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prove that if n,m are natural numbers and nm is even, then either n is even or m is even.

Proof: Assume n,m are natural numbers and nm is even, then either n or m is even.

Case 1: Assume n,m are natual numbers and nm is even, then n and m are even such that there exists natural numbers k and r so that n=2k and m=2k.

Consider nm: nm = (2k)(2r) = 4kr = 4(kr).

since k and r are natural numbers and 4 is even, then 4kr is even therefore nm is even.

Christian Soto
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  • You've proved the wrong direction. You've proved "If $n$ and $m$ are even, then $nm$ is even". – Anon Sep 09 '16 at 03:03
  • What do you mean? Should i do the contrapostive? – Christian Soto Sep 09 '16 at 03:10
  • Suppose $nm = 2k$ is even and $n = 2l+1$ is odd, then $2k = (2l+1)m$, i.e. $m = 2(k-lm)$ is even. Logically, A implies B or C is equivalent to A and not B implies C. – Weaam Sep 09 '16 at 03:23
  • He means you proved the wrong thing and should start over entirely. If you begin with nm is even, then ending with showing nm is even is pointless. You knew that from the very begining. You conclusion needs to be n or m is even. You concluded, completely incorrectly, that both n and m were even. And then continued to show what you already knew; that nm is even. You must start with nm being even and prove one or the other of n or m is even. THAT must be your conclusion. – fleablood Sep 09 '16 at 05:45

1 Answers1

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Suppose by contraposition that n is odd and m is odd.

Then n = 2k + 1, m = 2j + 1 for some integers k, j.

Then nm = (2k + 1)(2j+ 1) = ...

You'll find that nm is odd.

jeff
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