Proof that Pi is Irrational

Suppose \(\pi = a/b\). Define

\[f(x)=\frac{x^n(a-bx)^n}{n!}\]

and

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

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

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

whence

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

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.


Ben Lynn blynn@cs.stanford.edu 💡