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Here is the theorem they prove:

$$|ab| = \operatorname{lcm}(a,b) \operatorname{gcd}(a,b)$$

where

  • $\operatorname{lcm}(a,b)$ denotes the lowest common multiple of $a$ and $b$.
  • $\operatorname{gcd}(a,b)$ denotes the greatest common divisor of $a$ and $b$.

Proof 1 is below (in case the original site changes). I have highlighted the end part that I don't understand in red.

It is sufficient to prove that $\operatorname{lcm}(a,b) \operatorname{gcd}(a,b) = ab$ where $a,b \in \mathbb{Z}_{>0}$.

$d = \operatorname{gcd}(a,b)$

$\implies d | ab$

$\exists n \in \mathbb{Z}_{>0}: ab = dn$

$d | a \land d | b$

$\implies \exists u,v \in \mathbb{Z}:a =du \land b = dv$

$\implies dub = dn \land adv = dn$

$\implies n=bu \land n = av$

$\implies a | n \land b | n$

Now we have $a|m \land b | m \implies m = ar=bs$.

Also, by Bézout's Lemma we have $d = ax + by$.

So: \begin{align}md &= axm + bym \\ &= bsax + arby \\ &= ab(sx + ry) \\ &= dn (sx + ry) \end{align}

So:

$m = n(sx + ry)$

Thus:

$n|m \implies n \leq |m|$

while:

$\color{red}{ab = dn = \operatorname{gcd}(a,b) \operatorname{lcm}(a,b)}$

as required.

$\blacksquare$

I get that we are meant to conclude that $n = \operatorname{lcm}$, but I don't yet see why the workup to this point entails that $n$ must be the least common multiple.

Galen
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    We know by "⟹a|n∧b|n" that $n$ is a common multiple. And we just proved that for any other common multiple, $m$, that $n \le m$. That means that by definition $n$ is the least common multiple. – fleablood Jul 02 '22 at 23:57
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    The proof shows $,n := ab/d = {\rm \color{#c00}L\color{#0a0}CM}(a,b),$ by first showing that $,n,$ is a $\rm\color{#0a0}{Common}$ multiple of $,a,b,,$ necessarily the $\rm\color{#c00}{Least}$ common multiple since it divides every common multiple $,m.\ $ See here in the linked dupe for a clearer presentation. – Bill Dubuque Jul 03 '22 at 05:28

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