I read on Wikipedia about that zero is an even number. Oke, that it is through. But now the explanation they have given:
Why zero is even
The standard definition of "even number" can be used to directly prove that zero is even. A number is called "even" if it is an integer multiple of $2$. As an example, the reason that $10$ is even is that it equals $5 \times 2$. In the same way, zero is an integer multiple of $2$, namely $0 \times 2$, so zero is even."
What?? $0\times 2$ is $0$. Yes that is correct. But also $0\times 3 = 0$. So what kind of proof is this??
Totally wrong.
Or do I miss something?
Thank you
$\endgroup$ 43 Answers
$\begingroup$By definition, for a number $n$, if there exists some number $k\in\mathbb Z$ such that $n=2\cdot k$, then $n$ is even.
- For $0$, we know that $0=2\cdot 0$
- Therefore, if $k=0$, we know that $0=2\cdot k$,
- Therefore, it is true that there exists some $k\in\mathbb Z$ such that $0=2\cdot k$.
- Therefore, $0$ is even (because the statement "$0$ is even is equivalent to the statement in the point above).
That's all. That's how mathematical definitions work.
There is no part of the definition that says "but if $n=3\cdot k$, then $n$ is not even".
Your point is completely off-topic. So what if $0=3\cdot 0$? Does that have any effect on the truth of the statement $0=2\cdot 0$?
Does the fact that $12=3\cdot 4$ mean that $12$ is not an even number?
What you ar experiencing is pretty normal for people first experiencing mathematical proofs. You must understand that just because something seems "strange", that is no reason for it to be wrong. If you can prove a mathematical statement, then that statement is true, no matter how strange it may sound to you.
The way to prove a statement is by following strict definitions, and only the definitions. In your case, all multiples of $2$ are even, and that's all you know, so that's all you have to work with.
$\endgroup$ $\begingroup$They mean, the set of even numbers is $$\{2k\mid k \in \mathbb{Z}\}$$
This means, we can write every even number $x$ as $x = 2k$ for some integer $k$. Since $0 = 2*0$ ($0$ is an integer), it follows that $0$ is even.
You seem to be misunderstanding that an odd number can be divided by $3$, which is not the case (for example $5$ is odd), so there really is no contradiction here.
In fact, from the definition of odd numbers, we can deduce that $0$ is not odd.
An odd number $x$ is a number that can be written as $x = 2k - 1$ for an integer $k$.
Assume that $0$ is odd, then $0 = 2k - 1$, so $1 = 2k$, implying that $1$ is even. A contradiction.
$\endgroup$ 4 $\begingroup$An even number is an integer number, which divided by $2$.
Thus, $0$ is an even number.
If $0\cdot3=0$ it not says that $3$ is an even. It says that $0$ divided by $3$. It's all.
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