Some books say that even numbers start from $2$ but if you consider the number line concept, I think zero($0$) should be even because it is in between $-1$ and $+1$ (i.e in between two odd numbers). What is the real answer?
Elementary Number Theory – Is Zero Odd or Even?
elementary-number-theorymath-historymodular arithmeticparity
Related Solutions
Yes. The generalization is provided by modular arithmetic. The properties you are observing all come from the fact that taking the remainder modulo $n$ respects addition and multiplication, and this generalizes to any $n$. More generally in abstract algebra we study rings and their ideals for the same reasons.
The notion of evenness and oddness of functions is closely related, but it is somewhat hard to explain exactly why. The key point is that there is a certain group, the cyclic group $C_2$ of order $2$, which is behind both concepts. For now, note that the product of two even functions is even, the product of an even and odd function is odd, and the product of two odd functions is even, so even and odd functions under multiplication behave exactly the same way as even and odd numbers under addition.
There are also huge generalizations depending on exactly what you're looking at, so it's hard to give a complete list here. You mentioned chessboards; there is a more general construction here, but it is somewhat hard to explain and there are no good elementary references that I know of. Once you learn some modular arithmetic, here is the modular arithmetic explanation of the chessboard idea: you can assign integer coordinates $(x, y)$ to each square (for example the coordinate of the lower left corner), and then you partition them into black or white squares depending on whether $x + y$ is even or odd; that is, depending on the value of $x + y \bmod 2$. Then given two points $(a, b)$ and $(c, d)$ you can consider the difference $c + d - a - b \bmod 2$, and constraints on this difference translate to constraints on the movement of certain pieces. This idea can be used, for example, to prove that certain chessboards (with pieces cut out of them) cannot be tiled with $1 \times 2$ or $2 \times 1$ tiles because these tiles must cover both a white square and a black square. Of course there are generalizations with $2$ replaced by a larger modulus and larger tiles.
As for matrices and vectors, let's just say that there are a lot of things this could mean, and none of them are straightforward generalizations of the above concept.
The basic problem with your argument is that the concept of half of an infinite set is not well-defined. It’s possible to pair up the even integers with the odd integers with none left over in either set, and if we were talking about finite sets, that would be a demonstration that each was half of the whole set of integers. However, it’s also possible to pair up the multiples of $100$, say, with all the rest of the integers with none left over in either set, and the multiples of $100$ are obviously only part of the set of even numbers. Clearly, then, this kind of pairing argument cannot lead to any very useful notion of half of the set of integers.
There is a notion of asymptotic density of a set of positive integers that does a pretty good job of capturing many people’s intuitive sense of what half (or any other fraction) of the set of positive integers should mean. Let $A\subseteq\mathbb{Z}^+$; for each $n\in\mathbb{Z}^+$, $$\frac{|A\cap\{1,2,\dots,n\}|}n$$ is the fraction of the first $n$ positive integers that are in the set $A$. If this fraction approaches a limit $\alpha$ as $n$ increases, i.e., if $$\lim_{n\to\infty}\frac{|A\cap\{1,2,\dots,n\}|}n=\alpha\;,$$ the set $A$ is said to have asymptotic density $\alpha$. (Note that the limit need not exist: not all sets of positive integers have asymptotic densities. It’s not hard to show that if $A$ is the set of even integers or the set of odd integers, its asymptotic density is $1/2$, so it’s meaningful to say that each of these sets comprises half of the positive integers in terms of asymptotic density. Similarly, for each positive integer $n$, the set of multiples of $n$ can easily be shown to have density $1/n$, exactly as one would wish.
Added: In the comments Srivatsan points out an excellent example: the perfect squares can be paired up one-for-one with the integers, but their asymptotic density is $0$: they get sparser and sparser as you get further away from $0$. Thus, in one sense there are just as many perfect squares as there are integers, and yet in another sense almost every positive integers is a non-square.
Best Answer
For that, we can try all the axioms formulated for even numbers. I'll use only four in this case.
Note: In this question, for the sake of my laziness, I will often use $N_e$ for even, and $N_o$ for odd.
Test 1:
We know that if $x,y\in \mathbb{Z}$ and $\dfrac{x}{y} \in \mathbb{Z},$ then $y$ is a divisor of $x$ (formally $y|x$).
Yes, both $0,2 \in \mathbb{Z}$ and yes, $\dfrac{0}{2}$ is $0$ which is an integer. Passed this one with flying colors!
Test 2:
Let's try an even number here, say $2$. If the answer results in an even number, then $0$ will pass this test. $\ \ \ \ \underbrace{2}_{\large{N_e}} + 0 = \underbrace{2}_{N_e} \ \ \ $, so zero has passed this one!
Test 3:
$0 + \underbrace{1}_{N_o} = \underbrace{1}_{N_o}$
Passed this test too!
Test 4:
We know that $2$ is even, so $2 - 2$ or $0$ is also even.