The Irrationality of \(\pi\)

We give a short proof that \(\pi\) is irrational, communicated to me by Vesselin Dimitrov.

Theorem. The constant \(\pi\) is irrational.

One way to show that a constant \(\gamma\) is irrational is to exhibit many good rational approximations. This amounts to finding integer linear polynomials \(aX+b\) for which \(a\gamma+b\) is small. We will proceed similarly by exhibiting many integer polynomials \(p(X)\) for which \(p(\pi)\) is small. These polynomials will be found using the coefficients of the power series of \[f(X):=\cos\big(2\pi\sqrt{1+4X}\big).\] (The \(2\pi\) and \(1+4X\) provide a convenient normalization; they are not essential.) The theorem will follow by understanding the coefficients of the Taylor expansion \(\sum_na_nX^n\) of \(f(X)\) around \(X=0\). Let’s collect the needed facts now, in order of difficulty: Lemma shows there are infinitely many nonzero coefficients, Lemma shows that the coefficients are small, and Lemma gives an integrality property involving \(\pi\).

Lemma. The coefficient \(a_n\) is nonzero for infinitely many nonnegative integers \(n\).

Proof. Suppose for the sake of contradiction that \(a_n\ne0\) for only finitely many \(n\). This means that all sufficiently high derivatives of \(f(X)\) vanish at \(X=0\). However, \(\cos(4\pi X)=f\big(X^2-1/4\big)\), so all sufficiently high derivatives of \(\cos(4\pi X)\) must vanish at \(X=1/2\). But the sequence of derivatives of \(\cos(X)\) is \(4\)-periodic and not constantly zero at \(X=2\pi\), so we have a contradiction. \(\blacksquare\)

Lemma. There is an absolute constant \(c>0\) such that for all sufficiently large integers \(n\), \[\left|a_n\right|<\frac{c^n}{(2n)!}.\]

Proof. This is a direct calculation. All series we consider will converge absolutely for \(X\in[-1/8,1/8]\) because factorials dominate exponentials. This justifies our rearrangements; we will not comment on this again. Using the Taylor expansion of \(\cos\) and the Binomial theorem, we have \[\begin{aligned}f(X) &= \sum_{m=0}^\infty(-1)^m\frac{(2\pi)^{2m}(1+4X)^m}{(2m)!} \\{} &= \sum_{n=0}^\infty\left(\sum_{m=n}^\infty(-1)^m\frac{(2\pi)^{2m}\cdot\binom{m}{n}4^n}{(2m)!}\right)X^n.\end{aligned}\] It remains to estimate this “series expansion” of \(a_n\). By the triangle inequality and the esimate \(\binom{m}{n}\le2^m\), \[\left|a_n\right|\le\frac{\left(32\pi^2\right)^n}{(2n)!}\sum_{m=n}^\infty\frac{\left(8\pi^2\right)^{m-n}}{(2m)!/(2n)!}\] The series converges and is decreasing in \(n\), so any \(c>32\pi^2\) will work. \(\blacksquare\)

Lemma. For each nonnegative integer \(n\), we have \(n!a_n\in\operatorname{span}_{\mathbb Z}\{1,\pi,\ldots,\pi^n\}\).

Proof. The main claim is that \[(1+4X)f''(X)+2f’(X)+16\pi^2f(X)\stackrel?=0.\] Verifying this is a direct calculation. Here is a convenient execution: by taking real parts, it is enough to show the same claim for \(g(X):=\exp(2\pi i\sqrt{1+4X})\). But \[\sqrt{1+4X}\cdot g’(X)=4\pi ig(X).\] Taking another derivative and rearranging proves the claim.

We now use the claim to complete the proof. For brevity, define \(b_n:=n!a_n\) so that \(f(X)=\sum_nb_n\frac{X^n}{n!}\). For any nonnegative integer \(n\), taking the \(X^n\)-coefficient in the main claim shows that \[(b_{n+2}+4nb_{n+1})+2b_{n+1}+16\pi^2b_n=0.\] The lemma now follows by induction upon computing the base cases \(b_0=f(0)=1\) and \(b_1=f’(0)=0\). \(\blacksquare\)

Remark. This main claim proves that \(f(X)\) is holonomic. This indicates that \(f(X)\) comes from the study of arithmetic holonomy.

Remark. Another way to prove Lemma is to write \[f(X)=\cos\big(2\pi\big(\sqrt{1+4X}-1\big)\big)\] and plug the Taylor expansion of \(\sqrt{1+4X}-1\) into the Taylor expansion of \(\cos\).

Thus, the coefficients \(n!a_n\) provide many integer polynomials which are small at \(\pi\). This is enough to deduce irrationality.

Proof of Theorem. Suppose for the sake of contradiction that \(\pi\) is rational. We will use Lemma and Lemma to show that \(a_n=0\) for sufficiently large \(n\), which contradicts Lemma.

Write \(\pi=p/q\) where \(p\) and \(q\) are positive integers. On one hand, by Lemma, \(q^nn!a_n\in\mathbb Z\) for all nonnegative integers \(n\). Thus, it is enough to show that \(\left|q^nn!a_n\right|<1\) for sufficiently large \(n\). On the other hand, by Lemma, there is an absolute constant \(c>0\) such that \[\left|q^nn!a_n\right|<q^nn!\cdot\frac{c^n}{(2n)!}\] for all sufficiently large integers \(n\). We conclude because \((cq)^nn!/(2n)!\to0\) as \(n\to\infty\). \(\blacksquare\)

Remark. Dimitrov has suggested a refinement of this argument (using André’s algebraicity criterion) which can show that \(\pi\) is trascendental.

AI Disclosure. ChatGPT was used to locate technical errors. It also suggested using induction to prove Lemma.