Hilbert Proves that e Is Transcendental

David Hilbert, 1862-1943 (photo 1912)

It's easy to prove that \( e \) is irrational, but takes more work to prove that it's transcendental, meaning not the root of any equation:

\[ \begin{equation}{a_0 + a_1 x + \cdots + a_{n-1} x^{n-1} + a_n x^n = 0, \hspace{16pt} a_i \in \mathbb{Z}.}\tag{1} \end{equation} \]

Put otherwise, it is never the case that \( a_n e^n + a_{n-1} e^{n-1} + \cdots + a_1 e + a_0 = 0 \; \) for any choice of integers \( a_0, a_1, \cdots, a_{n-1}, a_n \). We already have the result for \( n = 0 \) since \( e \) is not an integer, and for \( n = 1 \) since \( e \) is not the root of any equation \( bx - a = 0 \) for integers \( a \) and \( b \) (simply restating that \( e \) is irrational). In 1840, Liouville proved that \( e \) is not the root of a quadratic equation with integer coefficients, but the final object is to prove that \( e \) never satisfies \( (1) \) for any choice of integers \( a_i \) and any \( n = 0, 1, 2, 3, 4, \ldots \).

Hermite first proved that \( e \) is transcendental in 1873^[1] and Hilbert refined the proof in 1893^[2]. There are straightforward versions of Hermite's proof available, but Hilbert's proof may be more transparent. I struggled to understand it, so am going to recapitulate my thinking in case it's helpful to anyone else. Hilbert's original paper is concise to a fault, but of course has all the elements of the proof. The Wikisource people in Germany have done a fantastic job in transcribing Hilbert's paper — have Google translate it into English in the browser for a little help if your German is as bad as mine. The link [1] just above and to the left of the title at the Wikisource page leads to another page with the transcription opposite the original paper. The Wikipedia page on transcendental number is another resource, with an account following Hilbert closely. Felix Klein and Michael Spivak have nice write-ups as well.

The key to the proof is this rather forbidding looking integral:

\[ \begin{equation}{I_{n,p} = \int_{0}^{\infty} x^{p-1}\left[(x-1)(x-2)\cdots (x-n)\right]^p \cdot e^{-x} \; dx, \hspace{20pt} n, p \in \mathbb{N}.}\tag{2} \end{equation} \]

\( n \) is going to be the degree of the polynomial \( (1) \) purportedly satisfied by \( e \) (the proof will be by contradiction). The first step in understanding this integral is to look at the simpler integral:

\[ \Gamma(q) = \int_{0}^{\infty} x^{q-1} e^{-x} \; dx, \hspace{20pt} q \in \mathbb{N}. \]

Obviously:

\[ \Gamma(1) = \int_{0}^{\infty} x^{0} e^{-x} \; dx = \int_{0}^{\infty} e^{-x} \; dx = 1, \]

and a simple integration by parts establishes that:

\[ \Gamma(q) = (q-1) \cdot \Gamma(q-1). \]

\[ \begin{align*}
\therefore \Gamma(1) &= 1\\
\Gamma(2) &= 1 \cdot \Gamma(1) = 1 \cdot 1 = 1!\\
\Gamma(3) &= 2 \cdot \Gamma(2) = 2 \cdot 1! = 2!\\
\Gamma(4) &= 3 \cdot \Gamma(3) = 3 \cdot 2! = 3!\\
&\vdots\\
\Gamma(q) &= (q-1) \cdot \Gamma(q-1) = (q-1) \cdot (q-2)! = (q-1)!\\
\end{align*} \]

This shows that:

\[ \Gamma(q) = \int_{0}^{\infty} x^{q-1} e^{-x} \; dx = (q-1)!, \hspace{20pt} q \in \mathbb{N}. \]

