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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    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. – Mankind May 12 '15 at 18:59
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    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. – Gregory Grant May 12 '15 at 19:00
  • Can i do this by contradiction? –  May 12 '15 at 19:02
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    Related: [Can decimal numbers be considered “even” or “odd”?](http://math.stackexchange.com/questions/92451/can-decimal-numbers-be-considered-even-or-odd) – Martin Sleziak May 12 '15 at 19:34
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    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. – bof May 12 '15 at 21:42
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    I have seen an extension of oddness and evenness to include irrational numbers. Odd is simply defined as 'not even,' where even refers to whole numbers divisible by 2. Under this definition, all irrational numbers are odd. – A. Thomas Yerger May 13 '15 at 06:37
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    @GregoryGrant "No harm done as far as I can tell." That the sum of two "even" (in that sense) numbers may not be "even", very much seems like some harm done. – Did May 13 '15 at 06:40
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    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](https://en.wikipedia.org/wiki/Partition_of_a_set) 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*). – CiaPan May 13 '15 at 12:11
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    Irrationals are definitely odd numbers for most of the people. – Mher May 13 '15 at 14:06
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    The answer to the question in the title is 'no'. – Michael Albanese May 13 '15 at 21:34
  • @MichaelAlbanese Agree. So post your answer as an answer. – CiaPan May 14 '15 at 11:41
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    @CiaPan, you should talk to the French. To them `la brique` is definitely female :-) –  May 15 '15 at 01:48
  • As others have said, you cannot "prove" a definition; but you can use proof-like steps (induction and deduction) to _motivate_ a definition, as in celtschk's answer. – Kyle Strand May 15 '15 at 01:53
  • @paxdiablo I don't need to talk to the French, nouns in my language have grammatical genders, too (and a brick is female) :) If you look closer, also some English nouns have it, altough they don't display it in declension, just in pronouns: a dog is generally 'he' while a cat and a dove is 'she'. Also ships (esp. sailing ships) are female, possibly with an exception of navy ones. But I meant rather a biological sex here (and I'm sure you know it and your comment is kind of joke, not a misunderstanding). Anyway this is not a place for a discussion like this, so I suggest EOT. :) – CiaPan May 15 '15 at 07:36
  • it's like asking if you have stopped beating your wife. – chaohuang May 20 '15 at 14:09
  • You might find the *answer* of @JDH to *[this related question](http://math.stackexchange.com/questions/49034/is-infinity-an-odd-or-even-number)* about extensions of the concepts of even/odd numbers interesting. – epi163sqrt May 21 '15 at 10:56
  • @CiaPan But how to decide if question has or hasn't an answer? I mean I can say "every triangle has radius 8". This statement is false or meaningless? "Radius" is defined for circles. So can we say that there is a category error and the "statement" is meaningless? – user599310 Jul 28 '20 at 18:54
  • @user599310 Just use the definition. There is a definition of an even number, whose prerequisite is 'a number is integer'. If you _have_ a definition of a radius of a triangle, just use it. Otherwise the question is meaningless. – CiaPan Jul 28 '20 at 21:32
  • @CiaPan So is the function $f(x)=x^{2}$ an even number? How do we answer such questions? True, false, meaningless (undefined terms)? Because I can say that this function is not a member of the set "even numbers" so the above statement is false or I can say that even number is defined for integers so the above sentence is not a statement. – user599310 Sep 07 '20 at 14:07

10 Answers10



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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    "It seems to be an odd extension" - I see what you did there. – fluffy May 13 '15 at 02:28
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    +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". – Hal May 14 '15 at 11:27
  • 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? – user7530 May 18 '15 at 21:04

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.

Chris Culter
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    This is a wonderful answer! – Vincent May 13 '15 at 14:56
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    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.) – ShadSterling May 14 '15 at 23:00

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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  • Can i do this by contradiction? –  May 12 '15 at 19:07
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    Do what by contradiction? – quid May 12 '15 at 19:08
  • I mean prove this that irrationals are neither odd nor even by contradiction –  May 12 '15 at 19:09
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    What is your **definition** for odd and even? – quid May 12 '15 at 19:11
  • We say that a number is odd if it of the form 2n+1 and even if it is of the form 2n –  May 12 '15 at 19:13
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    What type of number is $n$? – quid May 12 '15 at 19:14
  • Couldn't we say that a circle is not a number therefore is not even? – user599310 Jul 06 '20 at 23:18
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    @user599310 I suppose you could, but then it is not odd either. If you say a circle is not an even integer then everybody will agree. – quid Jul 07 '20 at 09:17
  • Thanks for the answer. I have asked this question in other forums and I got mixed answers with others saying it is non-sense and others saying is false. I have asked a question [Philosophy SE](https://philosophy.stackexchange.com/questions/73965/how-to-contradict-a-general-statement) and I would be grateful if you can solve my misunderstanding. Thanks in advance. – user599310 Jul 07 '20 at 10:31

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.

Mark Bennet
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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!

Sudoku Polo
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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 $2$.

Sasho Nikolov
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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


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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All odd numbers on the pascal triangle are in a Sierpinski triangle, which is a fractal.

enter image description here

The binomial coefficient $\binom{n}{k}$ is defined for integers, but it can be extended to any real number.

$$\binom{\alpha }{k}= \frac{\alpha ^{\underline{k}}}{k!} $$

Another way is by using the Gamma function: $$\binom{\alpha }{k}=\frac{\Gamma \left ( \alpha +1 \right )}{\Gamma \left ( k +1 \right )*\Gamma \left ( \alpha-k+1 \right )}$$

The thing with the Sierpinski triangle, is that since it is a fractal, it has infinite detail, if you zoom in.

So the concept of being odd can be extended to all numbers $\binom{n}{k}$ whose coordinates (n,k) fall in the Sierpinski triangle, in the sense that they share a property of odd numbers.

I don't know if the white triangles (made of even numbers on the pascal triangle) also contain odd numbers for non integer coordinates, but if they do, and also conform a fractal, the same concept apply.

Raxi Ral
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