## Hermite’s proof that exp(x) is irrational for rational non-zero x

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 ∫ehtFn(t)dt with h a positive integer and Fn(t) a polynomial in t with integer coefficients, and whose degree is 2n.  If we integrate repeatedly by parts we obtain

$h^{2n+1}\int e^{ht}F_n(t)dt=e^{ht}\sum_{k=0}^{2n}(-1)^kh^{2n-k}F^{(k)}_n(t)$

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

$h^{2n+1}\int^1_0 e^{ht}F_n(t)dt=e^h\sum_{k=0}^{2n}(-1)^kh^{2n-k}F^{(k)}_n(1)-\\ \indent\sum_{k=0}^{2n}(-1)^kh^{2n-k}F^{(k)}_n(0)=e^hk_n-h_n$

with hn, kn integers.

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

$h\frac{(h^2)^n}{n!}\int^1_0e^{ht}F_n(t)dt=e^hk^\prime_n+h^\prime_n$

with k’n(h) = kn(h)/n! and h’n(h) = hn(h)/n! integers.  The expression on the left is greater than zero.  Since the integrand is positive but less than eh, the integral is also positive but less than eh.  Suppose n1 is the smallest integer greater than h2.  Therefore for n > n1 the expression on the left is equal to ABn 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, eh is irrational if h is a positive integer.

We can easily extend this result to show that ex 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 ex = ea/b is rational, then (ea/b)b = ea is rational so by the counterpositive, if ea is irrational then ex is irrational.  If e-x = 1/ex is rational then ex is rational so by the counterpositive, if ex 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) = ex 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 $F_n(t)=t^n(1 - t)^n$.  Then $F^{(k)}_n(0)$ and $F^{k)}_n(1)$ are divisible by n!

Proof:  In the first case, since the term of least degree in $F_n(t)$ is tn, $F^{(k)}_n(0)= 0$ when k < n.  When 2n ≥ k ≥ n, $F^{(k)}_n(t) = a_kk! + f(t)$, where f(t) is either zero or has a factor t.  Thus $F^{(k)}_n(0)$ is divisible by k! and hence by n!

In the second case, since $F_n(t)=(1-t)^n(1-(1-t))^n$ and $F^{(k)}_n (t)=(-1)^k\frac{d^k}{d(1-t)^k}F_n(t)$, we have $F^{(k)}_n(1)=0$ when k < n.  When 2n ≥ k ≥ n, $F^{(k)}(t) = b_kk!+f(1-t)$ with f(1 – t) either zero or having a factor 1 – t.  Hence $F^{(k)}(1)$ is divisible by n!