To find the least number which two given numbers measure.
Let A and B be the two given numbers. It is required to find the least number which they measure. Now either A and B are relatively prime or they are not.
First, let A and B be relatively prime. Multiply A by B to make C. Then B multiplied by A makes C. Therefore A and B measure C.
I say next that it is also the least number they measure. If not, then A and B measure some number D less than C
Let there be as many units in E as the times that A measures D, and as many units in F as the times that B measures D.
Then A multiplied by E makes D, and B multiplied by F makes D. Therefore the product of A and E equals the product of B and F. Therefore A is to B as F is to E.		VII.Def.15 VII.19
But A and B are relatively prime, primes are also 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, therefore B measures E as the consequent the consequent.		VII.21 VII.20
And, since A multiplied by B and by E makes C and D, therefore B is to E as C is to D. But B measures E, therefore C also measures D, the greater the less, which is impossible.		VII.17
Therefore A and B do not measure any number less than C. Therefore C is the least that is measured by A and B.
Next, let A and B not be relatively prime. Take F and E, the least numbers of those which have the same ratio with A and B. Therefore the product of A and E equals the product of B and F.		VII.33 VII.19
Multiply A by E to make C. Then B multiplied by F makes C. Therefore A and B measure C. I say next that it is also the least number that they measure. If not, then A and B measure some number D less than C.
Let there be as many units in G as the times that A measures D, and as many units in H as the times that B measures D.
Then A multiplied by G makes D, and B multiplied by H makes D. Therefore the product of A and G equals the product of B and H. Therefore A is to B as H is to G.		VII.19
But A is to B as F is to E. Therefore F is to E as H is to G.		(V.11)
But F and 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, therefore E measures G.		VII.20
And, since A multiplied by E and by G makes C and D, therefore E is to G as C is to D.		VII.17
But E measures G, therefore C also measures D, the greater the less, which is impossible. Therefore A and B do not measure any number less than C. Therefore C is the least that is measured by A and B.
Q.E.D.

Summary of the proof

Case 1. Suppose a and b are relatively prime. An indirect proof shows that their least common multiple is their product ab. If not, then there is a smaller number d which both a and b divide. Since a:b = (d/b):(d/a), and a:b is in lowest terms (since a and b are relatively prime), therefore b divides d/a. Also, b:(d/a) = ab:d, so ab divides d, but d is smaller than ab, a contradiction. Thus, when a and b are relatively prime, their least common multiple is their product.

Case 2. Suppose a and b are not relatively prime. Reduce the ratio a:b to its lowest terms f:e using the previous proposition VII.33. Then ae = bf. Let c denote this product. (Note that f = a/GCD(a, b), and e = b/GCD(a, b), so c = ab/GCD(a, b).) Both a and b divide c, therefore c is a common multiple of a and b. Suppose that it's not the least common multiple. Then there is a smaller number d which both a and b divide. Now

and f:e is in lowest terms, therfore e divides d/a. But e:(d/a) = ae:d, therefore ae also divides d. But c = ae, and d is less than c, a contradiction. Thus, LCM(a, b) = ab/LCM(a, b).

Use of this proposition

© 1996, 2002
D.E.Joyce
Clark University