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  1. Is zero even?
  2. Is zero a multiple of ANY number?

I did some reading, and I found the following:

for 1) I see a whole Wiki article saying that this is true, but I just can't understand why?

for 2) I think it is, but I don't have a precise definition, and the only thing I remember is that in the junior classes, I used to write that the smallest multiple of any number is the number itself and it was approved by my teachers...so a proof on 1) and 2) would be highly appreciated... Thanks a Ton!


marked as duplicate by user33321, J. M. is a poor mathematician, ShreevatsaR, rschwieb, Amzoti May 30 '13 at 16:52

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  • 1
    $\begingroup$ 1. $0/2=0$, so... 2. Since a number is certainly a multiple of itself... $\endgroup$ – J. M. is a poor mathematician May 30 '13 at 16:28
  • $\begingroup$ 1. Yes. 2. Yes. $\endgroup$ – Abhimanyu Pallavi Sudhir Jul 17 '13 at 7:11
  • $\begingroup$ Dividing a number by two, we find the number of pairs. The number Zero is neither odd nor even. See this: Number Zero parity | zero tinyurl.com/oexhr3k $\endgroup$ – user103028 Oct 26 '15 at 14:26

$1$. It depends on your definition of even number. If you define the set of even numbers as $$\mathbb{E} = \{x \in \mathbb{R}: x = 2n, \text{ where } n \in \mathbb{Z}^+\}$$ then $0$ is not even. However, if you define the set of even numbers as $$\mathbb{E} = \{x \in \mathbb{R}: x = 2n, \text{ where } n \in \mathbb{Z}\}$$ then it is.

$2$. For the second question again it depends on your definition of multiple of a number. If you define, the set of multiples of $x$ as $$M_x = \{y \in \mathbb{R}: y = z \times x, \text{ where }z \in \mathbb{Z} \}$$ then $0$ is indeed a multiple of any number since $0 = 0 \times x$ for any $x$ and in fact, $0$ is the only number present in all the sets, i.e., $\cap_{x \in \mathbb{R}} M_x = \{0\}$, i.e., $0$ is the only number that is a multiple of all numbers.


An integer $a$ is a multiple of an integer $b$ if there exists some third integer $c$ such that $ a = b \times c$. Hence since $0=m \times 0$, for any integer $m$, $0$ is a multiple of any number. In particular, a number is even if it is a multiple of $2$, so $0$ is even.


Remember that $k$ is even if and only if there is some $m$ such that $k=m+m$. No one said that $m$ cannot be equal to $k$. For $k=0$ we indeed take $m=0$ and we have $0=0+0$.

Similarly $0=0\cdot0=0\cdot1=0\cdot2=\ldots$, and as the definition of "$k$ is a multiple of $m$" says that there exists $n$ such that $k=n\cdot m$, we can take any $m$ whatsoever and $n=0$.


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