Divisor of One of Coprime Numbers is Coprime to Other/Proof 1

Theorem
Let $a, b \in \N$ be numbers such that $a$ and $b$ are coprime:
 * $a \perp b$

Let $c > 1$ be a divisor of $a$:
 * $c \mathop \backslash a$

Then $c$ is not a divisor of $b$:
 * $c \nmid b$

Proof
Let $a \perp b$ and $c > 1: c \mathop \backslash a$.

Suppose $c \not \perp b$.

So by definition of not coprime:


 * $\exists d > 1: d \mathop \backslash c, d \mathop \backslash b$.

But from Divides is Transitive:
 * $d \mathop \backslash c, c \mathop \backslash a \implies d \mathop \backslash a$

So $d$ is a common divisor of both $a$ and $b$.

So, by definition, $a$ and $b$ are not coprime.

The result follows by Proof by Contradiction.