Proofs that pi and e are irrational

Two short and well-known proofs of irrationality.

Timothy Chow wrote a great paper A well-motivated proof that pi is irrational, where he explains a proof of irrationality of \(e\) (attributed to Fourier) and a proof of irrationality of \(\pi\), which originally appeared in the I. Niven’s brilliant one-page article.

In this post I’ll just recall both proofs, so I can remember them: original articles will be more informative.

Irrationality of \(e\)

Suppose that \(e=a/b\) for positive integers \(a\) and \(b\). We can write \[ e = \sum_{n=0}^{\infty} \frac{1}{n!} \]

and reach a contradiction by studying an integer:

\[ 0 < b!\cdot e - b!\sum_{n=0}^b \frac{1}{n!} = \sum_{k=1}^{\infty} \frac{b!}{(b+k)!} < \sum_{k=1}^{\infty} \frac{1}{(b+1)^k} = \frac{1}{b} \le 1. \]

Hence, this integer would lie strictly between \(0\) and \(1\), what is impossible.

Irrationality of \(\pi\)

Suppose that \(\pi=a/b\) for positive integers \(a\) and \(b\). Let \(n\) be be a positive integer and define \[ f(x)=\frac{x^n(a-bx)^n}{n!} \]

The main idea of I. Niven’s proof is to show that the definite integral \[ I := \int \limits_0^\pi f(x) \sin x\, \mathrm{d}x \]

is an integer. But for \(0 < x < \pi\) there are bounds \[ 0 < f(x) \sin x < \frac{\pi^n\cdot (a-0)^n}{n!} = \frac{ (\pi a)^n }{n!}, \]

so that for sufficiently large \(n\) we have \(0 < I < 1\).

Hence, the whole difficulty lies in proving that \(I \in \mathbb Z\).

Define \[ F(x) = f(x) - f''(x) + f^{(4)}(x) - \cdots + (-1)^n f^{(2n)}(x), \]

where \(f^{(k)}\) is the \(k\)-th derivative of \(f\). We have

\[\begin{align*} \frac{\mathrm{d}}{\mathrm{d} x}\left( F'(x) \sin x - F(x) \cos x \right) &= F''(x) \sin x + F'(x) \cos x - F'(x) \cos x + F(x) \sin x\\ &= (F(x) + F''(x)) \sin x = f(x) \sin x, \end{align*}\] so that \[ \int\limits_0^\pi f(x) \sin x\, \mathrm{d}x = \big[F'(x) \sin x - F(x) \cos x\big]_0^\pi = F(0) + F(\pi). \]

We can prove that \(F(0) + F(\pi)\) is an integer by proving that all derivatives \(f^{(k)}_n(0)\) and \(f^{(k)}_n(\pi)\) are integers.

First, note that \(f_n(x) = f_n(a/b - x) = f_n(\pi - x)\), so that \(f^{(k)}_n(0) = \pm f^{(k)}_n(\pi)\). To prove that \(f^{(k)}_n(0)\) is an integer, observe that each summand in \(f_n\) has a form \(u x^{n+r} / n!\), where \(u\) is an integer and \(r \ge 0\). If we differentiate it \(k\) times and evaluate it at \(x=0\) we will either have \(0\) (for \(k \neq n+r\)) or \(u \cdot (n+r)! / n!\) for \(k=n+r\), which is an integer.

Links

• I. Niven, A simple proof that \(\pi\) is irrational, 1947: a one-page proof that \(\pi\) is irrational.
• Kostya_I, Exposition of Niven’s proof, 2023: a short note connecting Niven’s proof to orthogonal polynomials.
• T. Chow, A well-motivated proof that pi is irrational, 2024: proof of irrationality of \(\pi\) and \(e^r\) for nonzero rational \(r\).

Appendix: some theorems of transcendental number theory

I often forget the names of major results of transcendental number theory. Let’s quickly recall them:

Lindemann–Weierstrass theorem: if \(\alpha_1, \dotsc, \alpha_n\) are algebraic numbers linearly independent over \(\mathbb Q\), then \(e^{\alpha_1}, \dotsc, e^{\alpha_n}\) are algebraically independent over \(\mathbb Q\).

Gelfond–Schneider theorem: let \(a \in \mathbb C\setminus \{0, 1\}\) be a complex algebraic number. If \(b \in \mathbb C\setminus (\mathbb Q\times \{0\})\) is an irrational complex algebraic number, then \(a^b = \exp(b\log a)\) is transendental (independently on the branches of the functions used).