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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 $n \ge 1$.

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    $\begingroup$ I wonder... If you could assume $\sqrt2$ rational and from there arrive at a solution to Fermat's last theorem. $\endgroup$
    – Arthur
    Commented Jun 3, 2015 at 22:42
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    $\begingroup$ Hmm...I'm afraid you'll have to elaborate on the $3^2-2\cdot2^2 = 1$ proof. I'm too obtuse to follow without some exposition. $\endgroup$
    – Brian Tung
    Commented Jun 3, 2015 at 22:54
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    $\begingroup$ @Arthur: Yes, because a false statement implies anything. $\endgroup$
    – Asaf Karagila
    Commented Jun 3, 2015 at 22:58
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    $\begingroup$ It does not make a lot of sense to ask «What is the most unusual proof you know?» and then pick one as an accepted answer... $\endgroup$ Commented Jul 21, 2015 at 7:05
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    $\begingroup$ It's.... Irrational $\endgroup$
    – user351797
    Commented Dec 21, 2017 at 5:44

20 Answers 20

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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.

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    $\begingroup$ This is indeed a strange proof. $\endgroup$ Commented Jun 3, 2015 at 22:49
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    $\begingroup$ Isn't this just dressing up the usual proof with infinite descent? $\endgroup$
    – Asaf Karagila
    Commented Jun 3, 2015 at 23:02
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    $\begingroup$ @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 $\endgroup$ Commented Jun 3, 2015 at 23:12
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    $\begingroup$ @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 for further discussion of such denominator and conductor ideals. $\endgroup$ Commented Jun 3, 2015 at 23:42
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    $\begingroup$ And this generalizes very nicely to all non-square integers. $\endgroup$ Commented Jun 4, 2015 at 2:07
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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 $\sqrt2\notin\Bbb Q$.

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    $\begingroup$ 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). $\endgroup$
    – Asaf Karagila
    Commented Jun 3, 2015 at 22:57
  • $\begingroup$ Very nice use of the ultrapower construction. There should be a contest for answering questions outside logic/model theory using Łoś's Theorem :) $\endgroup$ Commented Jun 3, 2015 at 23:05
  • $\begingroup$ 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 $Q_p$. $\endgroup$
    – rondo9
    Commented Jun 3, 2015 at 23:06
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    $\begingroup$ @Jonas: I think that Terry Tao has a blog post about connecting hard analysis with soft analysis using ultraproducts. $\endgroup$
    – Asaf Karagila
    Commented Jun 3, 2015 at 23:09
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    $\begingroup$ @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. $\endgroup$
    – user14972
    Commented Aug 16, 2015 at 8:18
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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 $\mathbb{Z}^2\cap V=\emptyset$.

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  • $\begingroup$ Isn't this assuming the result in the last step? Why can't this contraction happen to map a $\mathbb{Z}^2$ to another $\mathbb{Z}^2$? $\endgroup$
    – Mark Hurd
    Commented Jun 9, 2015 at 2:58
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    $\begingroup$ Suppose $p\in \mathbb{Z}^2\cap V$. There exists $k$ such that $A^k(p)\in B(0, 1/2)\setminus\{0\}$. Absurd. $\endgroup$
    – user72870
    Commented Jun 9, 2015 at 11:11
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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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    $\begingroup$ The really funny part: FLT isn't powerful enough to prove the irrationality of $\sqrt 2$. $\endgroup$ Commented Jun 15, 2015 at 8:00
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    $\begingroup$ 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 ... $\endgroup$
    – WhatsUp
    Commented Jul 1, 2021 at 21:11
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    $\begingroup$ 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. $\endgroup$ Commented Jul 3, 2021 at 8:31
  • $\begingroup$ We can also prove $\sqrt 2$ is irrational similarly by FLT, e.g. see here $\endgroup$ Commented Apr 28, 2022 at 14:00
  • $\begingroup$ Thanks for the extension and references. $\endgroup$
    – PolyaPal
    Commented Apr 28, 2022 at 19:44
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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.

Proof.

  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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    $\begingroup$ This has to be the funniest math joke I've ever read. $\endgroup$
    – user286485
    Commented Mar 26, 2018 at 12:11
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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 $(*)$.

