Pi is irrational

Assume that

$\displaystyle \pi = \frac{a}{b}$

Consider

$\displaystyle f(x)=\frac{x^{n}(a-b x)^{n}}{n !}$

$\displaystyle F(x)=f(x)-f^{(2)}(x)+f^{(4)}(x)-\ldots+(-1)^{n} f^{(2 n)}(x)$

$\displaystyle f(x)=f(\frac{a}{b}-x)$

Note that ${F(0), F(\pi) }$ are both integers. (Small derivatives at 0 vanish and larger derivatives cancel the denominators from the coefficients coming after taking derivatives. And then use symmetry of f for $\displaystyle \pi = \frac{a}{b}.$ )

$\displaystyle \frac{d}{d x}\left(F^{\prime}(x) \sin x-F(x) \cos x\right)=F^{\prime \prime}(x) \sin x+F(x) \sin x=f(x) \sin x$

$\displaystyle \int_{0}^{\pi} f(x) \sin x ~~dx=\left[F^{\prime}(x) \sin x-F(x) \cos x\right]_{0}^{\pi}=F(\pi)+F(0) \in \mathbb{Z}$

But

$\displaystyle \int_{0}^{\pi} f(x) \sin (x) d x \leq \pi \frac{(\pi a)^{n}}{n !} \rightarrow_{n\rightarrow \infty} 0$

Contradiction!

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