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For, since each of the numbers AB, BC, CD, DE is odd, if an unit be subtracted from each, each of the remainders will be even; [VII. Def. 7] so that the sum of them will be even. [IX. 21]

But the multitude of the units is also even.

Therefore the whole AE is also even. [IX. 21] Q. E. D.


PROPOSITION 23.

If as many odd numbers as we please be added together, and their multitude be odd, the whole will also be odd.

For let as many odd numbers as we please, AB, BC, CD, the multitude of which is odd, be added together; I say that the whole AD is also odd.

Let the unit DE be subtracted from CD; therefore the remainder CE is even. [VII. Def. 7]

But CA is also even; [IX. 22] therefore the whole AE is also even. [IX. 21]

And DE is an unit.

Therefore AD is odd. [VII. Def. 7] Q. E. D. 1


PROPOSITION 24.

If from an even number an even number be subtracted, the remainder will be even.

For from the even number AB let the even number BC be subtracted: I say that the remainder CA is even.

For, since AB is even, it has a half part. [VII. Def. 6]

1 3. Literally “let there be as many numbers as we please, of which let the multitude be odd.” This form, natural in Greek, is awkward in English.

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