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I am having some trouble with the following homework problem:

Use "proof by contradiction" to explain why no integer can be both odd and even.

So, far I think that I need to start with a counterexample which would be any integer $2k+1$ for odd and $2n$ for even. So my question is, what does it mean when it says "and"?

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    $\begingroup$ If $2k + 1 = 2n$ it seems logical that $2(k-n) = 1$, so that $k - n = \frac{1}{2}$ is an integer. Does this seem like a problem to you? $\endgroup$ – David Wheeler Oct 4 '15 at 17:14
  • $\begingroup$ I don't understand where you got the 2(k-n) = 1 part??? $\endgroup$ – Blaine Miller Oct 4 '15 at 17:17
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    $\begingroup$ Let's say your "even and odd" integer is $t$. Since $t$ is odd, $t = 2k + 1$ for some integer $k$, Since $t$ is even, $t = 2n$ for some integer $n$. But $t = t$, so $2k + 1 = 2n$. And...oops, I made a mistake, it should be: $2(n-k) = 1$. $\endgroup$ – David Wheeler Oct 4 '15 at 17:21
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    $\begingroup$ Anyway, $2k ++ 1 = 2n$ means $1 = 2k + 1 - 2k = 2n - 2k = 2(n-k)$. Sorry for the mix-up. $\endgroup$ – David Wheeler Oct 4 '15 at 17:22
  • $\begingroup$ Great. I was just making it too hard. Thanks a lot. $\endgroup$ – Blaine Miller Oct 4 '15 at 17:26
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If $x$ is odd then $x=2k+1 $ , $k\in \mathbb {Z}$ . For hip. $x$ is also even, then $2$ divide $x$, but $2$ divide $2k$ , therefore $2$ divide $1$. Absurd!

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