# Prime not Divisor implies Coprime

## Theorem

Let $p, a \in \Z$.

If $p$ is a prime number then:

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

where:

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

- $p \divides q \implies p = q$
- $p \ne q \implies p \perp q$.

## Proof 1

Let $p \in \Bbb P, p \nmid a$.

We need to show that $\gcd \left\{{a, p}\right\} = 1$.

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

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

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

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

$\blacksquare$

## Proof 2

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:

- $\exists c \in \Z_{>1}: c \mathrel \backslash p, c \mathrel \backslash a$

where $\backslash$ denotes divisibility.

But then by definition of prime:

- $c = p$

Thus:

- $p \mathrel \backslash a$

The result follows by Proof by Contradiction.

$\blacksquare$

## Sources

- 1971: George E. Andrews:
*Number Theory*... (previous) ... (next): $\text {2-2}$ Divisibility: Example $\text {2-11}$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 12.1$: Highest common factors and Euclid's algorithm