Existence of Greatest Common Divisor/Proof 2

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Let $a, b \in \Z$ be integers such that $a \ne 0$ or $b \ne 0$.

Then the greatest common divisor of $a$ and $b$ exists.


By definition of greatest common divisor, we aim to show that there exists $c \in \Z_{>0}$ such that:

\(\ds c\) \(\divides\) \(\ds a\)
\(\ds c\) \(\divides\) \(\ds b\)


$d \divides a, d \divides b \implies d \divides c$

Consider the set $S$:

$S = \set {s \in \Z_0: \exists x, y \in \Z: s = a x + b y}$

$S$ is not empty, because by setting $x = 1$ and $y = 0$ we have at least that $a \in S$.

From the Well-Ordering Principle, there exists a smallest $c \in S$.

So, by definition, we have $c > 0$ is the smallest such that $c = a x + b y$ for some $x, y \in \Z$.

Let $d$ be such that $d \divides a$ and $d \divides b$.

Then from Common Divisor Divides Integer Combination:

$d \divides a x + b y$

That is:

$d \divides c$

We have that:

\(\ds \exists t, u \in \Z: \, \) \(\ds a\) \(=\) \(\ds c t + u:\) \(\ds 0 \le u < c\) Division Theorem
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds a x t + b y t + u\) Definition of $c$
\(\ds \leadsto \ \ \) \(\ds u\) \(=\) \(\ds r \paren {1 - x t} + b \paren {-y t}\) rearranging
\(\ds \leadsto \ \ \) \(\ds u\) \(=\) \(\ds 0\) as $u < c$ and the definition of $c$
\(\ds \leadsto \ \ \) \(\ds c\) \(\divides\) \(\ds a\) Definition of Divisor of Integer

Mutatis mutandis:

$c \divides b$

Now suppose $c'$ is such that:

\(\ds c'\) \(\divides\) \(\ds a\)
\(\ds c'\) \(\divides\) \(\ds b\)


$d \divides a, d \divides b \implies d \divides c'$

Then we have immediately that:

$c' \divides c$

and by the same coin: $c \divides c'$

and so:

$c = c'$

demonstrating that $c$ is unique.