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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you want to make it longer, you can add an intermediate step to prove that if \(nm\) and \(mk\) then \(nk\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Although I don't see any good reason for doing that! :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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 \(nm\), 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 \(a\equiv b\mod n\). \(\blacksquare\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Both of them look good to me. The second one is more rigorous as you said, which is better I think.
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