# 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.

## Proof

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

Repeated integration by parts gives:

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

The remaining integral equals zero since ${\displaystyle f^{(2n+1)}\!(x)}$  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 ${\displaystyle A_{n}}$  is a positive integer.

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

{\displaystyle {\begin{aligned}&0

hence ${\displaystyle 0 . But for sufficiently large ${\displaystyle n}$  we get ${\displaystyle 0 . A contradiction.

Therefore, ${\displaystyle \pi }$  is an irrational number.