$\operatorname{lcm}(a,b)=ab$ if and only if $\gcd(a,b)=1$

Question

I have a quick question about (https://math.stackexchange.com/q/921109)'s answer. I asked on there but since the question is old, I feel like I might not get a reply. Anyway, my question was: Why is $\operatorname{lcm}(a,b)\leq drs$ like they said?

Answer 1

In this statement it is assumed that gcd(a,b) = d, and d > 1. This means that the greatest common divisor between a and b is greater than 1, therefore they are not relatively prime (they share a factor, namely d).

Since we know that d divides both a and b, we can re-write a and b as

a = d⋅r, b = d⋅s for some integers r and s.

Considering our statement earlier, that a and b share a common factor d, we know that the least common multiple [lcm(a,b)] of a and b is some multiple of d, because both factors share a multiple d. The worst case scenario for the lcm(a,b) is going to be when lcm(a,b) = d⋅r⋅s because this will always be a factor of our two numbers a and b which share a common multiple.

Since we are talking about the LEAST common multiple, there can also be other common factors between them, but only if they are LESS than d⋅r⋅s. Hence the ≤.

We also know that since our lcm(a,b) is at most d⋅r⋅s, and d > 1, that d⋅(d⋅r⋅s) should surely be greater than d⋅r⋅s. Therefore lcm(a,b) ≤ d⋅r⋅s < d⋅(d⋅r⋅s) which we know is equal to a⋅b