Congruent to Zero iff Modulo is Divisor
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Theorem
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:
- $z \divides a \iff a \equiv 0 \pmod z$
Proof
\(\ds \exists k \in \Z: \, \) | \(\ds a\) | \(=\) | \(\ds k z\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists k \in \Z: \, \) | \(\ds a\) | \(=\) | \(\ds 0 + k z\) |
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 $z$.
$\blacksquare$