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

Theorem 1.

$\pi$ and ${\pi }^{\mathrm{2}}$ 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 we have

(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

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

We have that $f(0)=0$ and $f^{(m)}(0)=0$ if or $m>2n$. But, if $n\le m\le 2n$, 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]$$=$$[G^{\prime \prime }(x)+{\pi }^{2}G(x)]\sin \pi x$

$=$$b^n{\pi }^{2n+2}f(x)\sin \pi x$

$=$${\pi }^2{a}^n\sin \pi xf(x).$

Hence

$$\pi {\int }_0^1{a}^n\sin \pi xf(x)𝑑x={[\frac{G^{\prime }(x)\sin \pi x}{\pi }-G(x)\cos \pi x]}_0^1$$

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

witch is an integer. But by equation 1,

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.

See also

• The MacTutor History of Mathematics Archive,http://www-gap.dcs.st-and.ac.uk/ history/HistTopics/Pi_through_the_ages.htmlA history ofPi

• The MacTutor History of Mathematics Archive,http://www-history.mcs.st-andrews.ac.uk/Mathematicians/Lambert.htmlJohann HeinrichLambert

• http://numbers.computation.free.fr/Constants/Miscellaneous/irrationality.htmlIrrationality proofs