Integers Divided by GCD are Coprime/Proof 1

Theorem

Any pair of integers, not both zero, can be reduced to a pair of coprime ones by dividing them by their GCD:

$\gcd \set {a, b} = d \iff \dfrac a d, \dfrac b d \in \Z \land \gcd \set {\dfrac a d, \dfrac b d} = 1$

That is:

$\dfrac a {\gcd \set {a, b} } \perp \dfrac b {\gcd \set {a, b} }$

Proof

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

By definition of divisor:

$d \divides a \iff \exists s \in \Z: a = d s$
$d \divides b \iff \exists t \in \Z: b = d t$

So:

 $\ds \exists m, n \in \Z: \,$ $\ds d$ $=$ $\ds m a + n b$ Bézout's Identity $\ds \leadstoandfrom \ \$ $\ds d$ $=$ $\ds m d s + n d t$ Definition of $s$ and $t$ $\ds \leadstoandfrom \ \$ $\ds 1$ $=$ $\ds m s + n t$ dividing through by $d$ $\ds \leadstoandfrom \ \$ $\ds \gcd \set {s, t}$ $=$ $\ds 1$ Bézout's Identity $\ds \leadstoandfrom \ \$ $\ds \gcd \set {\frac a d, \frac b d}$ $=$ $\ds 1$ Definition of $s$ and $t$

$\blacksquare$