Congruent to Zero if Modulo is Divisor

Theorem

Let $a, z \in \R$.

Then $a$ is congruent to $0$ modulo $z$ iff $a$ is an integral multiple of $z$.

$\exists k \in \Z: k z = a \iff a \equiv 0 \pmod z$

If $z \in \Z$, then further:

$z \backslash a \iff a \equiv 0 \pmod z$

Proof

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \exists k \in \Z: a\)	\(=\)	\(\displaystyle k z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle \exists k \in \Z: a\)	\(=\)	\(\displaystyle 0 + k z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Thus by definition of congruence, $a \equiv 0 \pmod z$ and the result is proved.

If $z$ is an integer, then by definition of divisor:

$z \backslash a \iff \exists k \in \Z: a = k z$

Hence the result for integral $z$.

$\blacksquare$