# One of my Favorite Proofs

Proof that $\pi$ is irrational.

Assume $\pi$ is rational, that is, assume it is of the form $\frac{a}{b}$ where $a$ and $b$ are both positive integers. Let

\begin{align} f(x) &= \frac{x^n (a-bx)^n}{n!} \\ F(x) &= f(x) + \cdots + (-1)^j f^{[2j]}(x) + \cdots + (-1)^n f^{[2n]}(x) \end{align}

where $f^{[k]}$ denotes $k$-th derivative of $f$.

1. $f(x)$ has integer coefficients except $\frac{1}{n!}$
2. $f(x) = f(\pi - x)$
3. $0 \leq f(x) \leq \frac{\pi^n a^n}{n!}$ for $0 \leq x \leq \pi$
4. For $0 \leq j < n$, the $j$-th derivative of $f$ equals 0 at 0 and $\pi$
5. For $j \geq n$, the $j$-th derivative of $f$ is integer at 0 and $\pi$ (from 1. above)
6. $F(0)$, $F(\pi)$ is integer (from 4., 5. above)
7. $F(x) + F''(x) = f(x)$
8. $(F'(x) \sin x - F(x) \cos x)' = f(x) \sin x$ (from 7. above)
9. $\int_0^\pi f(x) \sin x$ is an integer
10. For large $n$, this integral is between 0 and 1 (from 3. above)

Contradiction. So $\pi$ is irrational.