Product of GCD and LCM/Proof 2

Proof
Let $a = g m$ and $b = g n$, where $g = \gcd \set {a, b}$ and $m$ and $n$ are coprime.

The existence of $m$ and $n$ are proved by Integers Divided by GCD are Coprime.

It follows that $a = g m \divides g m n$ and $b = g n \divides g m n$.

So $g m n$ is a common multiple of $a$ and $b$.

Let's assume there is a number $g k \leq g m n$ that is divisible by both $a$ and $b$

$a = g m, b = g n \implies gm , gn \divides g k \implies m , n \divides k$

As $g k \leq g m n$, $k \leq m n$. But $m, n$ are relatively prime so $LCM(m, n) = m n \implies k \not \lt m n \implies k = m n$

So $g k = g m n$ = $LCM(a, b)$

Then it follows that:
 * $\lcm \set {a, b} \times \gcd \set {a, b} = g m n \times g = g m \times g n = \size {a b}$