Euclid's Lemma/Proof 2

Theorem
Let $a, b, c \in \Z$.

Let $a \mathop \backslash b c$, where $\backslash$ denotes divisibility.

Let $a \perp b$, where $\perp$ denotes relative primeness.

Then $a \mathop \backslash c$.

Proof
Let $a, b, c \in \Z$.

We have that $a \perp b$.

That is:
 * $\gcd \left\{{a, b}\right\} = 1$

where $\gcd$ denotes greatest common divisor.

From Bézout's Lemma, we may write:
 * $a x + b y = 1$

for some $x, y \in \Z$.

Upon multiplication by $c$, we see that:
 * $c = c \left({a x + b y}\right) = c a x + c b y$

Now note that $c a x + c b y$ is an integer combination of $a c$ and $b c$.

So, since:
 * $a \mathop \backslash a c$

and:
 * $a \mathop \backslash b c$

it follows from Common Divisor Divides Integer Combination that:
 * $a \mathop \backslash \left({c a x + c b y}\right)$

However:
 * $c a x + c b y = c \left({a x + b y}\right) = c \cdot 1 = c$

Therefore:
 * $a \mathop \backslash c$

Also see

 * Euclid's Lemma for Prime Divisors