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

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• Jun 26th 2010, 04:46 PM
dwsmith
a \not\equiv b (mod m), then b \not\equiv a (mod m)
$a \not\equiv b \ \mbox{(mod m)}$, then $b \not\equiv a \ \mbox{(mod m)}$.

Is this all there is to this one:

$m\nmid (a-b)\rightarrow mx\neq (-1)(b-a)\rightarrow m(-x)\neq (b-a)\rightarrow m\nmid (b-a)$
• Jun 26th 2010, 05:12 PM
Ackbeet
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.
• Jun 26th 2010, 05:14 PM
dwsmith
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.
• Jun 26th 2010, 05:16 PM
Ackbeet
Oh, ok. Different authors define congruence differently, I guess.
• Jun 26th 2010, 06:21 PM
chiph588@
You have the right idea. $m\mid a-b \iff m\mid b-a$
• Jul 5th 2010, 02:44 AM
melese
Quote:

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.