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This question already has an answer here:

Some people say that 0 is neither even nor odd. I say that 0 is even.

Is there a simple way to convince people that 0 is even and the statement that "0 is neither even nor odd" is false.

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marked as duplicate by Daniel W. Farlow, Zev Chonoles, JMoravitz, Community Apr 4 '15 at 4:58

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  • $\begingroup$ Most people who practice mathematics regularly take $0$ to be an even integer, since it is divisible by $2$: $2 \times 0 = 0$. Furthermore, taking $2$ to be even makes a lot of statements a lot more concise. Call it odd, call it even if'n ya wanna, but I vote for $2$ even!!! Cheers! $\endgroup$ – Robert Lewis Apr 4 '15 at 4:20
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    $\begingroup$ en.wikipedia.org/wiki/Parity_of_zero $\endgroup$ – Todd Wilcox Apr 4 '15 at 4:21
  • $\begingroup$ "Believe"? This is mathematics, not some touchy-feely subject like poetry. $\endgroup$ – Jonathan Hebert Apr 4 '15 at 4:22
  • $\begingroup$ @JonathanHebert: Sorry, "think" is better? $\endgroup$ – user172675 Apr 4 '15 at 4:24
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    $\begingroup$ @JonathanHebert That is hardly necessary here. $\endgroup$ – Daniel W. Farlow Apr 4 '15 at 4:24
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An integer, $x$, is defined to be even whenever it can be written in the form $x=2k$ where $k$ is some integer.

Examples: $6=2\cdot 3,~~ 10 = 2\cdot 5,~~ 2218 = 2\cdot 1109,~~ -4 = 2\cdot (-2)$

An integer, $x$, is defined to be odd whenever it can be written in the form $x=2k+1$ where $k$ is some integer.

Examples: $-3 = 2\cdot (-2) + 1,~~~ 9 = 2\cdot 4 + 1,~~~ 1001 = 2\cdot 500 + 1$

Remember that $0$ is itself an integer, and that $0 = 2\cdot \color{red}{0}$, which is in the form $0=2k$ with $k = \color{red}{0}$, therefore $0$ is even.

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  • $\begingroup$ Same as my proof. Thanks you. $\endgroup$ – user172675 Apr 4 '15 at 4:23
  • $\begingroup$ @user172675 This is hardly a proof--if someone accepts that $0$ is actually a number, furthermore an integer, then the fact that $0$ is even is a trivial conclusion. Your question is far more philosophical than you may think, for the concept of zero goes way way back. $\endgroup$ – Daniel W. Farlow Apr 4 '15 at 4:28

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