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My sister just asked this question to me: "Is an irrational number odd or even?" I told her that decimals are not odd or even and that does imply that not recurring and non repeating decimals will also not be odd or even. But I want a rigorous math proof for this. Can someone help me with this proof? Also, can I do this proof by contradiction?

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    $\begingroup$ Hey. You are correct in that it does not make sense to talk about if a decimal number is odd or even. Only the whole numbers $\ldots,-3,-2,-1,0,1,2,\ldots$ can be odd or even. Decimal numbers like $2.71$ or $0.333333\ldots$ are neither odd nor even. You cannot prove this: It is by definition of the terms odd and even that they only apply to whole numbers. $\endgroup$ – Mankind May 12 '15 at 18:59
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    $\begingroup$ The concept of even/odd is defined only on the integers. So you are welcome to define an irrational number as even or odd if you want to. For example we could declare an irrational as even if its integer part is even. No harm done as far as I can tell. $\endgroup$ – Gregory Grant May 12 '15 at 19:00
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    $\begingroup$ Related: Can decimal numbers be considered “even” or “odd”? $\endgroup$ – Martin Sleziak May 12 '15 at 19:34
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    $\begingroup$ The question seems to imply that the answer is the same for all irrational numbers, that all irrationals have the same parity. This makes no sense; if $\pi$ is even then $\pi+1$ is odd. $\endgroup$ – bof May 12 '15 at 21:42
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    $\begingroup$ Similar questions: "Is a brick in the wall male or female?" "Is my left shoe wise or stupid?" The classes 'odd' and 'even' simply do not apply to fractions or irrational numbers, same as 'male' and 'female' do not apply to bricks or 'wise' and 'stupid' do not apply to shoes. Only whole numbers are partitioned into subsets of odd and even numbers; numbers outside the whole numbers set are not covered by the two classes, so the question has no answer (or the answer is: neither). $\endgroup$ – CiaPan May 13 '15 at 12:11
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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.

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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.

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    $\begingroup$ "It seems to be an odd extension" - I see what you did there. $\endgroup$ – fluffy May 13 '15 at 2:28
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    $\begingroup$ +1 for this answer: If I were a young person asking the question, I would be dissatisfied with the answer "it's defined that way". $\endgroup$ – Hal May 14 '15 at 11:27
  • $\begingroup$ One could call an algebraic number even/odd if the constant term of its minimal polynomial is even/odd. Does anything useful come from this definition? $\endgroup$ – user7530 May 18 '15 at 21:04
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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.

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    $\begingroup$ This is a wonderful answer! $\endgroup$ – Vincent May 13 '15 at 14:56
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    $\begingroup$ My solution was to use the sgn(ν₂(x)): positive is even, zero is odd, negative I called uneven. This is backward-compatible with the familiar rules for integers, and applies to any number for which you can devise a canonical prime factorization. It obviously includes the rationals, and also includes any integer roots of rationals. The results of some operations on unevens cannot be categorized without more information; ν₂ is enough except when adding numbers of equal ν₂. (It also gave a nice definition of a "totally irrational" number: the multiplicity of each prime factor is less than one.) $\endgroup$ – ShadSterling May 14 '15 at 23:00
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What you need first and foremost is not a proof but a definition.

Normally, one will say something like: one calls an integer even if it is divisible by $2$, and odd otherwise.

Then, being "even" and "odd" is a property of integers, and it makes no sense to ask it for other numbers. It also makes no sense to ask if a circle is even.

Now, you might have a different defintion of "even" and "odd" that makes sense for more numbers and the answer changes.

For example, if you extend the defintion as given above directly to the real numbers, then every number is even: in the real numbers every number is divisible by $2$, since $2X =a$ has a (real) solution for every $a$.

