Show if $a \equiv b \pmod{m}$ then $ac \equiv bc\pmod{mc}$ without restrictions?

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For this question here: Show if $a \equiv b \pmod{m}$ then $ac \equiv bc\pmod{mc}$

Does $c$ need to be greater than $0$, for instance can c be negative?, what about $m$ being $2$ or greater? In the proof it does not appear … Therefore, if these conditions are not given, is it invalid?

The definition of $a\equiv b\bmod m$ says $m\mid (b-a)$ for arbitrary integers $a,b,m$. But by convention we take $m$ positive. Therefore we want $m>0$ and $mc>0$, so $c>0$. — Dietrich Burde, Nov 02 '23 at 19:40

score 0 · Answer 1 · answered Nov 03 '23 at 13:28

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If $a\equiv b\pmod{m}$, it implies that $m\mid(a-b)$.
Since, $mc\mid(a-b)$ iff $m\mid(a-b)$, it must be true for all non-zero integer $c$. ($c$ cannot be $0$ since division by $0$ is absurd)
In conclusion, $c$ can be any non-zero integer.

answered Nov 03 '23 at 13:28

Lim Yun Zhe

64

1

For $c=0$: There are some definitions for $\equiv \pmod 0$ as a relation. While division by $0$ is absurd, determining whether $ac-bc = 0$ is a multiple of $0$ is possible (and always true). – peterwhy Nov 05 '23 at 04:01

score 0 · Answer 2 · answered Nov 03 '23 at 15:09

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$a \equiv b \pmod m \implies m|(a - b) \implies mk = a - b$

$ac - bc = c(a - b) = cmk \implies cm|(ac - bc)$

$\implies ac \equiv bc \pmod {mc}$

answered Nov 03 '23 at 15:09

Agent Smith

951

Show if $a \equiv b \pmod{m}$ then $ac \equiv bc\pmod{mc}$ without restrictions?

2 Answers2