Back to \( I_{n,p}, \) the daunting integral in \( (2) \). \( (x-1)(x-2)\cdots (x-n) \; \) is a just a polynomial in \( x \) with integral coefficients and that is still the case after taking that polynomial to the \( p^{\text{th}} \) power. Multiplying that polynomial by \( x^{p-1} \) and distributing results in still another polynomial with integer coefficients whose lowest term is an integral multiple of \( x^{p-1}. \) Breaking up the integral in turn results in a sum of integrals of the form \( \int_{0}^{\infty} x^s e^{-x} \; dx, \) each multiplied by an integer. That is, the integral evaluates to an integer! Furthermore, since each exponent \( s \geq p-1, \) each of those smaller integrals is divisible by \( (p-1)!, \) so the integral \( I_{n,p} \) is an integer divisible by \( (p-1)! \) as well. If not already clear, perhaps the example with \( n = 2 \) and \( p = 3 \) will help; if \( g(x) = x^2\left[(x-1)(x-2)\right]^3 \cdot e^{-x}, \) then:

\begin{align*}
I_{2,3} &= \int_{0}^{\infty} g(x) \; dx\\
&= \int_{0}^{\infty} x^2\left[(x-1)(x-2)\right]^3 \cdot e^{-x} \; dx\\
&= \int_{0}^{\infty} x^2\left(x^2-3x+2\right)^3 \cdot e^{-x} \; dx\\
&= \int_{0}^{\infty} x^2\left(x^6-9x^5+33x^4-63x^3+66x^2-36x+8\right) \cdot e^{-x} \; dx\\
&= \int_{0}^{\infty} \left(x^8-9x^7+33x^6-63x^5+66x^4-36x^3+8x^2\right) \cdot e^{-x} \; dx\\
&= \int_{0}^{\infty} x^8 e^{-x} \; dx - 9 \int_{0}^{\infty} x^7 e^{-x} \; dx + \cdots - 36 \int_{0}^{\infty} x^3 e^{-x} \; dx + 8 \int_{0}^{\infty} x^2 e^{-x} \; dx\\
&= \Gamma(9) - 9 \cdot \Gamma(8) + 33 \cdot \Gamma(7) - 63 \cdot \Gamma(6) + 66 \cdot \Gamma(5) - 36 \cdot \Gamma(4) + 8 \cdot \Gamma(3)\\
&= 8! - 9 \cdot 7! + 33 \cdot 6! - 63 \cdot 5! + 66 \cdot 4! - 36 \cdot 3! + 8 \cdot 2!\\
&= 40320 - 45360 + 23760 -7560 + 1584 -216 + 16\\
&= 12,544.\\
\end{align*}

Use WolframAlpha online to validate an indefinite integral like this if you like. Note that \( (p-1)! = 2! = 2 \) divides \( \int_{0}^{\infty} g(x) \; dx = 12,544, \) but \( p = 3 \) does not divide it. Neither fact is a coincidence and both are vitally important in proving the theorem. Obviously \( 2! | 12,544 \) in this case, but you could see that would be true even before finishing the calculations, because in the third line from the bottom, each summand contains a factorial and they are all greater than or equal to \( 2!. \) Look at the third line from the bottom again to establish that \( 3 \nmid 12,544 \). Each term but the last has a factor of \( 3! \) or greater, so \( 3 \) divides all those terms. But \( 3 \) does not divide the last term and therefore does not divide the entire sum, namely \( 12,544. \) To see that \( 3 \) does not divide the last term, look at the integrand generating it: \( x^2 \left((-1) \cdot (-2)\right)^3 \cdot e^{-x}, \) which evaluates to \( (2!)^3 \cdot \Gamma(3) = 8 \cdot 2 = 16, \) where \( 3 \) divides neither \( (2!)^3 \) nor \( \Gamma(3), \) so does not divide their product either — this is where \( p \) prime comes into play, also \( p \gt n \) (\(3\gt2\) here). With all those factorials, such integrals can get monstrously large quickly when increasing \( n \) and \( p; \) for example:

\[ I_{3,5} = \int_{0}^{\infty} x^4\left[(x-1)(x-2)(x-3)\right]^5 \cdot e^{-x} \; dx = 24,269,123,932,707,456. \]

As expected, this value is divisible by \( (p-1)! = 4! = 24 \) but is not divisible by \( p=5. \)

