第8卷命题 4 · 求给定比的最小连比例数

Let the given ratios in least numbers be that of A to B, that of C to D, and that of E to F; thus it is required to find numbers in continued proportion which are the least that are in the ratio of A to B, in the ratio of C to D, and in the ratio of E to F. Let G, the least number measured by B, C, be taken. [VII. 34] And, as many times as B measures G, so many times also let A measure H, and, as many times as C measures G, so many times also let D measure K. Now E either measures or does not measure K. First, let it measure it. And, as many times as E measures K, so many times let F measure L also. Now, since A measures H the same number of times that B measures G, therefore, as A is to B, so is H to G. [VII. Def. 20, VII. 13] For the same reason also, as C is to D, so is G to K, and further, as E is to F, so is K to L; therefore H, G, K, L are continuously proportional in the ratio of A to B, in the ratio of C to D, and in the ratio of E to F.

设最小数比A:B、C:D、E:F，取B和C的最小公倍数G，令H = (G/B)×A，K = (G/C)×D。

I say next that they are also the least that have this property. For, if H, G, K, L are not the least numbers continuously proportional in the ratios of A to B, of C to D, and of E to F, let them be N, O, M, P. Then since, as A is to B, so is N to O, while A, B are least, 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; therefore B measures O. [VII. 20] For the same reason C also measures O; therefore B, C measure O; therefore the least number measured by B, C will also measure O. [VII. 35] But G is the least number measured by B, C; therefore G measures O, the greater the less: which is impossible. Therefore there will be no numbers less than H, G, K, L which are continuously in the ratio of A to B, of C to D, and of E to F. Next, let E not measure K. Let M, the least number measured by E, K, be taken.

若E量尽K，则令L = (K/E)×F，则H、G、K、L成连比例，且比为A:B、C:D、E:F。假设存在更小的连比例数N、O、M、P，则B和C都量尽O，故G量尽O，矛盾。

And, as many times as K measures M, so many times let H, G measure N, O respectively, and, as many times as E measures M, so many times let F measure P also. Since H measures N the same number of times that G measures O, therefore, as H is to G, so is N to O. [VII. 13 and Def. 20] But, as H is to G, so is A to B; therefore also, as A is to B, so is N to O. For the same reason also, as C is to D, so is O to M. Again, since E measures M the same number of times that F measures P, therefore, as E is to F, so is M to P; [VII. 13 and Def. 20] therefore N, O, M, P are continuously proportional in the ratios of A to B, of C to D, and of E to F. I say next that they are also the least that are in the ratios A : B, C : D, E : F. For, if not, there will be some numbers less than N, O, M, P continuously proportional in the ratios A : B, C : D, E : F. Let them be Q, R, S, T.

若E不量尽K，取E和K的最小公倍数M，令N = (M/K)×H，O = (M/K)×G，P = (M/E)×F，则N、O、M、P成连比例。

Now since, as Q is to R, so is A to B, while A, B are least, and the least numbers measure those which have the same ratio with them the same number of times, the antecedent the antecedent and the consequent the consequent, [VII. 20] therefore B measures R. For the same reason C also measures R; therefore B, C measure R. Therefore the least number measured by B, C will also measure R. [VII. 35] But G is the least number measured by B, C; therefore G measures R. And, as G is to R, so is K to S: [VII. 13] therefore K also measures S. But E also measures S; therefore E, K measure S. Therefore the least number measured by E, K will also measure S. [VII. 35] But M is the least number measured by E, K; therefore M measures S, the greater the less: which is impossible.

假设存在更小的连比例数Q、R、S、T，则B和C都量尽R，故G量尽R；又K量尽S，E量尽S，故M量尽S，矛盾。