Product of GCD and LCM/Proof 3

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

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

Then we have:


 * $dl = (dj_1)b = a(dj_2)$

$\implies \; \; l = \; \; \; \; j_1b \thinspace = aj_2$

$\implies a | l$ 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 common 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, $ab = dl = (\gcd(a, b))\cdot(\operatorname{lcm}(a, b))$.