内容 第9卷 · 263
命题 Propositio IX.22
If as many odd numbers as we please be added together, and their multitude be even, the whole will be even.
任意多个奇数相加，如果它们的个数是偶数，则总和为偶数。
偶数个奇数 AB、BC、CD、DE 相加：每个减去单位后为偶数，余数和为偶；单位之和（个数为偶）亦为偶；故和 AE 为偶。
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分步证明Step-by-step proof
1 / 4For let as many odd numbers as we please, AB, BC, CD, DE, even in multitude, be added together; I say that the whole AE is even.
设任意多个奇数AB、BC、CD、DE，个数为偶数，相加得AE。
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.
由于每个奇数减去1后均为偶数（根据VII.定义7），故这些差之和为偶数（根据IX.21）。
[IX. 21] But the multitude of the units is also even.
减去的1的个数为偶数，因此这些1之和为偶数。
偶数加偶数得偶数，所以AE为偶数。
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