Examine these graphs to get a sense of what's going on geometrically in the example above. They're of the same function at different resolutions, but the aspect ratio is different, greatly accentuating the height of the bump just to the right of zero in the second graph. Functional values are close to \( 0 \) for \( 0 \leq x \leq 2 \), then start to rise as the cubic term dominates for \( x \gt 2, \) reaching a peak of \( 1600 \) or so around \( x = 10 \) then start declining as the factor of \( e^{-x} \) takes effect, finally hugging the y-axis asymptotically for \( x \gt 20. \) It's been established that the integral (that is, the area between the curve and the x-axis) is \( 12,544 \) and the great bulk of that is for \( x \geq 2. \) We can refine the estimate by looking at:

\[ \begin{align*}
e^2 \cdot \int_{2}^{\infty} g(x) \; dx &= e^2 \cdot \int_{2}^{\infty} x^2[(x-1)(x-2)]^3 \cdot e^{-x} \; dx\\
&= \int_{2}^{\infty} x^2[(x-1)(x-2)]^3 \cdot e^{-(x-2)} \; dx.
\end{align*} \]

The \( e^2 \) is introduced so it can be moved inside the integral, the integrand now being scaled by \( e^{-(x - 2)} \) instead of \( e^{-x} \), making the integral similar to those earlier. Introduce the substitution \( u = x - 2 \), so that \( du = dx \) and the limits of integration on \( u \) are \( u = 0 \) and \( \infty: \)

\[ \begin{align*}
e^2 \cdot \int_{2}^{\infty} g(x) \; dx &= \int_{u=0}^{\infty} (u+2)^2[(u+1)u)]^3 \cdot e^{-u} \; du\\
&= 92,688.
\end{align*} \]

It's notable that \( e^2 \) times the integral is an integer, and in this case:

\[ \int_2^\infty g(x) \; dx = {92,688 \over e^2} = 12,543.957. \]

Putting it together:

\[ \underbrace{\int_0^\infty g(x) \; dx}_{\large 12,544} = \underbrace{\int_0^2 g(x) \; dx}_{\large 0.043} + \underbrace{\int_2^\infty g(x) \; dx}_{\large 92,688/e^2 = 12,543.957}. \]

A similar calculation breaks the integral at \( 1 \) instead of \( 2: \)

\[ \underbrace{\int_0^\infty g(x) \; dx}_{\large 12,544} = \underbrace{\int_0^1 g(x) \; dx}_{\large 0.047} + \underbrace{\int_1^\infty g(x) \; dx}_{\large 34,098/e = 12,543.953}. \]

As mentioned above, \( 3 \) does not divide \( \int_{0}^{\infty} g(x) \; dx = 12,544 \), but notice that \( 3 \) does<\i> divide \( e^2 \cdot \int_{2}^{\infty} g(x) \; dx = 92,688 \) and \( e \cdot \int_{1}^{\infty} g(x) \; dx = 34,098. \)

We're now in a position to prove the theorem for \( n = 2 \), and in a way that will generalize to any \( n. \) Suppose that:

\[ a_0 + a_1 e + a_2 e^2 = 0, \hspace{20pt} a_0, a_1, a_2 \in \mathbb{Z}. \]

The object is to derive a contradiction, proving that this equation is impossible. Let:

\[ g_p(x) = x^{p-1}[(x-1)(x-2)]^p \cdot e^{-x}, \hspace{20pt} p \in \mathbb{N}, \]

where \( p \) is to be determined later. Multiply the quadratic equation in \( e \) by \( \int_{0}^{\infty} g_p(x) \; dx: \)

\[ a_0 \int_{0}^{\infty} g_p(x) \; dx + a_1 \cdot e \int_{0}^{\infty} g_p(x) \; dx + a_2 \cdot e^2 \int_{0}^{\infty} g_p(x) \; dx = 0. \]

Break each of those integrals into two pieces, leading to the sum of these two expressions for the left side of the equation above:

\[ \begin{alignedat}{2}
P_{1} &= a_0 \int_{0}^{\infty} g_p(x) \; dx \;+\; && a_{1} \cdot e \int_{1}^{\infty} g_p(x) \; dx + a_{2} \cdot e^2 \int_{2}^{\infty} g_p(x) \; dx,\\
P_{2} &= && a_{1} \cdot e \int_{0}^1 g_p(x) \; dx + a_{2} \cdot e^2 \int_0^2 g_p(x) \; dx.
\end{alignedat} \]

So \( P_1 + P_2 = 0 \) and like with the example of \( g(x) = g_3(x) \) fleshed out above, the integrals in \( P_1 \) contain almost all the area, the integrals in \( P_2 \) very little of it. Also as is the earlier example, each of the summands in \( P_1 \) is an integer, so \( P_1 \) too is an integer (a very large one for even modest values of \( p \)). It turns out that a single value \( p \) can be chosen so that both:

1) \( P_1 \neq 0. \)
2) \( |P_2| \lt 1/2. \)

This contradicts \( P_1 + P_2 = 0, \) because a small value added to a non-zero integer cannot be zero.

