# Irrationality of a common constant

It is rather embarrassing that, as a professional mathematician, that I didn't know how to prove the irrationality of $\pi$. The most common proof of this fact is this argument which PlanetMath cites to Hardy and Wright, but which I've seen also referred to in Bourbaki. I will discuss a slightly different proof here. The version that I read is from Zhou and Markov, though the idea is due originally to Lambert in 1761.

Theorem
If $r \in \mathbb{Q} \setminus \{0\}$ is a non-zero rational number, then the value $\tan(r)$ is irrational.
Before we continue to give the proof of the theorem, observe that the contrapositive of the theorem gives that if $\tan(x)$ is a rational number, then $x$ is either 0 or irrational. Since $\tan(\pi/4) = 1$, an immediate corollary is that $\pi/4$, and hence $\pi$, is irrational.

The following proof is an expanded version of the one given by Zhou and Markov.

Proof:

First we observe that implicit in the theorem is that $\tan(r)$ is defined for any rational $r$. On the other hand, we know that the tangent is not defined at $k\pi / 2$ where $k$ is an integer. But we note that if we can prove the theorem for $r \in \mathbb{Q} \setminus \{ k\pi / 2 \}$ where $k \in \mathbb{Z}$ (in other words, for rational numbers that are not the forbidden values), we have sufficient knowledge to show that $\pi/4$ is irrational. With this we can bootstrap our initial assumption: since $\pi/4$ is irrational, the set of all rational numbers not equal to $k\pi / 2$ is precisely the set of all rational numbers not equal to 0.

With this observation, we set out to prove the "weakened" statement assuming $r \in \mathbb{Q} \setminus \{ k\pi / 2 \}$. We prove by contradiction. Assume that $r = a/b$ is a fixed rational number with $a,b$ nonzero integers, and assume that $\tan(r) = p/q$ where $p,q$ are also integers.

Consider the following sequence of functions: $f_n(x) = \frac{(r^2 - x^2)^n}{2^n n!}$ and the associated integrals: $I_n = \int_0^r f_n(x) \cos(x) dx$

First we observe that \begin{equation} |b^n I_n| \leq \int_0^r \frac{(a^2 - b^2x^2)^n}{(2b)^n n!} |\cos(x)| dx \leq \int_0^r \frac{a^{2n}}{(2b)^n n!} dx = \frac{2a^{2n+1}}{(2b)^{n+1} n!} \end{equation} which implies that $\lim_{n\to\infty} b^nI_n = 0$. This is because of the strong factorial decay in the denominator of the definition of $f_n(x)$. Using the convention that $0! = 1$, we see that $f_0(x) = 1$ and $f_1(x) = (r^2 - x^2) / 2$. A computation then shows that \begin{aligned} I_0 &= \int_0^r \cos(x)dx = \sin(r) \newline I_1 &= \int_0^r \frac{r^2-x^2}{2} \cos(x) dx = \left.\frac{r^2-x^2}{2}\sin(x)\right]^r_0 + \int_0^r x\sin(x) dx \newline &= 0 - \left.x\cos(x)\right]^r_0 + \int_0^r\cos(x) dx = \sin(r) - r\cos(r) \end{aligned} where in the computation for $I_1$ we integrated by parts twice.

Secondly we can observe a functional recursion for the numbers $I_n$. Assume $n \geq 2$, then $\begin{gathered} f'_n(x) = \frac{(r^2 - x^2)^{n-1}}{2^n (n-1)!} \cdot (- 2x) = - x f_{n-1}(x) \newline f''_n(x) = (-x f_{n-1}(x))' = - f_{n-1}(x) + x^2 f_{n-2}(x) \end{gathered}$ Observing that $x^2f_{n-2}(x) = - (r^2 - x^2)f_{n-2}(x) + r^2 f_{n-2}(x) = - 2(n-1) f_{n-1}(x) + r^2f_{n-2}(x)$ we arrive at the equation $f''_n(x) = r^2f_{n-2}(x) - (2n-1) f_{n-1}(x) .$ We plug this expression into $I_n$ and integrate by parts \begin{equation} I_n = \int_0^r f_n \cos dx = f_n\sin ]_0^r - \int_0^r f'_n\sin dx = f'_n\cos ]_0^r - \int_0^r f''_n\cos dx. \end{equation} Noting that $f_n(r) = 0$ for $n > 0$, we have also that $f'_n(0) = f'_n(r) = 0$ for $n > 1$. Therefore if $n \geq 2$, we have that \begin{equation} I_n = (2n-1)I_{n-1} - r^2 I_{n-2} \end{equation} Comparing with the expressions already derived for $I_0,I_1$, the recursion relation implies immediately that $I_n = u_n \sin(r) + v_n \cos(r)$ where $u_n,v_n$ are polynomials of $r$ with integer coefficients, and degree at most $n$. (We can prove this by induction: the recursion relation implies that if the statement is true for $I_{n-1},I_{n-2}$, then it is true for $I_n$. The starting hypothesis is easily checked for $I_0,I_1$.)

A further implication of the recursion relation is that if two consecutive terms $I_{n}, I_{n-1}$ are both zero, the entire sequence must all vanish identically. So our knowledge that $I_0, I_1$ are not both vanishing implies that infinitely many of the terms $I_n$ are non-zero.

Now, for our fixed $r \neq k\pi / 2$, we know that $\cos(r) \neq 0$, so we can divide the expression $I_n / \cos(r) = u_n \tan(r) + v_n .$ By our initial hypothesis, if we multiply through by the integer $q$, we have $q I_n / \cos(r) = p u_n + q v_n.$ Lastly, using that $u_n,v_n$ are polynomials of $r$ of degree at most $n$, we see that if we multiply through by $b^n$, the expressions $b^nu_n,b^nv_n$ now are polynomials of $a,b$ of degree at most $n$ with integer coefficients, and so are integers themselves. In other words, we have shown that for any $n$, \begin{equation} \frac{q b^n I_n}{\cos(r)} \in \mathbb{Z}. \end{equation} On the other hand, observe that $q / \cos(r)$ is just a fixed constant. Since we know that $b^nI_n$ tends to zero as $n$ grows to infinity, we can choose some large $N$ such that for any $n > N$, we have $|b^n I_n| \leq |\cos(r) / (2q)|$. Furthermore, since the sequence $I_n$ has infinitely many non-vanishing terms, we can choose some number $m > N$ such that \begin{equation} 0 < \left|\frac{q b^m I_m}{\cos(r)}\right| < 1 \end{equation} which is clearly absurd as we've also shown that the term in the middle must be an integer! Therefore our initial assumption must be false, and the theorem is proved. ##### Willie WY Wong
###### Associate Professor

My research interests include partial differential equations, geometric analysis, fluid dynamics, and general relativity.