This post follows on from the previous one.

The following method was introduced by Hermite, in a more general form, in the lead-up to his proof that e is transcendental. It can be found on pp 153 and 154 of his Oeuvres III. Although Hermite’s proof postdates that of Lambert by more than a century, I give it first because it is somewhat simpler.

The idea is to produce an expression that enables us to use the following:

Let x be a real number. If for any given real number 1>ε>0, there exist integers h and k such that 0 <|kx-h|<ε, then x is irrational.

Consider the integral ∫e^{ht}F_{n}(t)dt with h a positive integer and F_{n}(t) a polynomial in t with integer coefficients, and whose degree is 2n. If we integrate repeatedly by parts we obtain

If we make this a definite integral from t = 0 to t = 1 we obtain the following expression, the right hand side of which has the desired form

with h_{n}, k_{n} integers.

Now suppose we take F_{n}(t) = t^{n}(1 – t)^{n}. This polynomial has the properties that 1 ≥ F_{n}(t) ≥ 0 if 1 ≥ t ≥ 0, and (see Lemma below) and are divisible by n!. Thus we have

with k’_{n}(h) = k_{n}(h)/n! and h’_{n}(h) = h_{n}(h)/n! integers. The expression on the left is greater than zero. Since the integrand is positive but less than e^{h}, the integral is also positive but less than e^{h}. Suppose n_{1} is the smallest integer greater than h^{2}. Therefore for n > n_{1} the expression on the left is equal to AB^{n} with A, B positive constants and B < 1. This expression can be made arbitrarily small by taking n large enough, so by the theorem above, e^{h} is irrational if h is a positive integer.

We can easily extend this result to show that e^{x} is irrational if x is non-zero and rational. Let x be positive and rational and equal to a/b, with a, b positive integers. Then if e^{x} = e^{a/b} is rational, then (e^{a/b})^{b} = e^{a} is rational so by the counterpositive, if e^{a} is irrational then e^{x} is irrational. If e^{-x} = 1/e^{x} is rational then e^{x} is rational so by the counterpositive, if e^{x} is irrational then e^{-x} is irrational.

By the following:

Let f(x) be a function whose argument is a set of real numbers and let it have an inverse function f ^{-1}(y) where y is the set of numbers f(x). Let f be such that if x is rational then f(x) is irrational. Then if y is rational f ^{-1}(y) is irrational

taking f(x) = e^{x} and the set x to be the set of real numbers excluding zero, f^{-1}(y) = ln y and y the set of real numbers greater than zero but excluding one, then ln y is irrational if y is rational.

Lemma: Let . Then and are divisible by n!

Proof: In the first case, since the term of least degree in is t^{n}, when k < n. When 2n ≥ k ≥ n, , where f(t) is either zero or has a factor t. Thus is divisible by k! and hence by n!

In the second case, since and , we have when k < n. When 2n ≥ k ≥ n, with f(1 – t) either zero or having a factor 1 – t. Hence is divisible by n!

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