Product of GCD and LCM/Proof 3

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

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

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 $a \divides l$ and $b \divides l$.

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 Identity:


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

So:

Thus:
 * $m = l \paren {b k_2 + a k_1}$

that is, $l \divides m$.

Hence by definition of the LCM:
 * $\lcm \set {a, b} = l$

In conclusion:


 * $a b = d l = \gcd \set {a, b} \cdot \lcm \set {a, b}$