Famous Theorems of Mathematics/π is irrational

The mathematical constant $\pi =3.141592\dots$ (the ratio of circumference to the diameter of the circle) is an irrational number.

In other words, it cannot be expressed as a ratio between two integers.

Proof

Let us assume that $\pi$ is rational, so there exist $a,b\in \mathbb{N}$ such that $\pi =\frac{a}{b}$.

For all $n\in \mathbb{N}$ let us define a polynomial

$f(x)=\frac{x^n(a-bx{)}^n}{n!}=\sum _{m=n}^{2n}\frac{c_m}{n!}{x}^m,:c_m\in \mathbb{Z}$

$f(x)=f(\pi -x)$ and so we get

$\begin{array}{cc}{f}^{(k)}(x)=(-1{)}^k{f}^{(k)}(\pi -x)& =\{\begin{array}{cc}\sum _{m=n}^{2n}\frac{k!}{n!}(\genfrac{}{}{0ex}{}{m}{k})c_m{x}^{m-k}& :0\le k\le n-1\\ \sum _{m=k}^{2n}\frac{k!}{n!}(\genfrac{}{}{0ex}{}{m}{k})c_m{x}^{m-k}& :n\le k\le 2n\end{array}\\ f^{(k)}(0)=(-1{)}^k{f}^{(k)}(\pi )& =\{\begin{array}{cc}0& :0\le k\le n-1\\ \frac{k!}{n!}{c}_k& :n\le k\le 2n\end{array}\end{array}$

Now let us define $A_n=\underset{0}{\overset{\pi }{\int }}f(x)\sin (x)dx$. The integrand is positive for all $x\in (0,\pi )$ and so $A_n>0$.

Repeated integration by parts gives:

$A_n=-f^{(0)}(x)\cos (x){|}_0^{\pi }+f^{(1)}(x)\sin (x){|}_0^{\pi }+f^{(2)}(x)\cos (x){|}_0^{\pi }-\cdots \pm f^{(2n)}(x)\cos (x){|}_0^{\pi }\mp \underset{0}{\overset{\pi }{\int }}{f}^{(2n+1)}(x)\cos (x)dx$

The remaining integral equals zero since $f^{(2n+1)}(x)$ is the zero-polynomial.

For all $0\le k\le 2n$ the functions $f^{(k)}(x),\sin (x),\cos (x)$ take integer values at $x=0,\pi$, hence $A_n$ is a positive integer.

Nevertheless, for all $x\in (0,\pi )$ we get

$\begin{array}{cc}& 0<x<\pi \\ & 0<a-bx<a\\ & 0<\sin (x)<1\end{array}$

hence $0<A_n<\pi \frac{(\pi a{)}^n}{n!}$. But for sufficiently large $n$ we get $0<A_n<1$. A contradiction.

Therefore, $\pi$ is an irrational number.