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This question is already answered on here. head over to that link.

I have been sitting here wondering if zero is even or odd and if it is even why is it even. I have no justification for why it is odd or even. Understanding that justification would be great.

It feels even, but no idea why.

enigma
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    is zero divisible by 2? – garserdt216 Jun 28 '17 at 01:07
  • $n $ is even if $n=2k $ fo some integer $k $. $0=20$. So it is even. $n $ is odd if $n=2k+1$ for some integer $k$. If $0=2k+1$ then $k=-1/2$ and that isn't an integer. So $0$ is not odd. There is no mystery or special 0 subtlety. $0$ is simply even, which shouldn't be any more surprising than $4$ is even or $5$ is odd. Why the heck wouldn't* 0 be even? – fleablood Jun 28 '17 at 05:37

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Note that the number before any odd number is even. For example, five is odd, and so 4 is even.

Note that 1 is odd. Therefore, zero is even.

Samuel Li
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  • That's quite convincing – matboy Jun 28 '17 at 01:29
  • fleablood I don't like the answer, because it never defines even, nor odd, but seems to express only an observation, not proof that by definition zero is even. But I object to using dumb to refer to an answer or justification, that is incorrect. Indeed, by your definition, your comment here is "really dumb" (and you never even bother to elaborate to any degree as to why conclude the answer is dumb and utterly unconvincing) and unconvincing. It's comments like yours, that are ultimately demeaning to users, that contribute to the stereotype of mathematicians as aloof and arrogant! – amWhy Jun 28 '17 at 13:32
  • Yeah. You're right. I felt really bad after writing it and was going to delete it but I was distracted by a shiny piece of metal and then completely forgot I had typed it... In greater detail, this answer is unacceptable because it states a single observation, provides no reason for the observation, extrapolates from a single instance a universal hypotheses with no verification or explanation, and no proof. That is simply not an acceptable argument. – fleablood Jun 29 '17 at 15:51
  • Note: even means divisible by two. If n = 2k then the number immediately after n; n+1 = 2k + 1 and the number immediately before; n-1 = 2k -1 = 2(k-1) + 1 will have a remainder of 1 and will not be even. The number before or after that; n+2 = 2k+2 =2(k+1) and $n-2 = 2k-2 = 2(k-1) will be even. So odd and even number alternate so the observation is true. ... However it's probably more direct to just observe the definition of even is divisible by 2. As 0 = 2x0, 0 is divisible by 2. So it is even. – fleablood Jun 29 '17 at 15:59