\(\large{1) \; P_1 \neq 0.} \) If \( p \) is a prime greater than \( a_0 \), then obviously \( p \nmid a_0. \) And we're going to show that for primes \( p \gt n, p \nmid \int_{0}^{\infty} g_p(x) \; dx. \) In that case, \( p, \) being a prime, would not divide their product, the first summand in \( P_1. \) Then we'll show that \( p \) does divide the other summands of \( P_1. \) It follows from these two facts that \( P_1 \not\equiv 0 \; (\text{mod } \; p) \) and therefore can't equal \( 0. \) That is:

\[ P_1 = \underbrace{a_0 \int_{0}^{\infty} g_p(x) \; dx}_{\large p \text{ does } \textit{not} \text{ divide this}}
+ \underbrace{a_{1} \cdot e \int_{1}^{\infty} g_p(x) \; dx + a_{2} \cdot e^2 \int_{2}^{\infty} g_p(x) \; dx.}_{\large p \text{ divides each integral and } \therefore \text{ the entire sum}} \]

Expand the integrand \( g_p(x) \) like was done in the example above, separating out the lowest power of \( x \):

\begin{align*}
\int_{0}^{\infty} g_p(x) \; dx &= \int_{0}^{\infty} x^{p-1}\left[(x-1)(x-2)\right]^p \cdot e^{-x} \; dx\\
&= \int_{0}^{\infty} \left( x^p \cdot \left( \text{some polynomial in } x \right) + (2!)^p x^{p-1} \right) \cdot e^{-x} \; dx\\
&= \int_{0}^{\infty} x^p \cdot (\text{some polynomial in } x) \cdot e^{-x} \; dx + (2!)^p \int_{0}^{\infty} x^{p-1} \cdot e^{-x} \; dx\\
&= A \cdot p! + (2!)^p (p-1)!, \hspace{20pt} A \in \mathbb{Z}.\\
\end{align*}

\( p \) does not divide the last expression, because it divides the first summand but not the second one, assuming \( p \gt 2. \) If the last step in the calculation is not clear, consider that the integrand with \( x^p \) evaluates to an integral sum of factorials, the smallest of which is \( p!, \) so their sum is an integral multiple of \( p!. \) Then for all the other integrals in \( P_1 \), a calculation like this obtains, substituting \( u=x-2: \)

\begin{align*}
e^2 \int_{2}^{\infty} g_p(x) \; dx &= e^2 \int_{2}^{\infty} x^{p-1}[(x-1)(x-2)]^p \cdot e^{-x} \; dx\\
&= \int_{u=0}^{\infty} (u+2)^{p-1}[(u+1) \cdot u]^p \cdot e^{-u} \; du\\
&= \int_{u=0}^{\infty} u^p \cdot (\text{some polynomial in } u) \cdot e^{-u} \; du\\
&= B \cdot p!, \hspace{20pt} B \in \mathbb{Z}.\\
\end{align*}

There is a factor of \( u^p \) in all those integrands, so each of the integrals is an integral multiple of \( p \) and so is their sum. QED.

\(\large{2) \; |P_2| \lt 1/2.} \) Write:

\begin{align*}
g_p(x) &= x^{p-1}[(x-1)(x-2)]^p \cdot e^{-x}\\
&= (x-1)(x-2) \cdot [x(x-1)(x-2)]^{p-1} \cdot e^{-x}\\
&= (x-1)(x-2) \cdot [h(x)]^{p-1} \cdot e^{-x},\\
\end{align*}

where \( h(x) = x(x-1)(x-2). \) Note that \( 0 \leq |(x-1)(x-2)| \leq 2 \) for \( 0 \leq x \leq 2; \) and as the graph here suggests, \( |h(x)| \leq 2 / (3 \sqrt{3}) = 0.385 \) for \( 0 \leq x \leq 2. \)

\[ \therefore |g_p(x)| \leq 2 \cdot \left({2 \over {3 \sqrt{3}}}\right)^{p-1} = 2 \cdot (0.385)^{p-1} \; \text{ for } \; 0 \leq x \leq 2. \]

It's clear now that each of the integrals in \( P_2 \) can be made as small as desired by increasing \( p. \) This is still true when the integrals are multiplied by fixed values like \( a_1 \cdot e \) and \( a_2 \cdot e^2. \) Recall that \( n \) is fixed, so if each of those scaled integrals can be made small, then so can their sum \( P_2. \) QED.

