Product of GCD and LCM/Proof 1

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$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$


$\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$
$\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$.


It is sufficient to prove that $\lcm \set {a, b} \times \gcd \set {a, b} = a b$, where $a, b \in \Z_{>0}$.

\(\ds d\) \(=\) \(\ds \gcd \set {a, b}\)
\(\ds \leadsto \ \ \) \(\ds d\) \(\divides\) \(\ds a b\)
\(\ds \leadsto \ \ \) \(\ds \exists n \in \Z_{>0}: \, \) \(\ds a b\) \(=\) \(\ds d n\)

\(\ds d \divides a\) \(\land\) \(\ds d \divides b\)
\(\ds \leadsto \ \ \) \(\ds \exists u, v \in \Z: \, \) \(\ds a = d u\) \(\land\) \(\ds b = d v\)
\(\ds \leadsto \ \ \) \(\ds d u b = d n\) \(\land\) \(\ds a d v = d n\)
\(\ds \leadsto \ \ \) \(\ds n = b u\) \(\land\) \(\ds n = a v\)
\(\ds \leadsto \ \ \) \(\ds a \divides n\) \(\land\) \(\ds b \divides n\)

Now we have:

$a \divides m \land b \divides m \implies m = a r = b s$

Also, by Bézout's Identity we have:

$d = a x + b y$


\(\ds m d\) \(=\) \(\ds a x m + b y m\)
\(\ds \) \(=\) \(\ds b s a x + a r b y\)
\(\ds \) \(=\) \(\ds a b \paren {s x + r y}\)
\(\ds \) \(=\) \(\ds d n \paren {s x + r y}\)


$m = n \paren {s x + r y}$


$n \divides m \implies n \le \size m$


$a b = d n = \gcd \set {a, b} \times \lcm \set {a, b}$

as required.