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 1
Let $d := \gcd(a, b)$. Then $\exists j_1, j_2 \in \Z$ such that $a = dj_1$ and $b = dj_2$.

Because $d$ divides both $a$ and $b$, it must divide their product. So, $\exists l \in \Z$ such that $ab = dl$.

Then we have $dl = (dj_1)b = a(dj_2) \implies l = j_1b = aj_2$.

By definition of divisibility, $a | l$ and $b | l$. That is, $l$ is a common divisor of $a$ and $b$.

Now it must be shown that $l$ is the least such number.

Let $m$ be any multiple of $a$ and $b$.

Then, $\exists k_1, k_2 \in \Z$ such that $m = ak_1 = bk_2$.

By Bézout's Lemma $\exists x, y \in \Z$ such that $d = a x + b y$.

So:

Thus $m = l(bk_2 + ak_1)$, which implies $l | m$.

But then $l \le |m|$, meaning $l$ is the least such multiple.

In conclusion, $ab = dl = (\gcd(a, b))(\operatorname{lcm}(a, b))$.