a \not\equiv b (mod m), then b \not\equiv a (mod m)

**dwsmith** · June 26th 2010, 04:46 PM

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

Is this all there is to this one:

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

**Ackbeet** · June 26th 2010, 05:12 PM

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.

**dwsmith** · June 26th 2010, 05:14 PM

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

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

**Ackbeet** · June 26th 2010, 05:16 PM

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

**chiph588@** · June 26th 2010, 06:21 PM

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

**melese** · July 5th 2010, 02:44 AM

Originally Posted by dwsmith

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

Is this all there is to this one:

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

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

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