Integers Divided by GCD are Coprime/Proof 1

From ProofWiki
Jump to navigation Jump to search

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

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 $t$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Integers_Divided_by_GCD_are_Coprime/Proof_1&oldid=591281"