A Proof That \(e\) Is Transcendental Number

The first proof that the base of the natural logarithms, \(e\), is transcendental dates from 1873. We will now follow the strategy of David Hilbert (1862–1943) who gave a simplification of the original proof of Charles Hermite. The idea is the following:

Assume, for purpose of finding a contradiction, that e is algebraic. Then there exists a finite set of integer coefficients \({c_0},{c_1},{\rm{ }}...,{c_n}\) satisfying the equation:

\({c_0} + {c_1}e + {c_2}{e^2} + ... + {c_n}{e^n} = 0,\) \({c_0},\,{c_n} \ne 0.\)

Now for a positive integer \(k\), we define the following polynomial:

\({f_k}(x) = {x^k}{\left[ {(x - 1) \cdots (x - n)} \right]^{k + 1}},\)

and multiply both sides of the above equation by

\(\int_0^\infty {{f_k}} {e^{ - x}}{\mkern 1mu} dx,\)


to arrive at the equation:

\({c_0}\left( {\int_0^\infty {{f_k}} {e^{ - x}}{\mkern 1mu} dx} \right) + {c_1}e\left( {\int_0^\infty {{f_k}} {e^{ - x}}{\mkern 1mu} dx} \right) + \cdots + {c_n}{e^n}\left( {\int_0^\infty {{f_k}} {e^{ - x}}{\mkern 1mu} dx} \right) = 0\).

This equation can be written in the form \(P + Q = 0\), where:

\(P = {c_0}\left( {\int_0^\infty {{f_k}} {e^{ - x}}{\mkern 1mu} dx} \right) + {c_1}e\left( {\int_1^\infty {{f_k}} {e^{ - x}}{\mkern 1mu} dx} \right) + {c_2}{e^2}\left( {\int_2^\infty {{f_k}} {e^{ - x}}{\mkern 1mu} dx} \right) + \cdots + {c_n}{e^n}\left( {\int_n^\infty {{f_k}} {e^{ - x}}{\mkern 1mu} dx} \right)\)

\(Q = {c_1}e\left( {\int_0^1 {{f_k}} {e^{ - x}}{\mkern 1mu} dx} \right) + {c_2}{e^2}\left( {\int_0^2 {{f_k}} {e^{ - x}}{\mkern 1mu} dx} \right) + \cdots + {c_n}{e^n}\left( {\int_0^n {{f_k}} {e^{ - x}}{\mkern 1mu} dx} \right)\)

Lemma 1. For an appropriate choice of \(k,{\textstyle{P \over {k!}}}\) is a non-zero integer.

Proof. Each term in \(P\) is an integer times a sum of factorials, which results from the relation

\(\int_0^\infty {{x^j}} {e^{ - x}}{\mkern 1mu} dx = j!\)

which is valid for any positive integer \(j\) (consider the Gamma Function).

It is non-zero because for every a satisfying \(0≤ a ≤ n\), the integrand in

\({c_a}{e^a}\int_a^\infty {{f_k}} {e^{ - x}}{\mkern 1mu} dx\)

is \({e^{ - x}}\) times a sum of terms whose lowest power of x is \(k+1\) after substituting \(x\) for \(x-a\) in the integral. Then this becomes a sum of integrals of the form

\(\int_0^\infty {{x^j}} {e^{ - x}}{\mkern 1mu} dx\)

with \(k+1≤ j\), and it is therefore an integer divisible by \((k+1)!\). After dividing by \(k!\), we get zero modulo \((k+1)\). However, we can write:

\(\int_0^\infty {{f_k}} {e^{ - x}}{\mkern 1mu} dx = \int_0^\infty {\left( {{{[{{( - 1)}^n}(n!)]}^{k + 1}}{e^{ - x}}{x^k} + \cdots } \right)} dx\)

and thus

\(\frac{1}{{k!}}{c_0}\int_0^\infty {{f_k}} {e^{ - x}}{\mkern 1mu} dx = {c_0}{[{( - 1)^n}(n!)]^{k + 1}}\) \(\bmod \,(k + 1)\).

By choosing \(k\) so that \(k+1\) is prime and larger than n and \(|{c_0}|\), we get that \({\textstyle{P \over {k!}}}\) is non-zero modulo \((k+1)\) and is thus non-zero. Lemma 2. \(\left| {{\textstyle{Q \over {k!}}}} \right| < 1\) for sufficiently large k. Proof. Note that:
 \({f_k}{e^{ - x}} = {x^k}{[(x - 1)(x - 2) \cdots (x - n)]^{k + 1}}{e^{ - x}} = \left( {{{[x(x - 1) \cdots (x - n)]}^k}} \right)\left( {(x - 1) \cdots (x - n){e^{ - x}}} \right)\)


Using upper bounds \(G\) and \(H\) for \(|x(x - 1)...(x - n)|\) and \(|(x - 1) ...(x - n){e^{ - x}}|\) on the interval \([0,n]\) we can infer that

\(|Q|<{G^k}H(|{c_1}|e + 2|{c_2}|{e^2} +...+ n|{c_n}|{e^n})\)

and since

\(\mathop {\lim }\limits_{k \to \infty } \frac{{{G^k}}}{{k!}} = 0\)

it follows that

\(\mathop {\lim }\limits_{k \to \infty } \frac{Q}{{k!}} = 0\)

which is sufficient to finish the proof of this lemma.

Noting that one can choose \(k\) so that both Lemmas hold we get the contradiction we needed to prove the transcendence of \(e\).