Prime not Divisor implies Coprime/Proof 2

Theorem

Let $p, a \in \Z$.

If $p$ is a prime number then:

$p \nmid a \implies p \perp a$

where:

$p \nmid a$ denotes that $p$ does not divide $a$

$p \perp a$ denotes that $p$ and $a$ are coprime.

It follows directly that if $p$ and $q$ are primes, then:

$p \divides q \implies p = q$

$p \ne q \implies p \perp q$

In the words of Euclid:

Any prime number is prime to any number which it does not measure.

Let $p$ be a prime number.

Let $a \in \Z$ be such that $p$ is not a divisor of $a$.

Aiming for a contradiction, suppose $p$ and $a$ are not coprime.

Then:

$\exists c \in \Z_{>1}: c \divides p, c \divides a$

where $\divides$ denotes divisibility.

But then by definition of prime:

$c = p$

Thus:

$p \divides a$

The result follows by Proof by Contradiction.

$\blacksquare$