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Prove that a even number + odd number = odd number

Let $x$ be the even number, let $y$ be the odd number.

From the definition of odd numbers, $y + 1$ is even.

Let:

$$x + y = z$$

Suppose $z$ is even.

$$x + y = z$$

It follows that:

$$x = x + y - y = z - y$$

But I cant think of any other contradiction here. The issue is with the assumption I suppose. Any help is appreciated!

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    $\begingroup$ A friendly reminder to watch for typos as you write. I'm sure several people look at this from the questions list and think "Prove even + odd is even!?" $\endgroup$ – JMoravitz Mar 3 '15 at 17:46
  • $\begingroup$ @JMoravitz I saw the title and expected to see the fake-proofs tag. $\endgroup$ – KSmarts Mar 3 '15 at 17:57
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An even number $n$ can always be written as $2k$. An odd number $m$ can always be written as $2h+1$.

Hence $n+m=2k+2h+1=2(h+k)+1$ which is odd.

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If $x$ is even, then $x$ can be written as $2a$ for some integer $a$. If $y$ is odd, then $y$ can be written as $2b + 1$ for some integer $b$.

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  • $\begingroup$ Be careful not to reuse variable names. If you say $x=2a$, then you should use a letter different than $a$ for describing $y$. It sounds as you have written it that you suggest $x+1=(2a)+1=2a+1=y$, which is not generally the case. $\endgroup$ – JMoravitz Mar 3 '15 at 17:49
  • $\begingroup$ Sloppy from my part, edited! $\endgroup$ – George Mar 3 '15 at 17:55

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