Famous Theorems of Mathematics/π is irrational

The mathematical constant $$\pi=3.141592\ldots$$ (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 $$a,b\in\N$$ such that $$\pi=\frac{a}{b}$$.

For all $$n\in\N$$ let us define a polynomial
 * $$f(x)=\frac{x^n(a-bx)^n}{n!}=\sum_{m\,=\,n}^{2n}\frac{c_m}{n!}x^m,\quad:c_m\in\Z$$

$$f(x)=f(\pi-x)$$ and so we get
 * $$\begin{align}f^{(k)}\!(x)=(-1)^kf^{(k)}\!(\pi-x)&=\begin{cases}\displaystyle\sum_{m\,=\,n}^{2n}\frac{k!}{n!}\binom{m}{k}c_mx^{m-k}&:0\le k\le n-1\\[5pt]\displaystyle\sum_{m\,=\,k}^{2n}\frac{k!}{n!}\binom{m}{k}c_mx^{m-k}&:n\le k\le2n\end{cases}\\[5pt]f^{(k)}\!(0)=(-1)^kf^{(k)}\!(\pi)&=\begin{cases}0&:0\le k\le n-1\\[5pt]\displaystyle\frac{k!}{n!}c_k&:n\le k\le2n\end{cases}\end{align}$$

Now let us define $$A_n=\int\limits_0^\pi f(x)\sin(x)dx$$. The integrand is positive for all $$x\in(0,\pi)$$ and so $$A_n>0$$.

Repeated integration by parts gives:
 * $$A_n=-f^{(0)}\!(x)\cos(x)\bigg|_0^\pi+f^{(1)}\!(x)\sin(x)\bigg|_0^\pi+f^{(2)}\!(x)\cos(x)\bigg|_0^\pi-\cdots\pm f^{(2n)}\!(x)\cos(x)\bigg|_0^\pi\mp\int\limits_0^\pi f^{(2n+1)}\!(x)\cos(x)dx$$

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

For all $$0\le k\le2n$$ 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
 * $$\begin{align}&0<x<\pi\\&0<a-bx<a\\&0<\sin(x)<1\end{align}$$

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.