# Product of GCD and LCM/Proof 2

## Theorem

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

where:

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

## Proof

Let $a = g m$ and $b = g n$, where $g = \gcd \set {a, b}$ and $m$ and $n$ are coprime.

The existence of $m$ and $n$ are proved by Integers Divided by GCD are Coprime.

It follows that:

$a = g m \divides g m n$

and:

$b = g n \divides g m n$

So $g m n$ is a common multiple of $a$ and $b$.

Hence there exists an integer $g k \le g m n$ that is divisible by both $a$ and $b$.

Then:

 $\ds a$ $=$ $\ds g m$ $\ds \leadsto \ \$ $\ds g m$ $\divides$ $\ds g k$ $\ds \leadsto \ \$ $\ds m$ $\divides$ $\ds k$

 $\ds b$ $=$ $\ds g n$ $\ds \leadsto \ \$ $\ds g n$ $\divides$ $\ds g k$ $\ds \leadsto \ \$ $\ds n$ $\divides$ $\ds k$

As $g k \le g m n$, it follows that:

$k \le m n$

But $m, n$ are coprime.

So:

 $\ds \map \lcm {m, n}$ $=$ $\ds m n$ $\ds \leadsto \ \$ $\ds k$ $\not <$ $\ds m n$ $\ds \leadsto \ \$ $\ds k$ $=$ $\ds m n$ $\ds \leadsto \ \$ $\ds g k$ $=$ $\ds g m n$ $\ds$ $=$ $\ds \map \lcm {a, b}$ $\ds \leadsto \ \$ $\ds \lcm \set {a, b} \times \gcd \set {a, b}$ $=$ $\ds g m n \times g$ $\ds$ $=$ $\ds g m \times g n$ $\ds$ $=$ $\ds \size {a b}$

$\blacksquare$