Product of GCD and LCM/Proof 1

Proof
It is sufficient to prove that $\lcm \set {a, b} \times \gcd \set {a, b} = a b$, where $a, b \in \Z_{>0}$.

Now we have $a \divides m \land b \divides m \implies m = a r = b s$.

Also, by Bézout's Lemma we have $d = a x + b y$.

So:

So:
 * $m = n \paren {s x + r y}$

Thus:
 * $n \divides m \implies n \le \size m$

while:
 * $a b = d n = \gcd \set {a, b} \times \lcm \set {a, b}$

as required.