Product of GCD and LCM/Proof 5

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

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

Let $m = \dfrac {a b} d$.

Then:
 * $m = a s = r b$

which makes $m$ a common multiple of $a$ and $b$.

Let $c \in \Z_{>0}$ be a common multiple of $a$ and $b$.

Let us say that:
 * $c = a u = b v$

From Bézout's Identity:
 * $\exists x, y \in \Z: d = a x + b y$

Then:

That is:
 * $m \divides c$

where $\divides$ denotes divisibility.

So by Absolute Value of Integer is not less than Divisors:
 * $m \le c$

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

In conclusion:


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

and the result follows.