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anonymous

  • 4 years ago

Am I a wisecrack or a fool? "Show that if \(a\), \(b\), \(m\), and \(n\) are integers such that \(m>0\), \(n>0\), \(n|m\), and \(a\equiv b (\mod m)\), then \(a\equiv b (\mod n)\)." Proof: By definition, we have that \(a\equiv b (\mod m)\implies m|(a-b)\). Since \(n|m\), it follows that \(n|(a-b)\). Therefore, \(a\equiv b (\mod n)\). \(\blacksquare\) This is too short to be true. What do you think?

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  1. anonymous
    • 4 years ago
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    Seems true to me.

  2. anonymous
    • 4 years ago
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    If you want to make it longer, you can add an intermediate step to prove that if \(n|m\) and \(m|k\) then \(n|k\).

  3. anonymous
    • 4 years ago
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    Although I don't see any good reason for doing that! :D

  4. anonymous
    • 4 years ago
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    I mean, I can always write a more rigorous proof, but I don't know if this one would be any better than the one above: Proof: By definition, we have that \(a\equiv b\mod m\implies a=b+km\), \(\exists k\in\mathbb{Z}\). Since \(n|m\), we have \(m=cn\), \(\exists c\in\mathbb{Z}\). Substituting yields \(a=b+km\implies a=b+kcn\). But because \(kc\in\mathbb{Z}\), it follows that \(a\equiv b\mod n\). \(\blacksquare\)

  5. anonymous
    • 4 years ago
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    Both of them look good to me. The second one is more rigorous as you said, which is better I think.

  6. anonymous
    • 4 years ago
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    Thank you. :)

  7. anonymous
    • 4 years ago
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    You're welcome! :)

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spraguer (Moderator)
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