Congruent to Zero iff Modulo is Divisor

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Let $a, z \in \R$.

Then $a$ is congruent to $0$ modulo $z$ if and only if $a$ is an integer multiple of $z$.

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

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

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


\(\ds \exists k \in \Z: \, \) \(\ds a\) \(=\) \(\ds k z\)
\(\ds \leadstoandfrom \ \ \) \(\ds \exists k \in \Z: \, \) \(\ds a\) \(=\) \(\ds 0 + k z\)

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:

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

Hence the result for integer $z$.