Integers Divided by GCD are Coprime/Proof 3
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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
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 $s$ |
$\blacksquare$