# Existence of Greatest Common Divisor/Proof 1

## Theorem

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.

## Proof

### Proof of Existence

This is proved in Greatest Common Divisor is at least $1$.

$\Box$

### Proof of there being a Largest

Without loss of generality, suppose $a \ne 0$.

First we note that from Absolute Value of Integer is not less than Divisors:

- $\forall c \in \Z: \forall a \in \Z_{>0}: c \divides a \implies c \le \size c \le \size a$

The same applies for $c \divides b$.

Now we have three different results depending on $a$ and $b$:

\(\ds a \ne 0 \land b \ne 0\) | \(\implies\) | \(\ds \gcd \set {a, b} \le \min \set {\size a, \size b}\) | ||||||||||||

\(\ds a = 0 \lor b = 0\) | \(\implies\) | \(\ds \gcd \set {a, b} = \max \set {\size a, \size b}\) | ||||||||||||

\(\ds a = b = 0\) | \(\implies\) | \(\ds \forall x \in \Z: x \divides a \land x \divides b\) |

So if $a$ and $b$ are *both* zero, then *any* $n \in \Z$ divides both, and there is no greatest common divisor.

This is why the proviso that $a \ne 0 \lor b \ne 0$.

So we have proved that common divisors exist and are bounded above.

Therefore, from Set of Integers Bounded Above by Integer has Greatest Element there is always a **greatest** common divisor.

$\blacksquare$