Existence of Greatest Common Divisor/Proof 1

Proof of Existence
From One Divides all Integers:


 * $\forall a, b \in \Z: 1 \divides a \land 1 \divides b$

so $1$ is always a common divisor of any two integers.

Proof of there being a Largest
As the definition of $\gcd$ shows that it is symmetric,, 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 $b$:

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.