Proof that Pi is Irrational

Suppose $\pi = a/b$. Define



\[F(x) = f(x) - f^{(2)}(x) + f^{(4)}(x) - ... + (-1)^n f^{(2n)}(x)\]

for every positive integer $n$.

First note that $f(x)$ and its derivatives $f^{(i)}(x)$ have integral values for $x = 0$, and also for $x = \pi = a/b$ since $f(x) = f(a/b - x)$.

We have

\[\frac{d}{d{x}}(F'(x)sin x - F(x) cos x) = F''(x)sin x + F(x) sin x = f(x) sin x\]


\[ \int^{\pi}_{0} f(x) sin x dx = [F'(x) sin x - F(x) cos x]^\pi_0 = F(\pi) + F(0) \in \mathbb{Z} \]

But for $0 \lt x \lt \pi$, we have

\[ 0 \lt f(x) sin x \lt \frac{\pi^n a^n}{n!} \]

which means we have an integer that is positive but tends to zero as $n$ approaches infinity, which is a contradiction.