Given two numbers, to find the least number which they measure.
给定两个数，求它们所能量尽的最小数。
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Let A, B be the two given numbers; thus it is required to find the least number which they measure. Now A, B are either prime to one another or not. First, let A, B be prime to one another, and let A by multiplying B make C; therefore also B by multiplying A has made C. [VII. 16] Therefore A, B measure C I say next that it is also the least number they measure. For, if not, A, B will measure some number which is less than C. Let them measure D. Then, as many times as A measures D, so many units let there be in E, and, as many times as B measures D, so many units let there be in F; therefore A by multiplying E has made D, and B by multiplying F has made D; [VII. Def. 15] therefore the product of A, E is equal to the product of B, F.
设A、B两数互素，令A乘B得C，则A、B能量尽C。
Therefore, as A is to B, so is F E. [VII. 19] But A, B are prime, primes are also least, [VII. 21] and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20] therefore B measures E, as consequent consequent. And, since A by multiplying B, E has made C, D, therefore, as B is to E, so is C to D. [VII. 17] But B measures E; therefore C also measures D, the greater the less: which is impossible. Therefore A, B do not measure any number less than C; therefore C is the least that is measured by A, B. Next, let A, B not be prime to one another, and let F, E, the least numbers of those which have the same ratio with A, B, be taken; [VII. 33] therefore the product of A, E is equal to the product of B, F.
若A、B能量尽某小于C的数D，则存在E、F使A乘E得D、B乘F得D，从而A比B等于F比E。
[VII. 19] And let A by multiplying E make C; therefore also B by multiplying F has made C; therefore A, B measure C. I say next that it is also the least number that they measure. For, if not, A, B will measure some number which is less than C. Let them measure D. And, as many times as A measures D, so many units let there be in G, and, as many times as B measures D, so many units let there be in H. Therefore A by multiplying G has made D, and B by multiplying H has made D. Therefore the product of A, G is equal to the product of B, H; therefore, as A is to B, so is H to G.
因A、B互素且最小，故B量尽E，进而C量尽D，矛盾。所以C为最小公倍数。
[VII. 19] But, as A is to B, so is F to E. Therefore also, as F is to E, so is H to G. But F, E are least, and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20] therefore E measures G. And, since A by multiplying E, G has made C, D, therefore, as E is to G, so is C to D. [VII. 17] But E measures G; therefore C also measures D, the greater the less: which is impossible. Therefore A, B will not measure any number which is less than C.
若A、B不互素，取与它们同比的最小数F、E，令A乘E得C，则A、B能量尽C。同理假设有更小的D，导出矛盾，故C为最小公倍数。