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  • $\begingroup$ Can i do this by contradiction? $\endgroup$ – user210387 May 12 '15 at 19:07
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    $\begingroup$ Do what by contradiction? $\endgroup$ – quid May 12 '15 at 19:08
  • $\begingroup$ I mean prove this that irrationals are neither odd nor even by contradiction $\endgroup$ – user210387 May 12 '15 at 19:09
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    $\begingroup$ What is your definition for odd and even? $\endgroup$ – quid May 12 '15 at 19:11
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    $\begingroup$ What type of number is $n$? $\endgroup$ – quid May 12 '15 at 19:14
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An irrational number only makes sense if we look at numbers beyond the rationals. In any extension of the rationals we will find the number $\frac 12$. And then any number is divisible by $2$.

We could say, in this context, that every number is even. The problem with this is that the concept simply ceases to be useful when used in that way. So we don't use it like that, and retain it for contexts where divisibility by $2$ makes a difference.

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    $\begingroup$ Great answer (and much better expressed than mine). $\endgroup$ – Sudoku Polo May 12 '15 at 19:17
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I guess that the original "motivation" for the odd/even distinction in human culture was to know if a number of things (like apples or stones) could be divided between two people in a fair way without having to cut or break a remainder one. And also to known if things could be paired: three women and three men was fine, but one woman and two men meant conflict!

The above distinction makes sense with natural numbers... But what motivation would we have to define odd/even for the rational or irrational for example? Any rational divided by two results in a rational, e.g. 1.5 / 2 = 0.75. If we accept cut or broken things, then we don't care about remainders!

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Here is one more perspective. Think of the congruence mod 2 relation $x \equiv y$, defined as "$x\equiv y$ if and only if $x-y$ is an integer multiple of $2$". (This is usually denoted $x \equiv y \pmod 2$, but I will omit the parenthesis because there will be no chance of confusion.)

It is not hard to verify that this is an equivalence relation: it is reflexive, symmetric, and transitive. The relation is well-defined over any subset of the real numbers, including, of course, the integers $\mathbb Z$. There are only two equivalence classes of $\mathbb{Z}$ modulo the "$\equiv$" relation: the even integers $[0] = \{0 + 2n: n \in \mathbb{Z}\}$ and the odd integers $[1] = \{1 + 2n: n \in \mathbb{Z}\}$.

Similarly, we can define the equivalence classes of the reals $\mathbb{R}$ modulo the "$\equiv$" relation. Here there are many more equivalence classes: except the odds $[1]$ and the evens $[0]$, there are also the equivalence classes $[0.5] =\{0.5 + 2n: n \in \mathbb Z\}$ and $[\sqrt{2}]=\{\sqrt{2} + 2n: n \in \mathbb{Z}\}$, for example. In fact there is a distinct equivalence class $[x] = \{x + 2n: n \in \mathbb{Z}\}$ for every $x \in [0, 2)$.

So in that sense, an irrational number, or in fact any number which is not an integer, is neither odd nor even, because it does not belong to either of the equivalence classes $[0]$ and $[1]$. This is because if $x$ is not an integer, then $x-0$ and $x-1$ are not integers either, and, therefore, neither of them is an integer multiple of $2$.

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Closely related to other answers but might be good for the purpose. By contradiction:

For a given irrational a, assume it is odd.
a/2 is also irrational; call it b
2b is obviously even since odd + odd = even
so a is even.
So all irrationals are even, 
which shows they cannot be split into two categories at least.
All rationals can be expressed as the product of two irrationals 
(which is  just saying that a rational divided by an irrational is irrational) 
so all rationals are even too.

∴ 1 is even

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An even number is any integer that is divisible by 2 so the equations for any even number is: $N_E=2n: n\in\Bbb Z$

An odd number is any integer that is not divisible by 2 so the equation for any odd number is: $N_O=2n-1: n\in\Bbb Z$

And if you put any integer (replaced by n shown on the equations above) to the equations shown above, none of the equations will give an irrational value so therefore an irrational number is not even and not odd, they are simply IRRATIONAL NUMBERS .

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protected by Daniel Fischer May 13 '15 at 16:29

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