So choose \( p \) to be a prime greater than \( n \) and \( a_0 \) (as required for condition 1 to be true) and sufficiently large so that \( |P_2| \lt 1/2 \) and the contradiction is established, invalidating the possibility that \( e \) can satisfy a quadratic equation with integer coefficients.

Now the integrand is:

\[ g(x) = g_{n,p}(x) = x^{p-1}[(x-1)(x-2) \cdots (x-n)]^p \cdot e^{-x}, \]

where the goal is to show that \( e \) satisfies no \( n^{\text{th}} \) degree polynomial, and \( P_1 \) and \( P_2 \) will have \( n+1 \) and \( n \) summands respectively. The proof that \( P_1 \) is a non-zero integer goes through as above, where now \( p \gt n \) and \( n! \) appears instead of \( 2!. \) The proof that \( P_2 \lt 1/2 \) for large enough \( p \) needs amendment, because \( |g_{n,p}(x)| \) is not generally bounded by \( 1. \) Divide \( P_1 \) and \( P_2 \) by \( (p-1)!: \)

\[ Q_1 = {P_1 \over (p-1)!}, \hspace{20pt} Q_2 = {P_2 \over (p-1)!}. \]

As discussed above, \( (p-1)! | P_1, \) so \( Q_1 \) too is a non-zero integer with \( Q_1 + Q_2 = 0 \) and we can prove that \( |Q_2| \lt 1/2 \) for large enough \( p. \) To this end, suppose that \( |(x-1)(x-2) \cdots (x-n)| \leq K \) for \( 0 \leq x \leq n, \) K a constant, and consider this calculation for the last summand \( S \) in \( Q_2: \)

\begin{align*}
|S| &= {|a_n| e^n \over (p-1)!} \cdot \left|\int_{0}^n g_{n,p}(x) \cdot e^{-x} \; dx \right|\\
&\leq {|a_n| e^n \over (p-1)!} \cdot \int_{0}^n |g_{n,p}(x)| \; dx\\
&= {|a_n| e^n \over (p-1)!} \cdot \int_{0}^n x^{p-1}|(x-1)(x-2) \cdots (x-n)|^p \; dx\\
&\leq {|a_n| e^n \over (p-1)!} \cdot \int_{0}^n n^{p-1} \cdot K^p \; dx\\
&= {|a_n| e^n \over (p-1)!} \cdot n \cdot (n^{p-1} \cdot K^p)\\
&= {|a_n| e^n \over (p-1)!} \cdot n^p \cdot K^p\\
&= {{|a_n| e^n(nK)^p} \over {(p-1)!}}\\
&\lt {1 \over 2} \text{ for large } p.
\end{align*}

Note that \( e^{-x} \) is dropped early in the calculation because it is less that one, also that everything in the second to last line is a constant except \( p \). Keep in mind as working through this calculation that the limits on the integral are \( 0 \) and \( n, \) so inside it, \( 0 \leq x \leq n. \) That expression on the second to last line can be made as small as desired by increasing \( p, \) since the factorial will eventually grow faster than the exponential to the power \( p, \) overwhelming the constant multipliers as well. The calculation for all the other summands is similar, so each of them can be made as small as desired, as can their sum \( Q_2. \) This contradiction proves that there can be no polynomial with integer coefficients satisfied by \( e; \) that is, \( e \) is transcendental. QED.

Mike Bertrand

March 3, 2018

^ 1. Sur la fonction exponentielle, by C. Hermite, Comptes rendus hebdomadaires des séances de l'Académie 77, 1873, pp. 18-24, 74-79; 226-233, 285-293. The original work was spread across four articles in Comptes rendus, as indicated. The entire article is in Volume 3 of Hermite's Oeuvres, pp. 150-181.

^ 2. Ueber die Transcendenz der Zahlen e und π, by David Hilbert, Mathematische Annalen, Vol. 43 (2-3), June 1893, pp. 216-219.