What is the most unusual proof you know that $\sqrt{2}$ is irrational?

Here is my favorite:

Theorem: $\sqrt{2}$ is irrational.

Proof: $3^2-2\cdot 2^2 = 1$.

(That's it)

That is a corollary of this result:

Theorem: If $n$ is a positive integer and there are positive integers $x$ and $y$ such that $x^2-ny^2 = 1$, then $\sqrt{n}$ is irrational.

The proof is in two parts, each of which has a one line proof.

Part 1:

Lemma: If $x^2-ny^2 = 1$, then there are arbitrarily large integers $u$ and $v$ such that $u^2-nv^2 = 1$.

Proof of part 1:

Apply the identity $(x^2+ny^2)^2-n(2xy)^2 =(x^2-ny^2)^2 $ as many times as needed.

Part 2:

Lemma: If $x^2-ny^2 = 1$ and $\sqrt{n} = \frac{a}{b}$ then $x < b$.

Proof of part 2:

$1 = x^2-ny^2 = x^2-\frac{a^2}{b^2}y^2 = \frac{x^2b^2-y^2a^2}{b^2} $ or $b^2 = x^2b^2-y^2a^2 = (xb-ya)(xb+ya) \ge xb+ya > xb $ so $x < b$.

These two parts are contradictory, so $\sqrt{n}$ must be irrational.

Two things to note about this proof.

First, this does not need Lagrange's theorem that for every non-square positive integer $n$ there are positive integers $x$ and $y$ such that $x^2-ny^2 = 1$.

Second, the key property of positive integers needed is that if $n > 0$ then $n \ge 1$.

Bart Michels
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marty cohen
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19 Answers19


Suppose that $\sqrt{2} = a/b$, with $a,b$ positive integers. Meaning $a = b\sqrt{2}$. Consider $$A = \{ m \in \Bbb Z \mid m > 0 \text{ and }m\sqrt{2} \in \Bbb Z \}.$$

Well, $A \neq \varnothing$, because $b \in A$. By the well-ordering principle, $A$ has a least element, $s$. And $s,s\sqrt{2} \in \Bbb Z_{>0}$. Then consider the integer: $$r= s\sqrt{2}-s.$$ We have $r =s(\sqrt{2}-1) < s$, and $r > 0$. But $r\sqrt{2} = 2s-s\sqrt{2}$ is again an integer. Hence $r \in A$ and $r < s$, contradiction.

Ivo Terek
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    This is indeed a strange proof. – Matt Samuel Jun 03 '15 at 22:49
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    Isn't this just dressing up the usual proof with infinite descent? – Asaf Karagila Jun 03 '15 at 23:02
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    @Asaf I'm not sure that it is the same as the usual one but I think it's pretty close philosophically. The usual uses the fundamental theorem of arithmetic but there's no mention of that here – Cameron Williams Jun 03 '15 at 23:12
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    @Matt The proof is very natural when viewed in proper light. It is simple Euclidean descent in the *ideal* of all possible denominators for the fraction $\,a/b = \sqrt{2}.\,$ See [this answer](http://math.stackexchange.com/a/2941/242) for further discussion of such *denominator* and *conductor* ideals. – Bill Dubuque Jun 03 '15 at 23:42
  • I added an answer which shows one way to generalize this to arbitrary sqrts. Readers may enjoy working out how the above is a special case. – Bill Dubuque Jun 04 '15 at 01:45
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    And this generalizes very nicely to all non-square integers. – marty cohen Jun 04 '15 at 02:07
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    The integers aren't well-ordered, but the natural numbers are. I think you want to explicitly exclude $0$ from $A$ as well. – Arthur Jun 04 '15 at 05:25
  • I find this proof more natural than that used "modulo". – Student Jun 24 '19 at 11:55
  • @AsafKaragila Ackshyually, this is taking Paramanand's proof-by-diagram below (https://math.stackexchange.com/a/1324600) and subtracting the diagram. – Rivers McForge Nov 24 '20 at 19:57
  • @RiversMcForge: I'd probably say that backwards... :-) – Asaf Karagila Nov 24 '20 at 20:28

Here is something that I just came up with.

If $\sqrt2$ were rational, it would have been in every field of characteristics $0$.

It is also well-known that there are infinitely many prime numbers $p$, such that $x^2-2$ has no root over $\Bbb F_p$. Let $P$ be the set of these primes, and let $U$ be a free ultrafilter over $P$. Now consider $F=\prod_{p\in P}\Bbb F_p/U$.

