Congruence relation possible typo? 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 $\pmod{m}$?
 A: 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 $$ac \equiv bc \pmod{mc} \iff mc\mid (ac - bc) \implies m\mid (ac - bc) \iff ac\equiv bc \pmod m$$
A: 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:
$$a \equiv b \pmod m \implies ac \equiv bc \pmod {mc}$$
A: $$a\equiv b\pmod m\iff a=b+n\cdot m$$ for some integer $n$
$$ac-bc=c(a-b)=c\cdot n\cdot m\equiv0\pmod {m\cdot c}$$ as $n$ is an integer
A: See $a\equiv b \pmod m\implies m\mid (a-b)\implies mc\mid(a-b)c\implies ac\equiv bc \pmod {mc}$
