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

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.

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.

• What does $P_n$ stand for? – Gerry Myerson Jul 3 '18 at 0:26
• I generalized an aspect of your answer. – marty cohen Jul 3 '18 at 3:33

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.

• If for every $e>0$ there exist $a,b\in \Bbb Z$ such that $0<|x-a/b|<e/|b|$ then $x\not \in \Bbb Q.$... Because if $x=c/d$ with $c,d\in \Bbb Z$ then $0<|x-a/b|$ $<e/|b|\implies 0<|bc-ad|\in$ $\Bbb Z\implies$ $1\leq |bc-ad|\implies$ $(1/|d|)/|b|\leq |x-a/b|,$ so we must have $e\geq 1/|d|.$ – DanielWainfleet Jul 3 '18 at 6:56
• Many thanks for your contribution, I'm going to take remarks in my notebook about the answer and your remarks. – user243301 Jul 3 '18 at 8:30