Using Los theorem we have that:

  1. $F$ is a field.
  2. $F$ has characteristics $0$.
  3. $\lnot\exists x(x^2-2=0)$.

This means exactly that we found a field which extends the rational numbers, but has no roots for $x^2-2$, which in other words means $\sqrt2\notin F$ and therefore $\sqrt2\notin\Bbb Q$.

Asaf Karagila
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    And there you have it friends, a proof using the axiom of choice! And in some models there are no free ultrafilters over a countable set, and in those models this method of proof fails (although this can be overcome by using all manners of absoluteness to repeat the proof in some inner model). – Asaf Karagila Jun 03 '15 at 22:57
  • Very nice use of the ultrapower construction. There should be a contest for answering questions outside logic/model theory using Łoś's Theorem :) – Jonas Gomes Jun 03 '15 at 23:05
  • Do you need this ultrafiler business? I thought you would say it would have to live in a p-adic field, but by Hensels lemma no solution over $F_p$ means no soultion in $Q_p$. – rondo9 Jun 03 '15 at 23:06
  • @rondo9: I guess that would work too. I am familiar with $\Bbb Q_p$, but not enough to know this fact that you quote. Łoś theorem, on the other hand, is something that just came up in my last lecture so it was on the tip of my mind (and spoiler for my students reading this now; finite fields might appear in the next exercise sheet). – Asaf Karagila Jun 03 '15 at 23:07
  • Yea, $\mathbb{Q}_p$ is characteristic zero, but (separable) polynomials over $\mathbb{Q}_p$ have roots if and only if they have roots over $\mathbb{F}_p$, and then we use your fact that we can always find a prime where $2$ is not a square. – rondo9 Jun 03 '15 at 23:09
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    @Jonas: I think that Terry Tao has a blog post about connecting hard analysis with soft analysis using ultraproducts. – Asaf Karagila Jun 03 '15 at 23:09
  • @rondo9: You should post this as an answer, then. – Asaf Karagila Jun 03 '15 at 23:09
  • Wouldn't $\sqrt 2 = a/b$ mean that $\sqrt 2$ exists in all fields except the ones where the characteristic divides $b$? This itself should be enough to contradict infinitely many primes where $2$ is not a square. – Asvin Jun 04 '15 at 05:12
  • @NaN: That's a number theoretic theorem, whose proof is nontrivial for itself. – Asaf Karagila Jul 14 '15 at 04:11
  • @AsafKaragila Do you have a reference for "It is also well-known that there are infinitely many prime numbers p, such that $x^2−2$ has no root over $\mathbb{F}_p$"? – Moya Jul 14 '15 at 05:11
  • @Moya: https://en.wikipedia.org/wiki/Quadratic_reciprocity#.C2.B12_and_the_second_supplement and now use the theorem there are infinitely many primes which satisfy this condition. – Asaf Karagila Jul 14 '15 at 08:15
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    @Asvin: The ultraproduct proof is roughly equivalent to the alternative you suggest, just done in a simpler way that requires less fiddly attention to details (assuming, of course, you're used to ultraproducts). Another way to phrase it is in terms of nonstandard analysis; there is an infinite prime $P$ such that $\sqrt{2} \in \mathbb{F}_P$, but externally, $\mathbb{F}_P$ is a field of characteristic zero. –  Aug 16 '15 at 08:18
  • @AsafKaragila, There is an easier way than absoluteness to transform this into a proof without the axiom of choice. Instead of the ultrapower, just use a countable non-standard model $M$ of $\mathbb{Z}$ which exist without choice. Use overspill to get a non-standard prime $p$ such that $x^2-2 \mod p$ has no root in $M$. Now, $M/pM$ is a field with characteristic $0$ but without a square root of $2$. – russoo May 01 '21 at 20:27
  • @russoo: Indeed. Or use compactness for countable languages, which is provable in ZF. – Asaf Karagila May 01 '21 at 21:08

Consider the linear application $A:\mathbb{R}^2\to \mathbb{R}^2$ given by $$A=\begin{pmatrix} -1&2 \\ 1&-1 \end{pmatrix} .$$ $A$ maps $\mathbb{Z}^2$ into itself and $V=\{y=\sqrt 2 x\}$ is an eigenspace relative to the eigenvalue $\sqrt 2-1$. But $A\mid_V$ is a contraction mapping, so $\mathbb{Z}^2\cap V=\emptyset$.

