Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\pmod{m}$?

share|cite|improve this question
Both conditions hold, so you really can't say whether this is a typo or not – Aang Jun 17 '13 at 18:12
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
If 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
up vote 2 down vote accepted

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 $$ac \equiv bc \pmod{mc} \iff mc\mid (ac - bc) \implies m\mid (ac - bc) \iff ac\equiv bc \pmod m$$

share|cite|improve this answer

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:

$$a \equiv b \pmod m \implies ac \equiv bc \pmod {mc}$$

share|cite|improve this answer
@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
apologies. I'llhenceforth leave your posts untouched. – amWhy Jun 17 '13 at 19:40

$$a\equiv b\pmod m\iff a=b+n\cdot m$$ for some integer $n$

$$ac-bc=c(a-b)=c\cdot n\cdot m\equiv0\pmod {m\cdot c}$$ as $n$ is an integer

share|cite|improve this answer

See $a\equiv b \pmod m\implies m\mid (a-b)\implies mc\mid(a-b)c\implies ac\equiv bc \pmod {mc}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.