Is zero odd or even?

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?

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:

An even number is always divisible by $2$.

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:

$N_e + N_e$ results in $N_e$

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:

$N_e + N_o$ results in $N_o$

$0 + \underbrace{1}_{N_o} = \underbrace{1}_{N_o}$

Passed this test too!

Test 4:

If $n$ is an integer of parity $P$, then $n - 2$ will also be an integer of parity $P$.

We know that $2$ is even, so $2 - 2$ or $0$ is also even.

Yes, the classification of naturals by their parity (= remainder modulo $2\:$) extends naturally to all integers: even integers are those integers divisible by $ 2,\,$ i.e. $\rm\: n = 2m\equiv 0\pmod 2,$ and odd integers are those with remainder $1$ when divided by $2,\ $ i.e. $\rm\ n = 2m\!+\! 1\equiv 1\pmod 2.\,$

The effectiveness of this parity classification arises from the fact that it is compatible with integer arithmetic operations, i.e. if $\rm\ \bar{a}\ :=\ a\pmod 2\ $ then $\rm\ \overline{ a+b}\ =\ \bar a + \bar b,\ \ \overline{a\ b}\ =\ \bar a\ \bar b.\: $ Iterating, we infer that equalities between expressions composed of these integer operations (i.e. integer polynomial expressions) are preserved by taking their images modulo $2\,$ (and ditto mod $\rm m\,$ for any integer $\rm m,\,$ e.g $ $ mod $ 9\,$ reduction yields $ $ casting out nines). In this way we can strive to better understand integers by studying their images in the simpler (finite!) rings $\rm\: \mathbb Z/m\, =\, $ integers modulo $\rm m.\:$

For example, if an integer coefficient polynomial has an integer root $\rm\ P(n) = 0\ $ then it persists as a root mod $2,\,$ i.e. $\rm\ P(\bar n)\equiv 0\ (mod\ 2),\,$ by the Polynial Congruence Rule. Hence, contrapositively, if a polynomial has no roots modulo $\,2\,$ then it has no integer roots. This leads to the following simple

Parity Root Test $\ $ A polynomial $\rm\:P(x)\:$ with integer coefficients has no integer roots
when its constant coefficient $\,\rm P(0)\,$ and coefficient sum $\,\rm P(1)\,$ are both odd.

Proof $\ $ The test verifies that $\rm\ P(0) \equiv P(1)\equiv 1\ \ (mod\ 2),\ $ i.e. that $\rm\:P(x)\:$ has no roots mod $2$, hence, as argued above, it has no integer roots. $\quad$ QED

So $\rm\, a x^2\! + b x\! + c\, $ has no integer roots if $\rm\,c\,$ is odd and $\rm\,a,b\,$ have equal parity $\rm\,a\equiv b\pmod 2$

Compare the conciseness of this test to the messy reformulation that would result if we had to restrict it to positive integers. Then we could no longer represent polynomial equations in the normal form $\rm\:f(x) = 0\:$ but, rather, we would need to consider general equalities $\rm\:f(x) = g(x)\:$ where both polynomials have positive coefficients. Now the test would be much messier - bifurcating into motley cases. Indeed, historically, before the acceptance of negative integers and zero, the formula for the solution of a quadratic equation was stated in such an analogous obfuscated way - involving many cases. But by extending the naturals to the ring of integers we are able to unify what were previously motley separate cases into a single universal method of solving a general quadratic equation.

Analogous examples exist throughout history that help serve to motivate the reasons behind various number system enlargements. Studying mathematical history will help provide one with a much better appreciation of the motivations behind the successive enlargements of the notion of "number systems", e.g. see Kleiner: From numbers to rings: the early history of ring theory.

Above is but one of many examples where "completing" a structure in some manner serves to simplify its theory. Such ideas motivated many of the extensions of the classical number systems (as well as analogous geometrical and topological completions concept, e.g. adjoining points at $\infty$, projective closure, compactification, model completion, etc). For some interesting expositions on such methods see the references here.

Note $\: $ Analogous remarks (on the power gained by normalizing equations to the form $\ldots = 0\:$) hold true more generally for any algebraic structure whose congruences are determined by ideals - so-called ideal determined varieties, e.g. see my post here and see Gumm and Ursini: Ideals in universal algebras. Without zero and negative numbers (additive inverses) we would not be able to rewrite expressions into such concise normal forms and we would not have available such powerful algorithms such as the Grobner basis algorithm, Hermite/Smith normal forms, etc.