# Congruence by Divisor of Modulus

## Theorem

Let $z \in \R$ be a real number.

Let $a, b \in \R$ such that $a$ is congruent modulo $z$ to $b$, that is:

$a \equiv b \pmod z$

Let $m \in \R$ such that $z$ is an integer multiple of $m$:

$\exists k \in \Z: z = k m$

Then:

$a \equiv b \pmod m$

### Integer Modulus

When $z$ is an integer, and therefore a composite number such that $z = r s$, this result can be expressed as:

Let $r, s \in \Z$ be integers.

Let $a, b \in \Z$ such that $a$ is congruent modulo $r s$ to $b$, that is:

$a \equiv b \pmod {r s}$

Then:

$a \equiv b \pmod r$

and:

$a \equiv b \pmod s$

## Proof

We are given that $\exists k \in \Z: z = k m$.

Thus:

 $\displaystyle a$ $\equiv$ $\displaystyle b$ $\displaystyle \pmod z$ $\displaystyle \implies \ \$ $\displaystyle \exists k' \in \Z: a$ $=$ $\displaystyle b + k' z$ Definition of Congruence $\displaystyle \implies \ \$ $\displaystyle a$ $=$ $\displaystyle b + k' k m$ $\displaystyle a$ $\equiv$ $\displaystyle b$ $\displaystyle \pmod m$ Definition of congruence: $k' k$ is an integer

$\blacksquare$