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Math Help - Modulo and congruence

  1. #1
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    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?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by dwsmith View Post
    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)
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  3. #3
    A Plied Mathematician
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    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.
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  4. #4
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    So it is just that easy than, thanks.
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