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$\pi$ 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:
$f=f(x)=\frac{x^{n}(1x)^{n}}{n!}=\frac{1}{n!}\sum_{{m=n}}^{{2n}}c_{m}x^{m}$ 
where $c_{m}$ are integers. For $0<x<1$ we have
$0<f(x)<\frac{1}{n!}$  (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
$G(x)=b^{n}[\pi^{{2n}}f(x)\pi^{{2n2}}f^{{\prime\prime}}(x)+\pi^{{2n4}}f^{{(4% )}}(x)...+(1)^{n}f^{{(2n)}}(x)].$ 
We have that $f(0)=0$ and $f^{{(m)}}(0)=0$ if $m<n$ 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 $f(1x)=f(x)$, 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 (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
See also

The MacTutor History of Mathematics Archive, A history of Pi

The MacTutor History of Mathematics Archive, Johann Heinrich Lambert
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Contains proof by mathwizard ✓
Last sentence by alozano ✓
portability by jac ✘
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Comments
\pi is irrational
To make it clearer that \pi is irrational because \pi^2 is irrational, have a look at http://planetmath.org/?op=getobj&from=objects&id=5779
Re: \pi is irrational
Sincerely, what we won with this Mr. Gunnar? I think that anyway we must read the Lambert's hard paper.
$\pi$ and $\pi^2$ are irrational
In general,
$\cos{x} \neq \pi\cos{\pix}$
and
$\sin{x} \neq \sin{\pix}$
Re: \pi is irrational
Well of course it's interesting to read that paper. I'll start googling for it.