Let a and b be natural numbers and LCM(a,b) = m.
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...