# $\pi$ and $\pi^{2}$ are irrational

###### Theorem 1.

$\pi$ and $\pi^{2}$ are irrational.

###### Proof.

For any strictly positive integer $n$ ,$x\in(0,1)$ we define:

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

where $c_{m}$ are integers. For $0 we have

 $0 (1)

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

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

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

We have that $f(0)=0$ and $f^{(m)}(0)=0$ if $m or $m>2n$. But, if $n\leq m\leq 2n$, then

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

an integer. Hence $f(x)$ and all its derivates take integral values at $x=0$.Since $f(1-x)=f(x)$, the same is true at $x=1$

so that $G(0)$ and $G(1)$ are integers. We have

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

Hence

 $\pi\int_{0}^{1}a^{n}\sin{\pi x}f(x)dx=[\frac{G^{\prime}(x)\sin{\pi x}}{\pi}-G(% x)\cos{\pi x}]_{0}^{1}$
 $=G(0)+G(1),$

witch is an integer. But by equation 1,

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

For a large enough $n$, we obtain a contradiction.

For any integer $n$, if $a^{n}$ is irrational then a is irrational http://planetmath.org/?op=getobj&from=objects&id=5779(proof), and since $\pi^{2}$ is irrational $\sqrt{\pi^{2}}=\pi$ is also irrational. ∎

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

## References

• 1 G.H.Hardy and E.M.Wright An Introduction to the Theory of Numbers, Oxford University Press, 1959

Title $\pi$ and $\pi^{2}$ are irrational piAndpi2AreIrrational 2013-03-22 14:44:00 2013-03-22 14:44:00 mathcam (2727) mathcam (2727) 15 mathcam (2727) Theorem msc 51-00 msc 11-00