I have checked the question upon this topic, but everytime the answer is the parity of zero is even because $0\times2=0$, and it is between two odd numbers, $-1,1$.
My question is, $0\times\text{(any odd number)}$ is also equal to zero, making the odd number a factor of zero, and thus $0$ could also be said to be odd.
Also, the logic that every even number lies between two odd numbers may be an exception for zero, as such exceptions occurs frequently in Number Theory.
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$\begingroup$An even number, say $k$, is a number where $$k\mod 2=0$$
Therefore, we set $k=0$ and note that $$0\mod 2=0$$ and therefore $0$ is even
We could also say that we know $1$ is odd, and therefore $$1\mod 2=1$$
We can note that $0=1-1$ and therefore \begin{align}0\mod 2&\equiv 1-1\mod 2\\ &\equiv (1\mod 2) - (1\mod 2)\\ &\equiv 1-1\\ &\equiv 0\end{align}
There is also a whole Wikiepdia page dedicated to this problem!
$\endgroup$ $\begingroup$An even number times an odd number is always even. For example, $2\cdot 3 = 6$. $3$ is a factor of $6$, but $6$ is still even. Being a multiple of an odd number doesn't make a number odd.
It's simply true by definition that $k\in \mathbb{Z}$ is even if there exists $n\in \mathbb{Z}$ such that $k=2n$. This applies to $0$, and so $0$ is an even number.
$\endgroup$ $\begingroup$You've got things backwards. If $a$ is a factor of $b$ and $b$ is odd, then $a$ is odd (e.g. since $9$ is odd and $3$ divides $9$, we know that $3$ is odd - although this is a bit of a silly way to find that out!). However, odd numbers can be factors of even numbers without a problem.
More to the point, by definition a number is even if it is of the form $2k$ for some integer $k$. Since $0=2\cdot 0$, this means $0$ is even.
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