PlanetMath: $\pi$ and $\pi^2$ are irrational

, $x\in (0,1)$ we define:

$\displaystyle f=f(x)=\frac{x^n(1-x)^n}{n!}=\frac{1}{n!}\sum_{m=n}^{2n}c_mx^m$

where

are integers. For

we have

$\displaystyle 0<f(x)<\frac{1}{n!}$

(1)

For a contradiction, suppose $\pi^2$ is rational, so that $\pi^2=\frac{a}{b}$ , where are positive integers.

For $x\in (0,1)$ let us define

$\displaystyle G(x)=b^n[\pi^{2n}f(x)-\pi^{2n-2}f''(x)+\pi^{2n-4}f^{(4)}(x)-...+(-1)^nf^{(2n)}(x)].$

We have that

and $f^{(m)}(0)=0$ if

. But, if $n \leq m \leq 2n$ , then

$\displaystyle f^{(m)}(0)=\frac{m!}{n!}c_m,$

an integer. Hence

and all its derivates take integral values at

.Since

, the same is true at

so that and are integers. We have

$\displaystyle \frac{d}{dx}[G'(x)\sin{\pi x}-\pi G(x)\cos{\pi x}]$	$\displaystyle =$	$\displaystyle [G''(x)+\pi^2G(x)]\sin{\pi x}$
	$\displaystyle =$	$\displaystyle b^n\pi^{2n+2}f(x)\sin{\pi x}$
	$\displaystyle =$	$\displaystyle \pi^2a^n \sin{\pi x}f(x).$

Hence

$\displaystyle \pi\int_0^1a^n \sin{\pi x}f(x)dx=[\frac{G'(x)\sin{\pi x}}{\pi}-G(x)\cos{\pi x}]_0^1$

$\displaystyle =G(0)+G(1),$

witch is an integer. But by equation 1,

$\displaystyle 0<\pi\int_0^1a^n \sin{\pi x}f(x)dx<\frac{\pi a^n}{n!}<1.$

For a large enough

, we obtain a contradiction.

For any integer , if is irrational then a is irrational (proof), and since $\pi^2$ is irrational $\sqrt{\pi^2}=\pi$ is also irrational. $\qedsymbol$

The irrationality of $\pi$ was Proved by Lambert in 1761. The above proof is not the original proof due to Lambert.

Bibliography