pi is irrational

 $\pi$ is irrational, due to Bourbaki.

 This proof uses the characterization of $\pi$ as the smallest positive zero of the sine function.

First of all, assume that $\pi$ is rational, i.e., $\pi=\frac{a}{b}$ for some integers $a$ and $b$, $b$ nonzero. Set $$ f(x)=\frac{x^{n}(a-bx)^{n}}{n!} $$ for $x\in \mathbb{R}$ and $$ A_{n}=\int_{0}^{\pi}f(x)\sin xdx=b^{n}\int_{0}^{\pi}\frac{x^{n}(\pi-x)^{n}}{n!}\sin xdx. $$

 The steps are given below.

Claim 1. $A_{n}$ is a positive integer.

Claim 2. $A_{n}$ converges to 0 as $n \to \infty$.

Claim 3. Therefore, it is absurd if $\pi$ is rational.

Fill the details, and complete the proof.

 You may consider for the first step that $$ A_{n}=\int_{0}^{\pi}f(x)\sin xdx=[-f(x)\cos (x)]_{0}^{\pi}-[-f^{(1)}(x)\sin (x)]_{0}^{\pi}+\cdots \pm [f^{(2n)}(x)\cos (x)]_{0}^{\pi} \pm \int_{0}^{\pi}f^{(2n+1)}(x)\cos xdx $$ and $f(x)$ is expanded with integral coefficients over $n!$. For the second step, use the inequality $$ x^{n}(\pi -x)^{n}\leq (\frac{\pi}{2})^{n}$$ for $x\in [0,\pi].$