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?
 A: YES! zero is an even number. Here is the Dr. Math's explanation. (Wayback Machine)

This seems to be a matter of confusion for many others around this planet, you may always like to ask  google for your confusion.
A: 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.
A: 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 Polynomial 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 this answer or this sci.math post 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 Gröbner basis algorithm, Hermite/Smith normal forms, etc.
A: Which numbers are you using? Positive integer? In this case, you don't have to consider zero (that's why, perhaps, some books says that even numbers start from 2). If 0 is in your number set, then yes, it is divisible by 2.
A: $0$ is even. The difference of two distinct even numbers is also even, for example: $32 - 20 = 12$, $20 - 12 = 8$, $12 - 8 = 4$, etc. Also, the sum of two distinct even numbers is again even: $-12 + 32 = 20$, $32 + 20 = 52$, etc.
But with, say, $32 - 32 = 0$, why would this difference suddenly become odd? The answer is simple, it doesn't, it's still even.
A: This problem arose e.g. during the Beijing even-odd car ban for the 2008 Olympics, where cars with odd numbered licence plates were banned one day, then even the next day.
The choice is between:


*

*0 is even and not odd,

*0 is odd and not even,

*0 is both even and odd,

*0 is neither even nor odd (like infinity or $\pi$) or

*0 is assigned a unique title (like how 1 called a "unit" -- neither prime nor composite).


This is a matter of definition, so while you could define 0 to be any of the above, it's best to choose the definition that will be the most consistent with the usage of "even" and "odd" for numbers other than 0.
Let $W=\{2,4,6,\ldots\}$, $V=\{1,3,5,\ldots,\}$ and lets look at the properties of even and odd numbers on these sets that we are familiar with.


*

*A number is either even or odd, and not both.

*If $w,x \in W$ then $w+x \in W$ (even + even = even).

*If $w \in W$ and $v \in V$ then $w+v \in V$ (even + odd = odd).

*If $y,v \in V$ then $y+v \in W$ (odd + odd = even).


[and probably many others I've forgotten to write here]
So it would be desirable that whichever definition we choose for 0, it preserves the above properties.  Now lets say we let 0 be odd (the second and third cases listed above).  Then our definition is not consistent with these properties.  So, if we choose to define 0 as odd, we should have some substantial benefits to outweigh the losses.  On the other hand, defining 0 to be even and not odd is consistent with the above properties.
The last two candidate definitions are essentially saying there is no consistent way of defining even or oddness to 0.  But in this case, there is -- 0 is even and not odd.
[Note: We also have the property that elements of W are all divisible by 2, but whether or not 0 is divisible by 2 is another matter of definition, for which we should again apply the "which is the most sensible definition" concept.]
A: I think zero is an even number because it's in between $1$ and $-1$. If they are odd numbers then the number in between must be even. Which means zero is an even number. 
A: The real answer depends on the definition, because there is math-history tag invoked there was a time that $1$ was not considered an odd numbers, $0$ and negative numbers for sure where not considered even or odd. Historically the concept was defined only for natural numbers. 
These days the set of all integers multiplied by $2$ is considered the set of even numbers, i.e. $\dots,-4,-2,0,2,4,\dots$ and all the integers not in that set are defined to be odd.
There is no real answer, it all depends on the definition, same way that the book is only dealing with natural numbers and not integers.
The concept was extended from naturals to integers, but there are uncountably many ways to define the even and odds beyond the natural numbers. Just make sure others know what definition you are using to label something even or odd. 
A: Is Zero Even? - Numberphile
PS : This is completely opposite to my previous post (and view) but it has many good points and we are here to learn and not to push one's own view of things.
A: As the history of mathematics shows, zero is a very odd number whose general acceptance is surprisingly recent. However, it is not odd. Perhaps the ambiguity of "odd" causes all the confusion.
A: Zero is even. An even integer is a number of the form $2n$ where $n$ is also an integer. A number is odd, in the mathematical sense, if it is of the form $2n + 1$ with $n$ an integer.
If zero was odd, we could solve the equation $2x + 1 = 0$ in integers. But the only possible solution is $x = -\frac{1}{2}$, which is a rational number but not an integer.
The set of integers is sometimes said to be "doubly infinite," stretching out to positive infinity in one direction and to negative infinity in the other. If you wanted to, you could say the even numbers start at $-2$ and continue with $-4$, then $-6$, then $-8$, etc.
A: The AP sequence
$$ (-6,-4,-2,0,2,4,6)$$
shows that it is even.
If $k$ is an even integer, then
$ (-1)^ k = 1 $
Division of $0$ by $2$ leaves no remainder.
&c. are further confirmations.
