Page:The Algebra of Mohammed Ben Musa (1831).djvu/40

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If the instance be: “ten and thing to be multiplied by ten,”^[1] 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,”^[2] 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,”^[3] 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;”^[4] 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

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1. $(10+x)\times 10=10\times 10+10x=100+10x$
2. $(10+x)(10+2)=10\times 10+10x+10x+x^{2}=100+20x+x^{2}$
3. $(10-x)\times (10-2)=10\times 10-10x-10x+x^{2}=100-20x+x^{2}$
4. $(1-{\tfrac {1}{6}})\times (1-{\tfrac {1}{6}})=1-{\tfrac {1}{3}}+{\tfrac {1}{6}}\times {\tfrac {1}{6}}={\tfrac {2}{3}}+{\tfrac {1}{6}}\times {\tfrac {1}{6}}{\text{; i.e}}\;{\tfrac {25}{36}}={\tfrac {2}{3}}+{\tfrac {1}{6}}\times {\tfrac {1}{6}}$