Let a and b be natural numbers and LCM(a,b) = m.(adsbygoogle = window.adsbygoogle || []).push({});

Prove that if GCD(a,b) = 1, then LCM(a,b) = ab.

What I got so far was that since b|LCM(a,b), then b*k = LCM(a,b) for some natural number k. I know I need to show that k </= a and k >/= a so I get a = k. And show that b*k = ba = LCM(a,b).

I can get a </= k:

Since LCM(a,b) = m, then a|m and b|m. Since LCM(a,b) = m = b*k, then a|bk. <And by the Euclid's Lemma, if a|bc and GCD(a,b) = 1, then a|c>. So

a|k. So a*j = k for some natural number j, and therefore a </= k.

I'm sure I have to derive that a >/= k by the antecedent. I'm trying to use the property that GCD(a,b) = 1 = a*x + b*y for some integers x and y by showing that k|a. But I'm stuck.

Any suggestions? Thanks

Edit: Maybe I should have posted this in Number Theory. Mod, can you move it if you think it should be there?

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# Suggestion for proof

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