Ivo Terek
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  • Isn't this assuming the result in the last step? Why can't this contraction happen to map a $\mathbb{Z}^2$ to another $\mathbb{Z}^2$? – Mark Hurd Jun 09 '15 at 02:58
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    Suppose $p\in \mathbb{Z}^2\cap V$. There exists $k$ such that $A^k(p)\in B(0, 1/2)\setminus\{0\}$. Absurd. – user72870 Jun 09 '15 at 11:11

Ever so slightly off-topic, but I can't resist reminding folks of the proof that $\sqrt[n]{2}$ is irrational for $n \ge 3$ using Fermat's Last Theorem:

Suppose that $\sqrt[n]{2} = a/b$ for some positive integers $a$ and $b$. Then we have $2 = a^n / b^n$, or $b^n + b^n = a^n$. But Andrew Wiles has shown that there are no nonzero integers $a, b$ satisfying the last equation. Thus $\sqrt[n]{2}$ must be irrational. [This proof is due to W. H. Schultz and appeared in the May, 2003 issue of the American Mathematical Monthly.]

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    The really funny part: FLT isn't powerful enough to prove the irrationality of $\sqrt 2$. – Martín-Blas Pérez Pinilla Jun 15 '15 at 08:00
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    But are you sure that this is not a circular reasoning, namely the irrationality of $\sqrt[n]2$ is not used in the proof of Fermat's Last Theorem by Andrew Wiles? Without having thoroughly read the proof, I dare not say that ... – WhatsUp Jul 01 '21 at 21:11
  • In fact the initial manipulations of the FLT proof, going from an FLT counterexample to a semistable modular elliptic curve which contradicts Ribet's theorem, freely uses the fact that if a^n+b^n=c^n is nontrivial then WLOG gcd(a,b,c)=1, so it's really just the same techniques as the proof that 2^{1/n} is irrational. – Kevin Buzzard Jul 03 '21 at 08:31
  • We can also prove $\sqrt 2$ is irrational similarly by FLT, e.g. [see here](https://math.stackexchange.com/a/4438248/242) – Bill Dubuque Apr 28 '22 at 14:00
  • Thanks for the extension and references. – PolyaPal Apr 28 '22 at 19:44

Consider $\mathbb Z[\sqrt{2}]= \{a + b\sqrt 2 ; a,b \in \mathbb Z\}$. Take $\alpha = \sqrt 2 - 1 \in \mathbb Z [\sqrt 2]$, then $$0 < \alpha < 1 \implies \alpha ^k \to 0 \,\,\text{as} \,\, k \to \infty \tag {*}$$

Say $\sqrt 2 = \frac{p}{q}$, since $\mathbb Z[\sqrt 2]$ is closed under multiplication and addition we have

$$\alpha^k = e + f \sqrt 2 = \frac{eq + fp}{q} \geq \frac{1}{q}$$

which contradicts $(*)$.

Aaron Maroja
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    [See here](http://mathforum.org/kb/thread.jspa?forumID=149&threadID=383333&messageID=1181901#1181901) for a generalization, contrasting *discretness* of ring of integers vs. *denseness* of rings of fractions. – Bill Dubuque Jun 03 '15 at 23:55
  • I'll definitely take a look at it. – Aaron Maroja Jun 04 '15 at 00:01
  • @AaronMaroja could you please explain the last inequality? Can't $eq+fp$ be negative? If it can, and you meant absolute value, why can't $\lvert eq+fp \rvert$ be zero? – vuur Aug 10 '15 at 11:32
  • Notice that $\alpha > 0$ and $\sqrt 2 = \frac{p}{q}$. So $$e + f \sqrt 2 = e + f\frac{p}{q} = \frac{eq + fp}{q} \geq \frac{1}{q}> 0 $$ – Aaron Maroja Aug 10 '15 at 13:11

I want to add the following figure. I might not call it unusual but perhaps not common enough to be widely available.

enter image description here

Update: I later checked the image from wikipedia article (linked in comments to the current question) and it presents the same proof in a bit complex fashion. There is no need for two arcs. Just note that if $m$ is hypotenuse of larger triangle and $n$ is one of the other sides then after this construction the length of side of smaller triangle is $(m - n)$ (this is the obvious part). The slightly hard part is to show that the hypotenuse of smaller triangle is $(2n - m)$ and for this wiki article draws two arcs.

However we can easily see that tangents drawn from external point to a circle are of equal length. Hence both the tangents are equal to side of smaller triangle i.e. $(m - n)$. Hence the hypotenuse of smaller triangle is $(n - (m - n)) = 2n - m$.

