The Irrationality of \(\pi\)

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

Theorem 1. The constant \(\pi\) is irrational.

All proofs of this result (that I am aware of) begin with some magic. For us, the magic lies in considering the function \[f(X):=\cos\big(2\pi\sqrt{1+4X}\big).\] The theorem will follow by understanding the coefficients of the Taylor expansion \(\sum_na_nX^n\) of \(f(X)\) around \(X=0\).

Lemma 2. 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 \(X\in[-1/4,\infty)\) 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. \(\blacksquare\)

It remains to find \(\pi\). We will do this by computing the Taylor expansion of \[f(X)=\cos\big(2\pi\big(\sqrt{1+4X}-1\big)\big).\]

Lemma 3. For each nonnegative integer \(n\), we have \(n!a_n\in\operatorname{span}_{\mathbb Z}\{1,\pi,\ldots,\pi^n\}\).

Proof. This is another direct calculation. As in Lemma 2, all series converge absolutely for \(X\in[-1/8,1/8]\), which justifies rearrangements. Recall the Taylor expansion \[\sqrt{1+4X}-1=\sum_{\ell=1}^\infty4^\ell\binom{1/2}{\ell}X^\ell.\] (This follows from the Generalized binomial theorem.) Importantly, \(4^n\binom{1/2}n\) is an integer. Plugging into \(\cos\) yields \[f(X)=\sum_{k=0}^\infty(-1)^k\frac{(2\pi)^{2k}}{(2k)!}\left(\sum_{\ell=1}^\infty4^\ell\binom{1/2}{\ell}X^\ell\right)^{2k}.\] We can now extract the coefficient of \(X^n\) by first truncating the series to \(2k\le n\) and \(\ell\le n\). (For example, one can expand the product \((?)^{2k}\) as a Cauchy product of series and then rearrange everything to get a power series in \(X\).) Thus, expanding with the Multinomial theorem shows that the \(X^n\)-term is an integer linear combination of terms of the form \(\frac{\pi^k}{k!}X^n\) where \(k\le n\). The lemma follows. \(\blacksquare\)

Remark 4. Another way to state Lemma 3 is to say that the Taylor expansion of \(f(X)\) lives in the set \[\left\{\sum_{n=0}^\infty\frac{a_n(\pi)}{n!}X^n:a_n\in\mathbb Z[T],\deg a_n\le n\right\}.\] This phrasing is perhaps convenient because this set is a ring. In fact, it is generated over \(\mathbb Z\) by the monomials \(\pi^nX^n/n!\), so it is a “divided powers” ring.

Comparing the two lemmas proves the theorem.

Proof of Theorem 1. Suppose for the sake of contradiction that \(\pi\) is a rational number. The main claim is that \(a_n=0\) for sufficiently large \(n\). This claim quickly produces contradiction: for example, \(f(X)\) is real analytic on \((-1/4,\infty)\), so \(f(X)\) must then be a polynomial on \((-1/4,\infty)\), which is false because \(f(X)\) is non-constant and has infinitely many roots.

It remains to show the main claim. Write \(\pi=p/q\) where \(p\) and \(q\) are nonzero integers. On one hand, by Lemma 3, \(q^nn!a_n\in\mathbb Z\) 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 2, there is an absolute constant \(c>0\) such that \[q^nn!\left|a_n\right|<\frac{c^n}{(2n)!}\] for all sufficiently large integers \(n\). We conclude because \((c/q)^n\cdot n!/(2n)!\) tends to \(0\) as \(n\to\infty\). \(\blacksquare\)

Remark 5. Dimitrov has suggested a refinement of this argument (using André’s algebraicity criterion) which can show that \(\pi\) is trascendental.