Congruence of Product

Theorem
Let $$a, b, z \in \R$$.

Let $$a$$ be congruent to $b$ modulo $z$, i.e. $$a \equiv b \left({\bmod\, z}\right)$$.

Then:
 * $$\forall m \in \Z: m a \equiv m b \left({\bmod\, z}\right)$$

Proof
Let $$m \in \Z$$ and $$a \equiv b \left({\bmod\, z}\right)$$.

Suppose $$m = 0$$. Then the RHS of the assertion degenerates to $$0 \equiv 0 \left({\bmod\, z}\right)$$ which is trivially true.

Otherwise, from Congruence by Product of Modulo, we have:
 * $$a \equiv b \left({\bmod\, z}\right) \iff m a \equiv m b \left({\bmod\, m z}\right)$$

As $$m \in \Z$$, it follows that $$m z$$ is an integral multiple of $$z$$.

Hence from Congruence by Divisor of Modulus, it follows that:
 * $$m a \equiv m b \left({\bmod\, m z}\right) \implies m a \equiv m b \left({\bmod\, z}\right)$$