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数学 / 几何原本 / Proposition VIII.1

第8卷命题 1 · 连比例中互质两端数为最小

If there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, the numbers are the least of those which have the same ratio with them.

若任意多个数成连比例,且其两端数互质,则这些数是具有相同比值的数中最小的。

A A_r B B_r C C_r D D_r E E_r F F_r G G_r H H_r
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分步证明Step-by-step proof
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  1. Let there be as many numbers as we please, A, B, C, D, in continued proportion, and let the extremes of them A, D be prime to one another; I say that A, B, C, D are the least of those which have the same ratio with them. For, if not, let E, F, G, H be less than A, B, C, D, and in the same ratio with them.

    设数A、B、C、D成连比例,且两端A与D互质。

  2. Now, since A, B, C, D are in the same ratio with E, F, G, H, and the multitude of the numbers A, B, C, D is equal to the multitude of the numbers E, F, G, H, therefore, ex aequali, as A is to D, so is E to H.

    假设存在更小的数E、F、G、H与A、B、C、D有相同比值。

  3. [VII. 14] But A, D are prime, primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent. [VII. 20] Therefore A measures E, the greater the less: which is impossible.

    由等比定理,A比D等于E比H;但A与D互质,故它们是最小的,且最小数能量尽同比值中的数。

  4. Therefore E, F, G, H which are less than A, B, C, D are not in the same ratio with them.

    因此A能量尽E,即大量能量尽小量,矛盾。故不存在更小的同比值数组。

本卷的比例和数列问题,可放回数论史长卷 math-meta/topic-number-theory-history 中看欧几里得数论的位置。