第5卷命题 13 · 比例传递性定理

For let a first magnitude A have to a second B the same ratio as a third C has to a fourth D, and let the third C have to the fourth D a greater ratio than a fifth E has to a sixth F; I say that the first A will also have to the second B a greater ratio than the fifth E to the sixth F. For, since there are some equimultiples of C, E, and of D, F other, chance, equimultiples, such that the multiple of C is in excess of the multiple of D, while the multiple of E is not in excess of the multiple of F, [V. Def. 7] let them be taken, and let G, H be equimultiples of C, E, and K, L other, chance, equimultiples of D, F, so that G is in excess of K, but H is not in excess of L; and, whatever multiple G is of C, let M be also that multiple of A, and, whatever multiple K is of D, let N be also that multiple of B.

设第一量A对第二量B之比等于第三量C对第四量D之比，且C对D之比大于E对F之比。

Now, since, as A is to B, so is C to D, and of A, C equimultiples M, G have been taken, and of B, D other, chance, equimultiples N, K, therefore, if M is in excess of N, G is also in excess of K, if equal, equal, and if less, less.

由定义7，存在C、E的等倍数G、H，以及D、F的等倍数K、L，使得G大于K，但H不大于L。

[V. Def. 5] But G is in excess of K; therefore M is also in excess of N.

取M为A的与G同倍数，N为B的与K同倍数。由定义5，因A:B=C:D，故M大于N当且仅当G大于K；已知G大于K，故M大于N。

But H is not in excess of L; and M, H are equimultiples of A, E, and N, L other, chance, equimultiples of B, F; therefore A has to B a greater ratio than E has to F.

但H不大于L，且M、H是A、E的等倍数，N、L是B、F的等倍数，由定义7，A对B之比大于E对F之比。