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I've seen similar posts here asking why $(ab)\bmod m = (a \bmod m)( b \bmod m)$, but none of them really answers my question or gives a detailed the mathematical proof.

I know that $a\equiv b \pmod m$, $c\equiv d \pmod m)$ implies $ac\equiv bd \pmod m$, and I was going to approach this by setting $a = k_am + a \pmod m$ and $b = k_bm + b \pmod m$ then multiply $a$ and $b$, but I got stuck.

Bernard
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PTN
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    By said congruence product rule, and $,(a^n\bmod m)\equiv a^n\pmod m,,$ you know they are congruent, say $y \equiv y'\pmod{m}.,$ So it suffces to show that this implies $,(y\bmod m) = (y'\bmod m),,$ i.e. congruent integers have the same remainder. Do you see how to do that? See here for another example. – Bill Dubuque Apr 04 '19 at 17:57

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It is a definition, you want define a product on $\mathbb{Z}/m\mathbb{Z}$ but you have only a product on $\mathbb{Z}$. A way to define product on $\mathbb{Z}/m\mathbb{Z}$ it is to prove that

if $a\cong c$ mod $m $ and $b\cong d$ mod m then

$ab\cong cd $ mod $m$ so it is well define the following operation on $\mathbb{Z}/m\mathbb{Z}$ :

($a$ mod $m$)($b$ mod $m$):= $ab$ mod $m$$

Federico Fallucca
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