Product of GCD and LCM/Proof 1

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

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

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

So:

So $m = n \left({s x + r y}\right)$.

Thus $n \mathrel \backslash m \implies n \le \left|{m}\right|$, while $a b = d n = \gcd \left\{{a, b}\right\} \times \operatorname{lcm} \left\{{a, b}\right\}$ as required.