Paramanand Singh
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$$\boxed{\text{If the boxed statement is true, then the square root of two is irrational.}}$$

Lemma. The boxed statement is true.

Proof. Assume for a contradiction that the boxed statement is false. Then it has the form "if $S$ then $T$" where $S$ is false, but a conditional with a false antecedent is true.

Theorem. The square root of two is irrational.


  1. The boxed statement is true. (By the Lemma.)

  2. If the boxed statement is true, then the square root of two is irrational. (This is the boxed statement itself.)

  3. The square root of two is irrational. (Modus ponens.)

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I just thought of this one:

Consider the equation $x^2-n=0$ for natural $n$. Evidently, $\sqrt n$ is a solution to the equation. Now the rational root theorem implies that for a root to be rational for that equation, it must be a factor of $n$ (up to sign). If $\sqrt n$ is a factor of $n$ ($n$ is a perfect square), then $\sqrt n$ is rational. If $\sqrt n$ is not a factor of $n$ ($n$ is not a perfect square), then $\sqrt n$ must be irrational.

Now just plug in $n=2$.

Pauly B
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Here is my favorite but perhaps a little too sophisticated. It make use of the theorem which says that if $K$ is an extension of some field $F$ then $[K:F]=1$ iff $K=F$.

From this fact consider the minimum polynomial of $\sqrt{2}$ over $\mathbb{Q}$ which is $m(x)=x^2-2$ by Eisenstein's Criterion. Thus, $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}] > 1$ and so by the contrapositive of the previous remarks $\mathbb{Q} (\sqrt{2}) \neq \mathbb{Q}$. That is, $\sqrt{2} \notin \mathbb{Q}$.

This uses the notion $[K:F]$ means the degree of the minimum polynomial of the extension(s) element(s).

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    Related: Proof by Gauss lemma is one line. In fact, Gauss lemma is one way to prove the Eisenstein criterion. – rondo9 Jun 04 '15 at 02:51

Below is a simple way to implement the (Euclidean) denominator descent implicit in Ivo Terek's answer (which was John Conway's favorite way to present this proof).

Theorem $\quad \rm r = \sqrt{n}\:$ is an $\rm\color{#c00}{integer}$ if rational,$\:$ for $\:\rm n\in\mathbb{N}$

Proof $\ \ \ $ Put $\ \ \displaystyle\rm r = \frac{A}B ,\;$ least $\rm\; B>0.\,$ $\ \displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{A}B = \frac{nB}A.\ $ Taking fractional parts: $\rm\displaystyle\ \frac{b}B = \frac{a}A\ $ for $\rm\ 0 \le b < B.\ $ If $\,\rm\displaystyle\ \color{#c00}{B\nmid A}\,$ then $\rm\ b\ne 0,\ $ so $\rm\,\ \displaystyle \frac{A}B = \frac{a}b,\ $ contra leastness of $\,\rm B$.

Remark $\ $ See here for a conceptual view of the proof (principality of denominator ideals). Though the proof may seem unusual to those who have not yet studied advanced number theory, it is quite natural once one learns about conductor and denominator ideals.

Bill Dubuque
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Define the $2$-adic valuation $\nu_2(r)$ of a nonzero rational number $r = \frac{p}{q}$ to be the number of times $2$ divides $p$ minus the number of times $2$ divides $q$. The $2$-adic valuation of the square of a rational number is even. But the $2$-adic valuation of $2$ is odd. Hence $2$ is not the square of a rational number.

This argument generalizes with no difficulty to the following: if $n$ is a positive integer, then $\sqrt[k]{n}$ is rational iff the $p$-adic valuation $\nu_p(n)$ of $n$ is always divisible by $k$.

Qiaochu Yuan
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    +1, but I'm not sure I'd call this proof unusual; it's exactly the proof I'd give if someone asked me now to prove that $\sqrt{2}$ is irrational. – anomaly Jun 14 '15 at 06:13

Tennenbaum gave a geometric proof of the irrationality of $\sqrt{2}$:

Tennenbaum's proof

The large square has side length $a$, while the light purple/blue square has side length $b$, with $a, b$ positive integers such that $({a\over b})^2=2$. But then it's easy to see that the blue square has twice the area of the pink square - that is, $({2b-a\over a-b})^2=2$. Since the numerator and denominator are each positive integers integers, and are less than $a$ and $b$ respectively, we have an infinite descent.

