Congruence of Product

Theorem

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

Let $a$ be congruent to $b$ modulo $z$, i.e. $a \equiv b \pmod z$.

Then:

$\forall m \in \Z: m a \equiv m b \pmod z$

Proof

Let $m \in \Z$ and $a \equiv b \pmod z$.

Suppose $m = 0$. Then the RHS of the assertion degenerates to $0 \equiv 0 \pmod z$ which is trivially true.

Otherwise, from Congruence by Product of Modulo, we have:

$a \equiv b \iff m a \equiv m b \pmod z$

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:

$m a \equiv m b \implies m a \equiv m b \pmod z$

$\blacksquare$

Sources

• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 14.4 \ \text{(i)}$