In 1873 Charles Hermite showed that e is transcendental. Liouville had earlier shown how to construct such numbers, but Hermite’s was the first proof that an ‘interesting’ mathematical constant was transcendental.

Hermite’s proof is long and complex. It has since been simplified considerably and in particular a short and elegant proof appears in Baker’s book ‘Transcendental Number Theory’. However Baker himself says that the motivation of the proof has been obscured by the process of simplification, and it is necessary to go back to Hermite’s original paper in order to understand it.

I think this is not entirely so but nevertheless in this post I give in outline a version of Baker’s proof that I preface with some references to ideas in Hermite’s paper and I hope in so doing I make the motivation clearer. A more detailed version of this outline is in a pdf here.

As discussed in an earlier post, by repeated integration by parts Hermite obtained an expression of the form

where F_{m}(t)=t^{m}(1-t)^{m} , m and i are positive integers and k_{i}, h_{i} are polynomials with integer coefficients. The expression on the left of the equation is not zero but can be made arbitrarily small in absolute value by taking m sufficiently large, thereby proving that e^{i} is irrational for any integer i.

This suggests that one could obtain in a similar manner a set of equations of the form

…………………………………………. (1)

where k(m) is a non-zero integer not dependent on i and the I_{i}(m) either decrease in absolute value or grow only slowly with increasing m.

Now let q_{0}, q_{1}, … , q_{n} be any set of integers (provided that n > 0 and q_{0}, q_{n} ≠0). If we multiply (1) by q_{i} and sum over i then we have

Then if we can find I_{i}(m) and h_{i}(m) and non-zero k(m) such that for sufficiently large m,

…………………………………… (2)

we have proved e is transcendental because then it is not possible to have

If we use in the integral e^{i-t} rather than e^{it} as used by Hermite we immediately obtain an expression in the desired form (1). Let

where f(t) is a polynomial of degree m with integer coefficients. For i > 0 repeated integration by parts gives

and the equation, which is identical in form to (1), is also correct in a formal sense for I_{0}=0. Now let

Then

and it is not hard to find that for sufficiently large p

where A is positive and independent of p. This puts an upper bound on the left hand side of (2). It is also not too hard to show (pdf) that

cannot be zero if we make p a prime and greater than n (which we are quite at liberty to do). Finally we can show that as long as p>|q_{0}| then

giving a lower bound to the right hand side of (2). But since (p-1)!>A^{p} for some large enough p, then (2) holds and we have proved e is transcendental.