Prime not Divisor implies Coprime

Theorem
Let $$p, a \in \Z$$.

If $$p$$ is a prime number and $$p \nmid a$$, then $$p \perp a$$.

It follows directly that if $$p$$ and $$q$$ are primes, then:
 * $$p \backslash q \Longrightarrow p = q$$;
 * $$p \ne q \Longrightarrow p \perp q$$.

Proof
Let $$p \in \mathbb{P}, p \nmid a$$. We need to show that $$\gcd \left\{{a, p}\right\} = 1$$.

Let $$\gcd \left\{{a, p}\right\} = d$$.

As $$d \backslash p$$, we must have $$d = 1$$ or $$d = p$$ by GCD with Prime.

But if $$d = p$$, then $$p \backslash a$$ by definition of greatest common divisor.

So $$d \ne p$$ and therefore $$d = 1$$.