To start with, it is rather easy to see (as did Fourier in 1815) that \(e = \sum _{k \ge 0} \frac{1}{k!}\) is irrational. Indeed, if we had \(e = \frac{a}{b}\) for integers \(a\) and \(b {\gt} 0\), then we would get

for every \(n \ge 0\). But this cannot be true, because on the right-hand side we have an integer, while the left-hand side with

decomposes into an integral part

and a second part

which is approximately \(\frac{b}{n}\), so that for large \(n\) it certainly cannot be integral: It is larger than \(\frac{b}{n+1}\) and smaller than \(\frac{b}{n}\), as one can see from a comparison with a geometric series:

Now one might be led to think that this simple multiply-by-\(n!\) trick is not sufficient to show that \(e^2\) is irrational. This is a stronger statement: \(\sqrt{2}\) is an example of a number which is irrational, but whose square is not. From John Cosgrave we have learned that with two nice ideas/observations (let’s call them “tricks”) one can get two steps further nevertheless: Each of the tricks is sufficient to show that \(e^2\) is irrational, the combination of both of them even yields the same for \(e^4\). The first trick may be found in a one page paper by J. Liouville from 1840 — and the second one in a two page “addendum” which Liouville published on the next two journal pages.

Why is \(e^2\) irrational? What can we derive from \(e^2 = \frac{a}{b}\)? According to Liouville we should write this as

substitute the series

and

and then multiply by \(n!\), for a sufficiently large even \(n\). Then we see that \(n! b e\) is nearly integral:

is an integer, and the rest

is approximately \(\frac{b}{n}\): It is larger than \(\frac{b}{n+1}\) but smaller than \(\frac{b}{n}\), as we have seen above.

At the same time \(n! a e^{-1}\) is nearly integral as well: Again we get a large integral part, and then a rest

and this is approximately \((-1)^{n+1} \frac{a}{n}\). More precisely: for even \(n\) the rest is larger than \(-\frac{a}{n}\), but smaller than

But this cannot be true, since for large even \(n\) it would imply that \(n! a e^{-1}\) is just a bit smaller than an integer, while \(n! b e\) is a bit larger than an integer, so \(n! a e^{-1} = n! b e\) cannot hold.

This lemma is not hard to show: \(\lfloor \frac{n}{2} \rfloor \) of the factors of \(n!\) are even, \(\lfloor \frac{n}{4} \rfloor \) of them are divisible by \(4\), and so on. So if \(2^k\) is the largest power of two which satisfies \(2^k \le n\), then \(n!\) contains the prime factor \(2\) exactly

times, with equality in both inequalities exactly if \(n = 2^k\).

In order to show that \(e^4\) is irrational, we now courageously assume that \(e^4 = \frac{a}{b}\) were rational, and write this as

We could now try to multiply this by \(n!\) for some large \(n\), and collect the non-integral summands, but this leads to nothing useful: The sum of the remaining terms on the left-hand side will be approximately \(b \frac{2^{n+1}}{n}\), on the right side \((-1)^{n+1} a \frac{2^{n+1}}{n}\), and both will be very large if \(n\) gets large.

So one has to examine the situation a bit more carefully, and make two little adjustments to the strategy: First we will not take an arbitrary large \(n\), but a large power of two, \(n = 2^m\); and secondly we will not multiply by \(n!\), but by \(\frac{n!}{2^{n-1}}\). Then we need the little lemma 8.3, a special case of Legendre’s theorem (see page 10).

Let’s get back to \(b e^2 = a e^{-2}\). We are looking at

and substitute the series

and

For \(r \le n\) we get integral summands on both sides, namely

where for \(r {\gt} 0\) the denominator \(r!\) contains the prime factor \(2\) at most \(r-1\) times, while \(n!\) contains it exactly \(n-1\) times. (So for \(r {\gt} 0\) the summands are even.)

And since \(n\) is even (we assume that \(n = 2^m\)), the series that we get for \(r \ge n+1\) are

resp.

These series will for large \(n\) be roughly \(\frac{4b}{n}\) resp. \(-\frac{4a}{n}\), as one sees again by comparison with geometric series. For large \(n = 2^m\) this means that the left-hand side of ?? is a bit larger than an integer, while the right-hand side is a bit smaller — contradiction!

Part (i) is clear.

Part (ii) is also clear.

For (iii) note that by (i) the \(k\)-th derivative \(f^{(k)}\) vanishes at \(x=0\) unless \(n \le k \le 2n\), and in this range \(f^{(k)}(0) = \frac{k!}{n!} c_k\) is an integer. From \(f(x) = f(1-x)\) we get \(f'(x) = (-1) f'(1-x)\) for all \(x\), and hence \(f^{(k)}(1) = (-1)^k f^{(k)}(0)\), which is an integer.

It suffices to show that \(e^s\) cannot be rational for a positive integer \(s\) (if \(e^{\frac{s}{t}}\) were rational, then \(\left(e^{\frac{s}{t}}\right)^t = e^s\) would be rational, too). Assume that \(e^s = \frac{a}{b}\) for integers \(a, b {\gt} 0\), and let \(n\) be so large that \(n! {\gt} a s^{2n+1}\). Put

where \(f(x)\) is the function of the lemma. \(F(x)\) may also be written as an infinite sum

since the higher derivatives \(f^{(k)}(x)\), for \(k {\gt} 2n\), vanish. From this we see that the polynomial \(F(x)\) satisfies the identity

Thus differentiation yields

and hence

This is an integer, since part (iii) of the lemma implies that \(F(0)\) and \(F(1)\) are integers. However, part (ii) of the lemma yields estimates for the size of \(N\) from below and from above,

which shows that \(N\) cannot be an integer: contradiction.

Assume that \(\pi ^2 = \frac{a}{b}\) for integers \(a, b {\gt} 0\). We now use the polynomial

which satisfies \(F''(x) = -\pi ^2 F(x) + b^n \pi ^{2n+2} f(x)\).

From part (iii) of the lemma we get that \(F(0)\) and \(F(1)\) are integers. Elementary differentiation rules yield

and thus we obtain

which is an integer. Furthermore \(N\) is positive since it is defined as the integral of a function that is positive (except on the boundary). However, if we choose \(n\) so large that \(\frac{\pi a^n}{n!} {\lt} 1\), then from part (ii) of the lemma we obtain

a contradiction.

We use the addition theorem

from elementary trigonometry, which for \(\alpha = (k+1)\varphi \) and \(\beta = (k-1)\varphi \) yields

For the angle \(\varphi _n = \arccos \left(\frac{1}{\sqrt{n}}\right)\), which is defined by \(\cos \varphi _n = \frac{1}{\sqrt{n}}\) and \(0 \le \varphi _n \le \pi \), this yields representations of the form

where \(A_k\) is an integer that is not divisible by \(n\), for all \(k \ge 0\). In fact, we have such a representation for \(k=0, 1\) with \(A_0 = A_1 = 1\), and by induction on \(k\) using ?? we get for \(k \ge 1\)

Thus we obtain \(A_{k+1} = 2A_k - n A_{k-1}\). If \(n \ge 3\) is odd, and \(A_k\) is not divisible by \(n\), then we find that \(A_{k+1}\) cannot be divisible by \(n\), either.

Now assume that

is rational (with integers \(k, \ell {\gt} 0\)). Then \(\ell \varphi _n = k \pi \) yields

Thus \(\sqrt{n}^\ell = \pm A_\ell \) is an integer, with \(\ell \ge 2\), and hence \(n | \sqrt{n}^\ell \). With \(\sqrt{n}^\ell | A_\ell \) we find that \(n\) divides \(A_\ell \), a contradiction.