Noah Schweber
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Piggy backing on Asaf's answer:

If $\sqrt{2}$ were rational it would have to live in every field of characteristic zero. There exist primes $p$ where $x^2-2$ has no roots. By Hensel's lemma, a polynomial with simple roots $f\in \mathbb{Q}_p[x]$ has a root in $\mathbb{Q}_p$ if and only if its reduction has a root in $\mathbb{F}_p$. For any $p$ such that $x^2-2$ has no roots, it then follows that $\sqrt{2}\notin \mathbb{Q}_p$, but the latter is a field of characteristic zero, contradiction.

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The irrationality of $\sqrt{2}$ can be deduced from the following

Theorem (Fermat, 1640): The number $1$ is not congruent.

Reasoning: If $\sqrt{2}$ were rational then $\sqrt{2},\sqrt{2}$,and $2$ would be the sides of a rational right triangle with area $1$. This is a contradiction of $1$ not being a congruent number.

A positive rational number $n$ is called a congruent number if there is a rational right triangle with area $n$: there are rational $a,b,c>0$ such that $$a^2+b^2=c^2\qquad\text{ and }\qquad\frac{1}{2}ab=n$$

A proof of this theorem based upon Fermat's method of descent is given in The congruent number problem Theorem 2.1 by Keith Conrad.

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  • First, I think you mean $\sqrt{2}$ and $1$, not $\sqrt{2}$ and $2$. Second, the proof about congruent numbers (which shows there are no solutions to $a^2+b^2 = c^4$) is harder than the proofs that $\sqrt{2}$ is irrational. – marty cohen Jun 20 '15 at 19:33
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    @martycohen: I'm referring to a right triangle with legs $\sqrt{2}$ and hypotenuse $2$. So, we get $(\sqrt{2})^2+(\sqrt{2})^2=2^2$ and $\frac{1}{2}\sqrt{2}\sqrt{2}=1$. Maybe you would also like to check the paper by Keith Conrad which I'm referring too. But I totally agree with your second argument! :-) – epi163sqrt Jun 20 '15 at 19:45
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    You are right. I read too fast. – marty cohen Jun 21 '15 at 15:49

Suppose $\sqrt2$ is arational, then there exist $p,q$ two natural numbers such that $$\color{Red}{p^2=q^2+q^2.}$$ Then by the parametric solution of Pythagoras Equation there exist two natural numbers $a,b$ such that $a\gt b\ge 1$ and $p=a^2+b^2,$ $$\color{Green}{q=a^2-b^2=2ab.}$$
Now, if $r=a+b$ and $s=a,$ then $$\color{Red}{r^2=s^2+s^2}$$ with $r\lt p,\ s\lt q.$ Hence, by the Infinite Descent there are no such $p$ and $q$ natural numbers.

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Here is my favorite one. Suppose for the sake of contradiction that $\sqrt{2} = \frac{a}b$ for integers $a,b$. Then $2b^2 = a^2$. Let $p$ be an odd prime that is not congruent to $\pm 1 \pmod 8$. Then by quadratic reciprocity, $$\left( \frac{2b^2}p \right) = \left( \frac{2}p \right) = -1 \ne \left( \frac{a^2}p \right) = 1.$$

Sandeep Silwal
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The final decimal digit of $a^2$ and the final decimal digit of $2b^2$ can't agree unless $a$ and $b$ are both multiples of five, leading to an infinite descent. (Check: the possible last digits of $a^2$ are 0,1,4,5,6,9 and the possible last digits of $2b^2$ are 0,2,8.)

James Propp
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Using the rational root theorem on $x^2 - 2 = 0$ is a very simple and elegant way of proving the irrationality of $ \sqrt{2} $. Peersonally, I like it beacuse it can be explained easily to high school students

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This proof will look much better if someone can add a diagram to it.

Suppose $\sqrt2=a/b$ with $a$ and $b$ positive integers chosen as small as possible. Draw a square of side $a$. In the upper left corner, place a square of side $b$, and another one in the lower right corner. The two $b$-squares have total area $2b^2=a^2$, the same as the area of the $a$-square. Thus, the overlap of the two $b$-squares, which is a square we'll call a $c$-square, must have area equal to that of the two corners of the $a$-square not covered by the $b$-squares. Those corners are squares, call them $d$-squares. Then $c^2=2d^2$, so $\sqrt2=c/d$, where $c$ and $d$ are integers (indeed, $c=2b-a$, $d=a-b$) and are less than $a$ and $b$. Contradiction!

You can see this, with diagram, here. It also appears, with many other proofs, here.

Gerry Myerson
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