Irrationality of a common constant

by Willie Wong

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 to 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. So 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. We will show that this is logically impossible.

The key is the following sequence of functions:

\displaystyle f_n(x) = \frac{(r^2 - x^2)^n}{2^n n!}

and the associated integrals:

\displaystyle I_n = \int_0^r f_n(x) \cos(x) dx

First we observe that

\displaystyle |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!}

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

\displaystyle I_0 = \int_0^r \cos(x)dx = \sin(r)
\displaystyle 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
\displaystyle \qquad = 0 - \left.x\cos(x)\right]^r_0 + \int_0^r\cos(x) dx = \sin(r) - r\cos(r)

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

\displaystyle f'_n(x) = \frac{(r^2 - x^2)^{n-1}}{2^n (n-1)!} \cdot (- 2x) = - x f_{n-1}(x)
\displaystyle f''_n(x) = (-x f_{n-1}(x))' = - f_{n-1}(x) + x^2 f_{n-2}(x)

Observing that

\displaystyle 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

\displaystyle 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

\displaystyle 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

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

\displaystyle I_n = (2n-1)I_{n-1} - r^2 I_{n-2}

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,

\displaystyle \frac{q b^n I_n}{\cos(r)} \in \mathbb{Z}

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

\displaystyle 0 < \left|\frac{q b^m I_m}{\cos(r)}\right| < 1

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.