# Integers Divided by GCD are Coprime

## Contents

## 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:

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

## Sources

- 1971: George E. Andrews:
*Number Theory*... (previous) ... (next): $\text {2-2}$ Divisibility: Example $\text {2-10}$