Famous Theorems of Mathematics/π is irrational

The mathematical constant ${\displaystyle \pi =3.141592\ldots }$ (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.

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

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

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

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

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

Now let us define ${\displaystyle N=\int \limits _{0}^{\pi }f(x)\sin(x)dx}$. The integrand is positive for all ${\displaystyle x\in (0,\pi )}$ and so ${\displaystyle N>0}$.

Repeated integration by parts gives:

${\displaystyle N=-f^{(0)}\!(x)\cos(x){\bigg |}_{0}^{\pi }+f^{(1)}\!(x)\sin(x){\bigg |}_{0}^{\pi }+f^{(2)}\!(x)\cos(x){\bigg |}_{0}^{\pi }-\cdots \pm f^{(2n)}\!(x)\cos(x){\bigg |}_{0}^{\pi }\mp \int \limits _{0}^{\pi }f^{(2n+1)}\!(x)\cos(x)dx}$

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

For all ${\displaystyle 0\leq k\leq 2n}$ the functions ${\displaystyle f^{(k)}\!(x),\sin(x),\cos(x)}$ take integer values at ${\displaystyle x=0,\pi }$, hence ${\displaystyle 1\leq N\in \mathbb {N} }$.

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

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

By the triangle inequality for integrals we get

${\displaystyle 1\leq N=\int \limits _{0}^{\pi }\sin(x)f(x)dx\leq \int \limits _{0}^{\pi }{\frac {(\pi a)^{n}}{n!}}dx=\pi {\frac {(\pi a)^{n}}{n!}}}$

But ${\displaystyle \lim _{n\to \infty }\pi {\frac {(\pi a)^{n}}{n!}}=0}$, hence for sufficiently large ${\displaystyle n}$ we get ${\displaystyle 1\leq N<1}$. A contradiction.

${\displaystyle \square }$

Conclusion: ${\displaystyle \pi }$ is an irrational number.

${\displaystyle \blacksquare }$