内容 第9卷 · 262
命题 Propositio IX.21
If as many even numbers as we please be added together, the whole is even.
任意多个偶数相加，其和必为偶数。
偶数 AB、BC、CD、DE 依次相加（A、B、C、D、E 共线）；每段皆有半，故和 AE 也有半，即 AE 为偶数。
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分步证明Step-by-step proof
1 / 4For let as many even numbers as we please, AB, BC, CD, DE, be added together; I say that the whole AE is even.
设任意多个偶数AB、BC、CD、DE相加，其和为AE。
For, since each of the numbers AB, BC, CD, DE is even, it has a half part; [VII. Def. 6] so that the whole AE also has a half part.
由于每个加数AB、BC、CD、DE均为偶数，故每个数都有半份。
因此，总和AE也有半份。
根据定义，有半份的数为偶数，故AE为偶数。
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