Irrationality of $\sqrt{2}$ invoking some fact of definite integrals

Question

Few months ago I tried to prove that $\sqrt{2}$ is irrational using some object/fact from (elementary, say us as simplest than is possible) integral calculus.

It is known that in some proofs of the irrationality of $\sqrt{2}$ is used the quantities $\sqrt{2}\pm a$, where $a$ is an integer (see 1).

Again I would like to ask about it (as an open question and I hope that the question is anserwerable).

Question. Is it possible to use any fact from integral calculus (you can to combine with the definition of Riemann integral, recurrence relations, calculation of areas, uses of the infinite...) to get a proof of the irrationaly of $\sqrt{2}$ (even if it is presented as more artificious than the ones that are in the literature)? Many thanks.

My last attempt was to write expressions as $$\int_0^1\frac{1}{\sqrt{x}\sqrt{2-\sqrt{x}}}dx=4(\sqrt{2}-1),$$ and (playing with this kind of integrals) defining the integer $4$ as $$\mathcal{J}=\int_0^1\frac{1}{\sqrt{x}\sqrt{1-\sqrt{x}}}dx$$ then $$\sum_{k=0}^{\mathcal{J}}\binom{\mathcal{J}}{k}\left(\int_0^1\frac{1}{\sqrt{x}\sqrt{\sqrt{\mathcal{J}}-\sqrt{x}}}dx\right)^k\cdot \mathcal{J}^{-k}=\mathcal{J}.$$

References:

1 Square root of 2 is irrational, Cut The Knot.

Answer 1

You may consider the integrals $$ I_n = \int_{0}^{1}\frac{P_n(2x-1)}{\sqrt{2-x}}\,dx $$ which by the generating function for Legendre polynomials fulfill $$ I_n = \frac{2}{2n+1}\left(\sqrt{2}-1\right)^n = \frac{2}{2n+1}(A_n\sqrt{2}-B_n),\qquad A,B\in\mathbb{Z}^+.$$ They provide approximations of $\sqrt{2}$ as $\frac{B_n}{A_n}$, and the coefficients $A_n$, given by A001653, fulfill the recurrence relation $A_{n+2}=6 A_{n+1}-A_n$, hence they behave like $\frac{2+\sqrt{2}}{4}(3+2\sqrt{2})^n$ for large values of $n$. In particular

$$ \left|\sqrt{2}-\frac{B_n}{A_n}\right|=\left|\left(n+\frac{1}{2}\right)\frac{I_n}{A_n}\right|\leq \frac{C}{A_n^{3/2}} $$ holds for infinite values of $n$ and $\sqrt{2}$ cannot be a rational number.

This is not really different from showing $\sqrt{2}=[1;2,2,2,2,\ldots]$, since the rational approximations of $\sqrt{2}$ found through $I_n$ are related to Pell numbers and to the convergents of such (infinite) continued fraction.

The argument above works also if applied to the sequence of rational approximations $\frac{b_n}{a_n}$ got by applying Newton's method to the polynomial $x^2-2$, with starting point $x_0=\frac{3}{2}$, for instance.

Answer 2

As usual, Jack D'Aurizio's answer is quite good.

I will try to generalize an aspect of his answer.

For a positive real $x$, suppose there a sequence of rationals $\frac{a_n}{b_n}$ with $b_n$ strictly increasing such that $|x-\frac{a_n}{b_n}| \lt \frac1{b_n^{1+c}} $ for all large enough $n$ for some $c > 0$.

Then $x$ is irrational or $x = \frac{a_n}{b_n}$ for all large enough $n$.

Proof.

Suppose $x$ is rational so $x = \frac{r}{s}$ for some positive integers $r$ and $s$. Then $|\frac{r}{s}-\frac{a_n}{b_n}| \lt \frac1{b_n^{1+c}} $. Multiplying by $sb_n$, this becomes $|rb_n-sa_n| \lt \frac{s}{b_n^c} $.

The left side of this is an integer. If we choose $n$ large enough so that $\frac{s}{b_n^c} \le 1$ (i.e. $b_n \ge s^{1/c} $), then the left side must be zero, so $x = \frac{a_n}{b_n}$ for all large enough $n$.

In Jack's answer, $\left|\sqrt{2}-\frac{B_n}{A_n}\right| =\left|\left(n+\frac{1}{2}\right)\frac{I_n}{A_n}\right| \ne 0$ with $c = \frac12$, so $\sqrt{2}$ is irrational.

Answer 3

Following is an alternative presentation of Jack D'Aurizio's answer that clarifies its relationship with Kostya_I's answer to a related question that I posed on MathOverflow. Kostya_I explained how Niven's proof of the irrationality of $\pi$ could be viewed as exploiting the fact that $\sin x$ is well approximated by Legendre polynomials. Similarly, Jack D'Aurizio's proof relies on the fact that $1/\sqrt{2-x}$ is well approximated by Legendre polynomials.

The Legendre polynomials $P_n(x)$ 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 Jack D'Aurizio and work over the interval $[0,1]$, which means we should consider $P_n(2x-1)$ rather than $P_n(x)$ itself. The $n$th coefficient in the Legendre polynomial expansion of $1/\sqrt{2-x}$ (up to a normalization factor) is $$I_n := \int_0^1 {P_n(2x-1)\over \sqrt{2-x}}dx.$$ The irrationality of $\sqrt{2}$ will follow from two Lemmas.

Lemma 1. $(4n+2) I_n = A_n\sqrt{2} + B_n$ for some integers $A_n$ and $B_n$.

Lemma 2. For sufficiently large $n$, $I_n$ is an exponentially small nonzero number.

For suppose $\sqrt{2} = p/q$ for positive integers $p$ and $q$. Then by Lemma 1, $q(2n+1)I_n$ is an integer. But by Lemma 2, 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 1. 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 1.

The proof of Lemma 2 proceeds along the same lines as in Kostya_I's answer (I spell out the calculation in more detail here), so I won't belabor it here.

(Note: It seems that D'Aurizio is using a different normalization of Legendre polynomials, because with my definitions, it is not the case that $I_n = 2(\sqrt{2}-1)^n\!/(2n+1)$, as D'Aurizio states. But this difference in normalization does not affect the main argument.)