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    $\begingroup$ See here for a generalization, contrasting discretness of ring of integers vs. denseness of rings of fractions. $\endgroup$ Commented Jun 3, 2015 at 23:55
  • $\begingroup$ I'll definitely take a look at it. $\endgroup$ Commented Jun 4, 2015 at 0:01
  • $\begingroup$ @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? $\endgroup$
    – vuur
    Commented Aug 10, 2015 at 11:32
  • $\begingroup$ Notice that $\alpha > 0$ and $\sqrt 2 = \frac{p}{q}$. So $$e + f \sqrt 2 = e + f\frac{p}{q} = \frac{eq + fp}{q} \geq \frac{1}{q}> 0 $$ $\endgroup$ Commented Aug 10, 2015 at 13:11
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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 $(n - (m - n)) = 2n - m$.

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  • $\begingroup$ Yep. That's the $\sqrt{2}-1$. $\endgroup$ Commented Jun 14, 2015 at 5:36
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    $\begingroup$ @martycohen: This is geometrical proof of the statement "If $\sqrt{2} = m/n$ then $\sqrt{2} = (2n - m)/(m - n)$" so that if $\sqrt{2}$ is a ratio then there is a ratio with smaller denominator which is also equal to $\sqrt{2}$. $\endgroup$
    – Paramanand Singh
    Commented Jun 14, 2015 at 5:43
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    $\begingroup$ Nice approach! G. H. Hardy uses this concept in his book "A Course of Pure Mathematics ". $\endgroup$
    – Learning
    Commented Jul 28, 2020 at 6:47
  • $\begingroup$ @Learning: yes I have read in Hardy's Pure Mathematics also. $\endgroup$
    – Paramanand Singh
    Commented Jul 28, 2020 at 11:31
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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 $n=2$.

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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 $k$.

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    $\begingroup$ +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. $\endgroup$
    – anomaly
    Commented Jun 14, 2015 at 6:13
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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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    $\begingroup$ Related: Proof by Gauss lemma is one line. In fact, Gauss lemma is one way to prove the Eisenstein criterion. $\endgroup$
    – rondo9
    Commented Jun 4, 2015 at 2:51
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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.

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    $\begingroup$ What does it mean for a number to be integral? $\endgroup$ Commented Jun 14, 2015 at 18:25
  • $\begingroup$ @YoTengoUnLCD Here integral means "an integer", $\endgroup$ Commented Jun 14, 2015 at 18:35
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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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  • $\begingroup$ For concreteness, $p=5$ works. Also $p=11$. $\endgroup$
    – Asaf Karagila
    Commented Jun 3, 2015 at 23:13
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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.

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  • $\begingroup$ A nice variation. $\endgroup$ Commented Oct 12, 2016 at 1:37
  • $\begingroup$ For me, this proof is entirely useless. Not only I'm color blind, I have problems thinking about math in terms of geometry... :-P $\endgroup$
    – Asaf Karagila
    Commented Dec 30, 2016 at 20:34
  • $\begingroup$ @AsafKaragila That's fair. I definitely find it less easily understood than the usual one, but I do think it's an interesting different take. $\endgroup$ Commented Dec 30, 2016 at 21:28
  • $\begingroup$ This is the proof I posted a year earlier, only my skills didn't extend to including a picture. $\endgroup$ Commented May 9, 2018 at 5:44
  • $\begingroup$ @GerryMyerson Sorry, I just saw this comment. You're absolutely right, I missed yours when I was checking to see if this had already been posted. Would you like me to delete this answer (I can also add the picture to yours)? $\endgroup$ Commented Jul 14, 2019 at 23:28
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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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  • $\begingroup$ 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. $\endgroup$ Commented Jun 20, 2015 at 19:33
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    $\begingroup$ @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! :-) $\endgroup$ Commented Jun 20, 2015 at 19:45
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    $\begingroup$ You are right. I read too fast. $\endgroup$ Commented Jun 21, 2015 at 15:49
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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.)

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  • $\begingroup$ I've recently been told that this proof is actually quite common in French textbooks. $\endgroup$ Commented Jul 6, 2021 at 13:06
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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, $$\left( \frac{2b^2}p \right) = \left( \frac{2}p \right) = -1 \ne \left( \frac{a^2}p \right) = 1.$$

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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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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.

