Set of Integer Combinations equals Set of Multiples of GCD

Theorem
The set of all integer combinations of $$a$$ and $$b$$ is precisely the set of all integer multiples of the GCD of $$a$$ and $$b$$:

$$\gcd \left\{{a, b}\right\} \backslash c \iff \exists x, y \in \mathbb{Z}: c = x a + y b$$

Proof

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

Then $$d \backslash c \Longrightarrow \exists m \in \mathbb{Z}: c = m d$$.

So:

Thus $$\gcd \left\{{a, b}\right\} \backslash c \Longrightarrow \exists x, y \in \mathbb{Z}: c = x a + y b$$.


 * Now suppose $$\exists x, y \in \mathbb{Z}: c = x a + y b$$.

From Common Divisor Divides Integer Combination, we have $$\gcd \left\{{a, b}\right\} \backslash \left({x a + y b}\right)$$.

It follows directly that $$\gcd \left\{{a, b}\right\} \backslash c$$ and the proof is finished.