Is it mathematically acceptable to use Prove if $n^2$ is even, then $n$ is even. to conclude since 2 is even then $\sqrt 2$ is even? Further more using that result to also conclude that $\sqrt [n]{2}$ is even for all n?

Similar argument for odd numbers should give $\sqrt[n]{k}$ is even or odd when k is even odd.

My question is does any of above has been considered under a more formal subject or it is a correct/nonsensical observation ?

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    I interpret this question as stating "Could I extend the notion of even number to $\sqrt 2$ and moreover any real number $x$ such that $x^2\in\mathbb Z$ and $2\mid x^2$?" Could you confirm whether this interpretation is correct? I have not found any contradictions in defining this rather useless extension. – Karl Kroningfeld Jun 06 '13 at 12:29
  • @user1 : Please edit and replace my question, yes you are correct. – jimjim Jun 06 '13 at 12:35
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    @Arjang: It would seem more natural that you edit your own question yourself. – Marc van Leeuwen Jun 06 '13 at 13:57
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    I vote against closing this question. – MJD Jun 06 '13 at 15:02
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    If you define *even* as *not an odd integer* then $\sqrt{\text{bicycle} + \text{fish}}$ is also even. – Kaz Jun 07 '13 at 04:48
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    @Kaz : I just realised that is the problem with defining things as not something! With the same definition $\sqrt {bicycle+fish}$ is also odd because it is *NOT* even, hence when something is both even and odd it does not fit with the idea of mutually exclusive equivalence classes. – jimjim Jun 09 '13 at 10:25
  • @Arjang That is true. Defining things negatively leads to problems. The thing to remember is that the set of objects that does **not** have some property usually does not constitute a meaningful class. Things are categorized by what they have, not by what they don't have. – Kaz Jun 09 '13 at 15:17
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    @Kaz - Integral domains are a subset of the set of rings, categorised by the fact that they do not have zero divisors. Absence of something can be a property in itself. Indeed, many things are defined negatively without a problem - you only arrive at a problem if the definitions become recursive (such as defining "even" to mean "not odd" and "odd" to mean "not even" simultaneously, or defining the elements of a set to be sets that do not contain themselves). – Glen O Jul 24 '13 at 16:29

4 Answers4


Yes, essentially analogous arguments work to extend the notion of parity from the ring of integers to many rings of algebraic integers such as $\,\Bbb Z[\sqrt[n]{k}].\ $

The key ideas are: one can apply parity arguments in any ring that has $\ \mathbb Z/2\ $ as an image, e.g. the ring of all rationals with odd denominator, or the Gaussian integers $\rm\:\mathbb Z[{\it i}\,],\:$ where the image $\rm\ \mathbb Z[{\it i}\,]/(2,{\it i}-\!1) \cong \mathbb Z/2\ $ yields the natural parity definition that $\rm\ a\!+\!b\,{\it i}\ $ is even iff $\rm\ a\equiv b\ \ (mod\ 2),\ $ i.e. $ $ if $\rm\ a+b\,{\it i}\ $ maps to $\:0\:$ via the above isomorphism, which maps $\rm\ 2\to 0,\ i\to 1\:$.

Generally, it is easy to show that if $\rm\:2\nmid f(x)\in \Bbb Z[x]\setminus \Bbb Z\:$ then the number of ways to define parity in the ring $\rm\ \mathbb Z[w] \cong \mathbb Z[x]/(f(x))\ $ is given by the number of roots of $\rm\: f(x)\: $ modulo $2\:.\ $ For suppose there exists a homomorphism $\rm\ h\, :\, \mathbb Z[w]\to \mathbb Z/2.\:$ Then $\rm\:w\:$ must map to a root of $\rm\:f(x)\:$ in $\rm\ \mathbb Z/2.\ $ Thus if $\rm\ f(0)\equiv 0\ (mod\ 2)\ $ then $\rm\: \mathbb Z[w]/(2,w) \cong \mathbb Z[x]/(2,x,f(x)) \cong \mathbb Z/2\ $ by $\rm\: x\mid f(x)\ (mod\ 2),\, $ and $\rm\, \!f(1)\equiv 0\ (mod\ 2) $ $\Rightarrow$ $\rm \mathbb Z[w]/(2,w\!-\!1) \cong \mathbb Z[x]/(2,x\!-\!1,f(x)) \cong \mathbb Z/2\, $ by $\rm\, x\!-\!1\,|\, f(x)\ (mod\ 2). $

