Is the following a typo? If $a \equiv b \pmod{m}$, then for some scalar $c>0$, $ac \equiv bc \pmod{mc}$
Or should it be $\pmod{m}$?
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Both conditions hold, so you really can't say whether this is a typo or not – Aang Jun 17 '13 at 18:12
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More importantly, the relationship goes the other way: if $c\ne 0$, then $a\equiv b\pmod{m}$ if and only if $ac\equiv bc\pmod{mc}$. – André Nicolas Jun 17 '13 at 18:12
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1If I were to revise the sentence you quoted, I'd leave "$\pmod{mc}$" alone, but I'd change "some scalar" to "every scalar". – Andreas Blass Jun 17 '13 at 18:40
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No typo in the congruence equations or implication:
Theorem: $$\text{If}\; a \equiv b \pmod{m},\;\text{ then for any scalar }\;c \neq 0,\; ac \equiv bc \pmod{mc} \tag{$\dagger$}$$
$$a \equiv b \pmod m \quad\iff (a - b) \equiv 0 \pmod m \quad\iff\; (a - b) = km, \;k\in \mathbb Z.$$
$$(a-b) = km \;\iff\; c(a-b) = c(km),\; (c\neq 0)\quad \iff \;(ac - bc) = k(mc),\;k\in \mathbb Z.$$
$$\iff (ac - bc)\equiv 0\pmod{mc} \quad \iff \;ac\equiv bc \pmod{mc}, mc \in \mathbb Z.$$
For your second question: "should it be $\pmod{m}$?"
That would certainly be true, as well, but is not as strong a statement. But we do indeed have that for $c\neq 0$: $$ac \equiv bc \pmod{mc} \implies ac\equiv bc \pmod m$$ since $$ac \equiv bc \pmod{mc} \iff mc\mid (ac - bc) \implies m\mid (ac - bc) \iff ac\equiv bc \pmod m$$
amWhy
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Recall the definition of $x \equiv y \pmod{z}$:
$$\exists k \in \Bbb Z: x - y = kz $$
Multiplying this equation by $c$, we immediately obtain that:
$$a \equiv b \pmod m \implies ac \equiv bc \pmod {mc}$$
Lord_Farin
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@amWhy Why did you do that? Modulo $0$ is simply the diagonal equivalence relation; since $0 = 0$, the consequent holds even if $c = 0$. – Lord_Farin Jun 17 '13 at 19:34
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$$a\equiv b\pmod m\iff a=b+n\cdot m$$ for some integer $n$
$$ac-bc=c(a-b)=c\cdot n\cdot m\equiv0\pmod {m\cdot c}$$ as $n$ is an integer
lab bhattacharjee
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