# Modulo and congruence

• June 26th 2010, 04:53 PM
dwsmith
Modulo and congruence
This one seems to easy.

if $a\not\equiv b \ \mbox{(mod m)}$, then $m\nmid (a-b)$

Does it suffice to just do this $a\not\equiv b \ \mbox{(mod m)}\rightarrow m\nmid (a-b)$ and now I am done?
• June 26th 2010, 05:08 PM
Also sprach Zarathustra
Quote:

Originally Posted by dwsmith
This one seems to easy.

if $a\not\equiv b \ \mbox{(mod m)}$, then $m\nmid (a-b)$

Does it suffice to just do this $a\not\equiv b \ \mbox{(mod m)}\rightarrow m\nmid (a-b)$ and now I am done?

$a\not\equiv b \ \mbox{(mod m)}$, then $\frac{a-b}{m} \neq k\in \mathbb{Z}$
So it is clear that no such $k$ exist, hence, $m\nmid (a-b)$
• June 26th 2010, 05:08 PM
Ackbeet
Perhaps you could insert the intermediate step using the definition of congruence:

$a\not\equiv b\;(\!\!\!\!\mod m)$ implies $\neg(\exists n\in\mathbb{Z})(mn=a-b)$ implies $m\not|(a-b)$.

[EDIT] Which is the same thing as AsZ's post.
• June 26th 2010, 05:10 PM
dwsmith
So it is just that easy than, thanks.