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 $a,b\in \mathbb {N}$ such that $\pi ={\frac {a}{b}}$.

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

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

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

${\begin{aligned}f^{(k)}\!(x)=(-1)^{k}f^{(k)}\!(\pi -x)&={\begin{cases}\displaystyle \sum _{m\,=\,n}^{2n}{\frac {k!}{n!}}{\binom {m}{k}}c_{m}x^{m-k}&:0\leq k\leq n-1\\[5pt]\displaystyle \sum _{m\,=\,k}^{2n}{\frac {k!}{n!}}{\binom {m}{k}}c_{m}x^{m-k}&:n\leq k\leq 2n\end{cases}}\\[5pt]f^{(k)}\!(0)=(-1)^{k}f^{(k)}\!(\pi )&={\begin{cases}0&:0\leq k\leq n-1\\[5pt]\displaystyle {\frac {k!}{n!}}c_{k}&:n\leq k\leq 2n\end{cases}}\end{aligned}}$

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 $0\leq k\leq 2n$ 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

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

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.