Product of GCD and LCM/Proof 3

Proof
Let $d := \gcd\left\{{a, b}\right\}$.

Then by definition of the GCD, there exist $j_1, j_2 \in \Z$ such that $a = dj_1$ and $b = dj_2$.

Because $d$ divides both $a$ and $b$, it must divide their product:


 * $\exists l \in \Z$ such that $a b = d l$

Then we have:

showing that $a \mid l$ and $b \mid l$.

That is, $l$ is a common multiple of $a$ and $b$.

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

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

Then there exist $k_1, k_2 \in \Z$ such that $m = a k_1 = b k_2$.

By Bézout's Lemma:


 * $\exists x, y \in \Z: d = a x + b y$

So:

Thus $m = l \left({b k_2 + a k_1}\right)$, that is, $l \mid m$.

Hence by definition of the LCM, $\operatorname{lcm}\left\{{a, b}\right\} = l$.

In conclusion:


 * $a b = d l = \gcd \left\{{a, b}\right\} \cdot \operatorname{lcm} \left\{{a, b}\right\}$