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Here's a proof that has the advantage that the same idea can be used to prove the irrationality of $\pi$ and the irrationality of $e^r$ for nonzero rational $r$. It follows the general paradigm of irrationality proofs whereby one assumes that $\alpha = p/q$ is rational, and deduces that $X=Y$, where $X$ is an integer and $Y$ is a nonzero number with $|Y| < 1$. To accomplish this, we typically seek a rapidly converging approximation to $\alpha$, whose fractional part lies between 0 and 1 even after scaling up by $q$ (or a suitable multiple of $q$).

The proof presented here is based on Jack D'Aurizio's answer to another question, in which he noted that the function $1/\sqrt{2-x}$ is well approximated by Legendre polynomials. The Legendre polynomials $P_n(x)$ form an orthogonal basis for the Hilbert space of square-integrable functions on a compact interval, and may be defined via the generating function $${1 \over \sqrt{1-2xt^2+t^4}} = \sum_{n=0}^\infty P_n(x)\, t^{2n}.$$ The Legendre polynomials usually live on the interval $[-1,1]$, but let us follow D'Aurizio and work over the interval $[0,1]$, which means we should consider $P_n(2x-1)$ rather than $P_n(x)$ itself. We consider the integral $$I_n := \int_0^1 {P_n(2x-1)\over \sqrt{2-x}}dx,$$ because (up to normalization) it is the $n$th coefficient in the Legendre polynomial expansion of $1/\sqrt{2-x}$. We should therefore not be too surprised to find that it is small; more specifically:

Lemma 1. For sufficiently large $n$, $I_n$ is an exponentially small nonzero number.

Lemma 1 (or something analogous) would hold for many other choices of orthogonal polynomials, but what's nice about Legendre polynomials is the following remarkable fact.

Lemma 2. $(4n+2) I_n = A_n\sqrt{2} + B_n$ for some integers $A_n$ and $B_n$.

Lemma 1 and Lemma 2 together imply that $\sqrt{2}$ is irrational. For suppose $\sqrt{2} = p/q$ for positive integers $p$ and $q$. Then by Lemma 2, $q(2n+1)I_n$ is an integer. But by Lemma 1, we can choose $n$ large enough that $q(2n+1)I_n$ is a nonzero number with absolute value less than 1, which is a contradiction.

Here is a sketch of a proof of Lemma 2, which is the trickier one. The generating function for Legendre polynomials tells us that $$\sum_{n=0}^\infty I_n t^{2n} = \int_0^1 {dx\over \sqrt{(2-x)(1-(4x-2)t^2+t^4}}.$$ With the help of a computer algebra package, we find that the integral on the right-hand side of the above equation equals $${1\over 2t}\ln\left({1+4t+2t^2-4t^3+t^4\over 1-4\sqrt{2} t + 10t^2-4\sqrt{2}t^3+t^4}\right).$$ In the Taylor expansion of the logarithm, the coefficient of $t^m$ will have the form $(a_m \sqrt{2} + b_m)/m$ for some integers $a_m$ and $b_m$; taking into account the $1/2t$ in front of the logarithm, we see that the coefficient of $t^{2n}$ (i.e., $I_n$) has the form $(a_{2n+1} \sqrt{2} + b_{2n+1})/(4n+2)$, which proves Lemma 2.

For a sketch of a proof of Lemma 1, we first note that $P_n(x)$ is orthogonal to all polynomials of degree less than $n$, so the value of the integral is unchanged if we replace $1/\sqrt{2-x}$ with $f_n(x)$, defined to be its Taylor series minus the terms with degree less than $n$. Then $I_n$ can be upper-bounded using Cauchy–Schwarz; the norm $||P_n(2x-1)||_2 = 1/\sqrt{2n+1}$ and the norm $||f_n(x)||_2$ is exponentially small.

Of course this is an incredibly convoluted way of proving that $\sqrt{2}$ is irrational, but the point is that if we replace $1/\sqrt{2-x}$ with $e^x$ (respectively, $\sin x$), then a very similar line of argumentation yields that $e^r$ (respectively, $\pi$) is irrational. I have written up the details of these latter two facts in an expository article that was inspired by Kostya_I's answer to a MathOverflow question.

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