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I'm having some trouble figuring out how to give a direct proof using the definitions of mod m and congruence modulo m, without using any theorem involving mod or congruence.

if $a, b, c$ & $m$ are integers such that $m \geq 2$, $c > 0$ and $a \equiv b \pmod{m}$, then $ac \equiv bc\pmod{mc}$

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  • $\begingroup$ What's your definition of congruence? In the common one it should follow from a simple multiplication. $\endgroup$ Nov 2, 2016 at 22:12

2 Answers 2

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$$a \equiv b \pmod{m}\iff b=a+km \implies cb=ca+kcm\implies ca \equiv cb \pmod{cm}.$$

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Hint $\,\ m\mid n\, \Rightarrow\, mc\mid nc.\ $ OP has $\ n = a-b$

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  • $\begingroup$ Thank you. Very helpful. $\endgroup$
    – Avv
    Mar 18, 2021 at 0:45

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