# Is 0 even, odd, or neither?

Is 0 even or odd?

**Why some people say it's even:** It's evenly divisible by 2.

**Why some people say it's odd:** It's not divisible by 2 and it's not a multiple of 2.

**Why some people say it's both:** Both of the top two arguments above are reasonable, so 0 is actually both even and odd!

**Why some people say it's neither:** Both of the arguments above are reasonable, but neither is completely true or sensible. Also, clearly, it can't be both.

The correct answer is that:

0 is \( \color{green}{\textbf{even}}\), and not odd.The definition of an even number:

Definition 1: A number is even if it is divisible by 2. (or "A number is even if it has 2 as a factor."

Definition 2: A number is even if it is a multiple of 2.

Proof that 0 is Even:There are several common definitions of multiple and divisor, but all of them make \(0\) an even number.

The definition of a divisor (or factor):

\(D\) is a

divisorof \(N\) if and only if \(\frac{N}{D}\) is an integer.By this definition, \(0\) is even because \(\frac{0}{2} = 0,\) which is an integer.

An alternative definition of a divisor:

\(D\) is a

divisorof \(N\) if and only if \(\frac{N}{D}\) has a remainder of \(0.\)By this definition, \(0\) is even because \(\frac{0}{2} = 0,\) with a remainder of \(0.\)

The definition of a Multiple:

An integer \(M\) is a multiple of an integer \(N\) if and only if there exists an integer, \(Z,\) such that \(N \times Z = M\).

By this definition, \(0\) is even because if we let \(Z\) be the integer \(0,\) then \(2 \times 0 = 0,\) therefore \(0\) is a multiple of \(2.\)

An interesting additional note is that, using the same logic, we can see that \(0\) is actually divisible by all integers other than itself (since \(\frac{0}{0}\) is undefined), and that \(0\) is a multiple of all integers.

The definition of an

odd number:

Definition 1:"A number is odd if it is equal to \(2n + 1\) for some integer, \(n.\)"

Common Notion:"A number is odd if it is an integer that is not even."

(Note: this common notion is true, but it's not considered the primary definition of "odd.")

Proof that 0 is NOT odd:If \(2n + 1 = 0,\) then, subtracting \(1\) from both sides, we see that \(2n = -1,\) and therefore \(n = -\frac{1}{2}.\) However, \(-\frac{1}{2}\) is not an integer, therefore \(0\) is not odd.

(This is a proof by contradiction.)

Rebuttal: \(2\) is not a factor of \(0\) because \(\frac{2}{0}\) is undefined.

Reply: You're mixing up the positions of the two variables in the definition of divisor when you set up that fraction. For example, by the same reasoning "\(5\) is not a factor of \(10\) because \(\frac{5}{10}\) is not an integer."The correct definition of a factor is that \(D\) is a factor of \(N\) if and only if \(\frac{N}{D}\) is an integer. Notice that the potential divisor is the number in the denominator of the fraction. Therefore, the fraction that we set up to test if \(2\) is a factor of \(0\) is: \(\frac{0}{2}.\) Since \(\frac{0}{2} = 0\) with remainder \(0\). \(2\) is a factor of \(0\).

Therefore, what you can conclude from your claim that \(\frac{2}{0}\) is undefined is that \(0\) is not a factor of \(2.\) However, that does not pertain to the question of whether or not \(0\) is even.

Rebuttal: This is crazy. We need to come up with new definitions if the ones that we have imply that \(0\) has infinitely many factors and is a multiple ofeverything.

Reply: While, in the proofs above, we only thought about this issue from the logical perspective of verifying the accepted definition of a prime number, it's also important to realize that the definitions are worded as they are to create a system which is as sensible and usable as possible. Including 0 as part of every set of multiples is actually very natural. For example, consider the visual representations of multiples of \(N\) pictured below. It's clear that including 0 in the set of multiples for each number completes the pattern, whereas omitting 0 would create a strange exception/irregularity in each set.Acknowledging and preserving this kind of pattern creates symmetry in the mathematics and makes it more likely that the theorems and proofs which use these definitions can be simply stated, without many exceptions and special cases. For example, consider the theorem, "The sum of any two multiples of a number is also a multiple of that number." If \(0\) were not a multiple of every number, this elegant theorem would have to be revised to, "The sum of any two multiples of a number is either 0 or a multiple of that number, and the sum of 0 and any multiple of a number is also a multiple of that number.".

This entire page is just a matter of definition. Mathematicians love to define things; they decide that \(0\) should be considered even because they can do so. But, of course, mathematicians also have reasons when defining things, and are not just making this decision at whim.

Want to make sure you've got this concept down? Try these problems:

What are the integral divisors of 0?

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**Clarification:** Given two integers \(N\) and \(M,\) \(N\) is an "integral divisor" of \(M\) if \(F = \frac{M}{N}\) is an integer.

**See Also**

**Cite as:**Is 0 even, odd, or neither?.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/is-0-even/