Let's consider some simple examples. Since $\rm\ x^2\!+1\ $ has the unique root $\rm\ x\equiv 1\ (mod\ 2),\:$ the Gaussian integers $\rm\ \mathbb Z[{\it i}\,]\cong \mathbb Z[x]/(x^2\!+1)\ $ have a unique definition of parity, with $\:i\:$ being odd. Since $\rm\ x^2\!+x+1\ $ has no roots modulo $\: 2,\: $ there is no way to define parity for the Eisenstein integers $\rm\ \mathbb Z[w] \cong \mathbb Z[x]/(x^2\!+x+1).\, $ Indeed since $\rm\ w^3 = 1\ $ we infer that $\rm\: w \equiv 1\ (mod\ 2)\ $ contra $\rm\ w^2\!+w+1 = 0.\ $ On the other hand $\rm\ \mathbb Z[w] \cong \mathbb Z[x]/(x^2\!+x+2)\ $ has two parity structures since both $\:0\:$ and $\rm\:1\:$ are roots of $\rm\ x^2\! + x + 2\ $ modul0 $\rm\:2,\:$ so we can define $\rm\:w\:$ to be either even or odd.

J. M. ain't a mathematician
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Key Ideas
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Note that that theorem begins with "Suppose $n$ is an integer, and that $n^2$ is even." So it does not hold when considering $\sqrt{2}$

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    Please consider the comment of user1. – jimjim Jun 06 '13 at 12:45
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    @Arjang -- That comment may explain your intent. What it does not explain is how you could prove $even(n^2) \implies even(n)$ without the stipulation that n is an integer. If you can only prove this for the case where n is an integer, then, yes, you can extend the definition of even, but, no, the usual definition of even does not apply to $\sqrt 2$. When you extend the definition, you are changing the language and anything which comprises the extended definition becomes an axiom of your new system, whether useful or not. – Heath Hunnicutt Jun 06 '13 at 18:08
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    ...what Arjang said. – Lucas Jun 06 '13 at 23:16
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    I haven't commented yet since Heath pretty much summed it up. Yes, there are various branches of mathematics in which similar theorems hold and $x^2$ is $P$ implies $x$ is $P$ (the algebraic integers for example) but in my opinion, this is far removed from the ideas of even and odd, which have very specific definitions based on the divides property. The OP asked if $\sqrt{2}$ was even and if you could show it using the stated theorem, to which the answer is a flat out no for the reason stated. Yes, there is plenty of interesting mathematics out there to consider, but I find that unrelated. – john Jun 08 '13 at 02:33

Other answerers are answering several slightly different interpretations of your question; but taking literally what you asked,

Is it mathematically acceptable to use “Prove if $n^2$ is even, then $n$ is even.” to conclude since 2 is even then $\sqrt 2$ is even?

the answer is a quite interesting “no, but also yes”.

Formally, the answer is definitely no: as @john explained, you can’t use that theorem to conclude what you suggest, since the theorem and proof start by assuming that n is an integer, so $\sqrt{2}$ is not a valid value for $n$.

However, it is excellent and very acceptable mathematical practice to do what you did, and take that theorem as inspiration for considering a new generalisation of the concepts involved. And, as @KeyIdeas describes, this can lead to some very nice theories.

TL;DR: you can’t conclude this from that theorem/proof, but you can certainly be inspired along these lines by them.

Peter LeFanu Lumsdaine
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The question you linked to starts (quite naturally) with "Suppose $n$ is an integer", which clearly excludes taking $n=\sqrt 2$. Therefore it cannot be used to conclude that $\sqrt2$ is even. Obviously $\sqrt2$ being even is absurd since even numbers in particular have to be integers. I would say your "observation" is sloppy (overlooking the stated conditions) and your following reasoning unfounded (or nonsensical if you prefer).

Marc van Leeuwen
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