# Thread: a \not\equiv b (mod m), then b \not\equiv a (mod m)

1. ## a \not\equiv b (mod m), then b \not\equiv a (mod m)

$\displaystyle a \not\equiv b \ \mbox{(mod m)}$, then $\displaystyle b \not\equiv a \ \mbox{(mod m)}$.

Is this all there is to this one:

$\displaystyle m\nmid (a-b)\rightarrow mx\neq (-1)(b-a)\rightarrow m(-x)\neq (b-a)\rightarrow m\nmid (b-a)$

2. If your other congruence post is logically prior to this one, then it looks good. Otherwise, you should interpose, twice, the intermediate step AsZ showed in the other post. Just for completeness.

3. In the book, it defines $\displaystyle m|(a-b) \ \mbox{as} \ a\equiv b \ \mbox{(mod m)}$.

The problem is it justed seemed to easy to be true though.

4. Oh, ok. Different authors define congruence differently, I guess.

5. You have the right idea. $\displaystyle m\mid a-b \iff m\mid b-a$

6. Originally Posted by dwsmith
$\displaystyle a \not\equiv b \ \mbox{(mod m)}$, then $\displaystyle b \not\equiv a \ \mbox{(mod m)}$.

Is this all there is to this one:

$\displaystyle m\nmid (a-b)\rightarrow mx\neq (-1)(b-a)\rightarrow m(-x)\neq (b-a)\rightarrow m\nmid (b-a)$
If you rephrase this, the problem is: If $\displaystyle m$ does not divide $\displaystyle a-b$, then also $\displaystyle m$ does not divide $\displaystyle b-a$.
Now, $\displaystyle a-b$ and $\displaystyle b-a$ have the same divisors. So if $\displaystyle m$ is not a divisor of $\displaystyle a-b$, it can't be a divisor of $\displaystyle b-a$.
This is essentially the same explantion "chiph588@" gave.