This one seems to easy.

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

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

Printable View

- Jun 26th 2010, 04:53 PMdwsmithModulo and congruence
This one seems to easy.

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

Does it suffice to just do this $\displaystyle a\not\equiv b \ \mbox{(mod m)}\rightarrow m\nmid (a-b)$ and now I am done? - Jun 26th 2010, 05:08 PMAlso sprach Zarathustra
- Jun 26th 2010, 05:08 PMAckbeet
Perhaps you could insert the intermediate step using the definition of congruence:

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

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