Integers Divided by GCD are Coprime

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

We have:

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

So:

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

$\blacksquare$

Also presented as

It can be expressed so as not to include fractions:

$\gcd \set {a, b} = d \iff \exists s, t \in \Z: a = d s \land b = d t \land \gcd \set {s, t} = 1$