac {}{4}}-50x-{\tfrac {5}{2}}x^{2}=5-49{\tfrac {3}{4}}x-2{\tfrac {1}{2}}x^{2}$
↑ $(10+x)(x-10)=(x+10)(x-10)=x^{2}+10x-10x-100=x^{2}-100$

[1] If $x$ is to be multiplied by $y,x$ is to be repeated as many times as there are units in $y$ .

[2] If $x\pm a$ is to be multiplied by $y\pm b$ , $x$ is to be multiplied by $y$ , $x$ is to be multiplied by $b$ , $a$ is to be multiplied by $y$ , and $a$ is to be multiplied by $b$ .

[3] ${\begin{aligned}{\text{In multiplying}}\;(x\pm a)\;{\text{by}}\;(y\pm b)\\+a\times +b=+ab\\-a\times -b=+ab\\+a\times -b=-ab\\-a\times +b=-ab\end{aligned}}$

[p38_r2-4] ${\begin{aligned}&(10+1)\times (10+2)\\&=10\times 10\ldots 100\\&+1\times 10\ldots 10\\&+2\times 10\ldots 20\\&+1\times 2\ldots 2\\&+132\end{aligned}}$

[5] ${\begin{aligned}&(10-1)(10-1)\\&=10\times 10\cdot \cdot +100\\&+1\times 10\cdot \cdot -10\\&+1\times 10\cdot \cdot -10\\&+1\times -1\cdot \cdot +1\\&+81\end{aligned}}$

[6] ${\begin{aligned}(10+2)&\times (10-1)=\\10\times 10&\ldots 100\\-1\times 10&\ldots -10\\+10\times 2&\ldots +20\\-1\times 2&\ldots -2\\&108\end{aligned}}$

[7] $(10-x)\times 10=10\times 10-10x=100-10x.$

[8] $(10+x)\times 10=10\times 10+10x=100+10x$

[9] $(10+x)(10+2)=10\times 10+10x+10x+x^{2}=100+20x+x^{2}$

[10] $(10-x)\times (10-2)=10\times 10-10x-10x+x^{2}=100-20x+x^{2}$

[11] $(1-{\tfrac {1}{6}})\times (1-{\tfrac {1}{6}})=1-{\tfrac {1}{3}}+{\tfrac {1}{6}}\times {\tfrac {1}{6}}={\tfrac {2}{3}}+{\tfrac {1}{6}}\times {\tfrac {1}{6}}{\text{; i.e}}\;{\tfrac {25}{36}}={\tfrac {2}{3}}+{\tfrac {1}{6}}\times {\tfrac {1}{6}}$

[12] $(10-x)(10+x)=10\times 10-10x+10x-x^{2}=100-x^{2}$

[13] $(10-x)\times x=10x-x^{2}$

[14] $(10+x)(x-10)=10x+x^{2}-100-10x=x^{2}-100$

[15] $(10+{\tfrac {x}{z}})({\tfrac {1}{2}}-5x)={\tfrac {10}{2}}+{\tfrac {}{4}}-50x-{\tfrac {5}{2}}x^{2}=5-49{\tfrac {3}{4}}x-2{\tfrac {1}{2}}x^{2}$

[16] $(10+x)(x-10)=(x+10)(x-10)=x^{2}+10x-10x-100=x^{2}-100$

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

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