Product of GCD and LCM

Theorem

 * $\operatorname{lcm} \left\{{a, b}\right\} \times \gcd \left\{{a, b}\right\} = \left|{a b}\right|$

where:
 * $\operatorname{lcm} \left\{{a, b}\right\}$ is the lowest common multiple of $a$ and $b$
 * $\gcd \left\{{a, b}\right\}$ is the greatest common divisor of $a$ and $b$.

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 $a \backslash m \land b \backslash m \implies m = a r = b s$.

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

So:

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

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