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If $x + y$ is even and $x$ is odd, then we want to show that $y$ must be odd.

Assume that $x + y$ is even and that $x$ is odd. Then, by the definition of even, $\exists k \in \mathbb{Z}$ such that $2k = x + y$. Similarly, by the definition of odd, $\exists m \in \mathbb{Z}$ such that $2m + 1 = x$.

Subtract the two equations to find that: $y = 2k - 2m - 1$. From here, I am confused.

Is it correct to say: Let $r = k - m - 1$. Substitute for $r$ and then find $y = 2r + 1$?

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    $\begingroup$ $y = (x+y)-x = \textrm{even} - \textrm{odd} = \textrm{odd}$ $\endgroup$ – MPW Aug 30 '16 at 21:14
  • $\begingroup$ "Is it correct to say: Let r=k−m−1. Substitute for r and then find y=2r+1? " Yes, it would. Clever. But it would be easier just to say y = 2(k-m) -1. .... Or if you like.... it'd be easier just to let r = k-m so y = 2r -1. $\endgroup$ – fleablood Aug 30 '16 at 21:29
  • $\begingroup$ This depends on precisely how odd and even were defined. It's probably worth proving the following. i) all integers are odd or even but never both. ii) all odd numbers can be written as $2k \pm 1$ and all numbers that can be written as $2k \pm 1$ (k an integer) are odd. $\endgroup$ – fleablood Aug 30 '16 at 22:20
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You basically did it. Just write $y = 2(k-m-1)+1$ and you're done!

This literally means that there exists an integer $a$ such that $y=2a+1$, which was your definition of odd.

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Maybe it is correct, but I definitely think it's needlessly confusing.

Back up just a bit to $y = 2k - 2m - 1$. Obviously $2k$ is even, and so is $2m$, therefore $2k - 2m$ is even as well (given that it can be rewritten as $2(k - m)$). When you subtract $1$ from an even number, you get an odd number, which means that $y$ is indeed odd. You're done.

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    $\begingroup$ I think the OP is very concerned with The Rulez that even means 2k and odd means 2k + 1 so you MUSTS gets it into the form 2k PLUS 1 and 2k - 1 isn't good enough. I think it's easier to simply note one time in your lifetime that that $x = odd \iff x = 2k + 1 \iff x = 2j - 1$ (and the proof is obvious; let j = k + 1) and just use "odd means $x = 2j \pm 1$, whichever is convenient". $\endgroup$ – fleablood Aug 30 '16 at 21:34
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$$y=(x+y)-x\equiv0-1\equiv1\mod 2$$

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