Integers Divided by GCD are Coprime/Proof 3

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:

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

if and only if:

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

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

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 $s$

$\blacksquare$