Is an irrational number odd or even?

Definition.

Let $n\in \mathbb{Z}$ be an integer. $n$ is called even if ...

According to the definition only integers are even or odd. It is not something that you have to prove.

The original concept of even and odd is defined on integers. However, one can ask: Is there a natural extension to all real numbers? And if so, what would be the evenness or oddness of irrational numbers?

Now, what would we demand of such an extension? Well, the most important demand is, of course, that integers are even under the extended definition if and only if they are even under the normal definition (that is, also our new definition should for example give us that $2$ is even and $5$ is odd).

Let's therefore try a few properties of even/odd integers:

• An number is even if can be written as $2x$.

  In the real numbers, every number can be written as $2x$. Thus this definition does not work.

• A number is even if it is twice an integer.

  This of course works on the real numbers (and makes all non-integers odd). But it seems to be an odd extension; it's certainly not very useful.

• The sum of two even or two odd numbers is even, the sum of an even and an odd number is odd; 1 is odd.

  Since $1=\frac12+\frac12$, $\frac12$ could be neither even nor odd. This definitiion could probably be made work by making three types of numbers: even, odd or neither. But there are many ways this could be done; the most natural one would be to declare all non-integers as neither even nor odd; but that's where we started anyway. For rational numbers, it would also work if a number is even if for the maximally cancelled form the numerator is even, and odd if both numerator and denominator are odd. But that definition has no natural extension to irrational numbers (except, again, all being neither even nor odd).

• Adding $1$ to an even number gives an odd number, and vice versa.

  Again, this gives not an unique definition. The most natural definition would be to use a rounding function $\mathbb R\to\mathbb Z$ (like rounding to nearest, rounding up, rounding down, or rounding towards zero) and defining $x$ to be even if $r(x)$ is even. However, which rounding function to choose?

So there are ways to extend even/odd to reals, however you'll definitely have to give up some properties of even/odd numbers, and it is not clear which of those that are possible should be chosen, especially given that the resulting definitions are not too useful anyway. None of them really capture the concept of even and odd numbers.

One useful approach to generalize odd and even numbers to the irrational numbers is to quantify the "amount of evenness" of a number. We will see that $12$ is "twice as even" as $6$. Then, instead of asking whether an irrational number is odd or even, ask how even it is.

First, start with Patrick Da Silva's answer to the linked question. He explains the basics much better than I could.

Still with me? Good! Now the challenge is to extend the 2-adic valuation $\nu_2$ from the rational numbers to the real numbers. This can indeed be done, and it gives us simple answers for some irrational numbers. For example, $\nu_2(\sqrt 2)=\frac12$. In this sense you could say that $\sqrt 2$ is "half even", although that's not standard terminology!

Sadly, there isn't a unique extension of $\nu_2$ to all real numbers. I can't tell you what $\nu_2(\pi)$ is, because it depends on the extension. Nonetheless, the mere fact that an extension exists turns out to be useful. For example:

• Monsky's theorem. It is not possible to dissect a square into an odd number of triangles of equal area.

That sounds like a classical result of ancient Greek geometry, right? Nope, it was proved in 1970 using the existence of an extension of $\nu_2$ to the real numbers! The proof hinges on an elaborate coloring of the plane that, at its core, and in layman's terms, does what you want: it calls irrational numbers odd or even.