# Integers Divided by GCD are Coprime

## Theorem

Let $a, b \in \Z$ be integers which are not both zero.

Let $d$ be a common divisor of $a$ and $b$, that is:

- $\dfrac a d, \dfrac b d \in \Z$

Then:

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

- $\gcd \set {\dfrac a d, \dfrac b d} = 1$

that is:

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

where:

- $\gcd$ denotes greatest common divisor
- $\perp$ denotes coprimality.

## Proof 1

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 $t$ |

$\blacksquare$

## Proof 2

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

We have:

- $(1): d \divides a \iff \exists s \in \Z: a = d s$

- $(2): d \divides b \iff \exists t \in \Z: b = d t$

We have to prove:

- $\gcd \set {s, t} = 1$

Aiming for a contradiction, suppose $\gcd \set {s, t} \ne 1$.

So:

- $(3): \exists k \in \N \setminus \set 1$ such that $k \divides s \land k \divides t$

So:

- $(4): \exists m, n \in \N: s = k m, t = k n$

Substituting from $(4)$ in $(1)$ and $(2)$:

- $a = d k m$, $b = d k n$

Therefore:

- $ d k \divides a \land d k \divides b$

From $(3)$ we have:

\(\ds \) | \(\) | \(\ds k \in \N \land k \ne 1\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\ds k > 1\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\ds d k > d\) |

As $d k$ is a common divisor of $a$ and $b$ greater than $d$, this contradicts $d = \gcd \set {a, b}$.

So our initial assumption that $\gcd \set {s, t} \ne 1$ is false.

Therefore, from Proof by Contradiction, we have:

- $\gcd \set {s, t} = 1 \implies \gcd \set {\dfrac a d, \dfrac b d} = 1$

$\blacksquare$

## Proof 3

Because $d$ is a common divisor of $a$ and $b$, we may form the expressions:

- $a = d r$
- $b = d s$

where $r, s \in \Z$.

Then:

\(\ds d\) | \(=\) | \(\ds \gcd \set {a, b}\) | by hypothesis | |||||||||||

\(\ds \) | \(=\) | \(\ds \gcd \set {d r, d s}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds d \gcd \set {r, s}\) | GCD of Integers with Common Divisor | |||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds 1\) | \(=\) | \(\ds \gcd \set {r, s}\) | dividing through by $d$ | ||||||||||

\(\ds \) | \(=\) | \(\ds \gcd \set {\dfrac a d, \dfrac b d}\) | Definition of $r$ and $s$ |

$\blacksquare$

## 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

- 1980: David M. Burton:
*Elementary Number Theory*(revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm: Problems $2.3$: $3$