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 . 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 is a non-zero rational number, then the value is irrational.
Before we continue to give the proof of the theorem, observe that the contrapositive of the theorem gives that if is a rational number, then is either 0 or irrational. Since , an immediate corollary is that , and hence , 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 is defined for any rational . On the other hand, we know that the tangent is not defined at where is an integer. But we note that if we can prove the theorem for where (in other words, for rational numbers that are not the forbidden values), we have sufficient knowledge to show that is irrational. With this we can bootstrap our initial assumption: since is irrational, the set of all rational numbers not equal to is precisely the set of all rational numbers not equal to 0.
With this observation, we set out to prove the “weakened” statement assuming . We prove by contradiction. So assume that is a fixed rational number with nonzero integers, and assume that where are also integers. We will show that this is logically impossible.
The key is the following sequence of functions:
and the associated integrals:
First we observe that
which implies that . This is because of the strong factorial decay in the denominator of the definition of . Using the convention that , we see that and . A computation then shows that
where in the computation for we integrated by parts twice.
Secondly we can observe a functional recursion for the numbers . Assume , then
we arrive at the equation
We plug this expression into and integrate by parts
Noting that for , we have also that for . Therefore if , we have that
Comparing with the expressions already derived for , the recursion relation implies immediately that where are polynomials of with integer coefficients, and degree at most . (We can prove this by induction: the recursion relation implies that if the statement is true for , then it is true for . The starting hypothesis is easily checked for .)
A further implication of the recursion relation is that if two consecutive terms are both zero, the entire sequence must all vanish identically. So our knowledge that are not both vanishing implies that infinitely many of the terms are non-zero.
Now, for our fixed , we know that , so we can divide the expression
by our initial hypothesis, if we multiply through by the integer , we have
Lastly, using that are polynomials of of degree at most , we see that if we multiply through by , the expressions now are polynomials of of degree at most with integer coefficients, and so are integers themselves. In other words, we have shown that for any ,
On the other hand, observe that is just a fixed constant. Since we know that tends to zero as grows to infinity, we can choose some large such that for any , we have . Furthermore, since the sequence has infinitely many non-vanishing terms, we can choose some number such that
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.