Liouville Proves That e Is Not the Root of a Quadratic Equation

More properly, \( e \) is not the root of a quadratic equation with integer coefficients, which is the same as saying rational coefficients, because denominators can be cleared. This is sometimes stated as \( e \) is not a quadratic irrationality. Liouville proved the theorem in his journal in 1840.^[1] It was a step towards proving that \( e \) is transcendental, meaning that it is not the root of any polynomial equation with integer coefficients. Apparently Liouville regarded it as such^[2], but couldn't push through the general proof. Hermite was the first to prove that \( e \) is transcendental in 1873.^[3] Hilbert simplified the proof^[4], among others. But let's get back to the quadratic case and Liouville's proof. Here it is in its entirety, translated into English:

One proves in the curriculum that the number \( e \), base of the Napierian logarithms, is not a rational number. We should, it seems to me, add that the same method also proves that \( e \) cannot be the root of a second-degree equation with rational coefficients, so that we cannot have \( ae + b / e = c \), where \( a \) is a positive integer and \( b, c \) are positive or negative integers. Indeed, if we replace \( e \) and \(1 /e \) or \( e^{-1} \) in this equation by their expansions derived from that of \( e^{x} \), then multiply each side of the equation by \( 1 \cdot 2 \cdot 3 \ldots n \), we easily find that: \[ {a \over {n+1}}\left(1 + {1 \over {n+2}} + \cdots \right) \pm {b \over {n+1}}\left(1 - {1 \over {n+2}} + \cdots \right) = \mu, \] \( \mu \) being an integer. We can always make sure that the factor \[ \pm {b \over {n+1}} \] is positive; it will suffice to suppose \( n \) is even if \( b \lt 0 \) and \( n \) is odd if \( b \gt 0 \); then taking \( n \) very large, the equation we just wrote leads to a contradiction; because the left side is positive and very small, being between \( 0 \) and \( 1 \), and so cannot be equal to an integer \( \mu \). So, etc.

This is pretty straightforward down to the equation; let's flesh it out past there, starting by adding one more term to those sums to make the pattern clear:

\[ {a \over {n+1}}\left(1 + {1 \over {n+2}} + {1 \over {(n+2)(n+3)}} + \cdots \right) \pm {b \over {n+1}}\left(1 - {1 \over {n+2}} + {1 \over {(n+2)(n+3)}} - \cdots \right) = \mu. \]

The denominators are tails of factorials, because we've multiplied through by \( n! \), leaving products past that point intact in every summand. All the signs in the first sum are positive, whereas the signs in the second sum alternate, considering that it derives from the expansion of \( e^{-x} \). Put:

\[ \begin{align*}
\sigma_n &= {1 \over {n+1}}\left(1 + {1 \over {n+2}} + {1 \over {(n+2)(n+3)}} + \cdots \right),\\
\tau_n &= {1 \over {n+1}}\left(1 - {1 \over {n+2}} + {1 \over {(n+2)(n+3)}} - \cdots \right),
\end{align*} \]

making the key equation:

\[ \begin{equation}{a \sigma_n \pm b \tau_n = \mu.}\tag{1} \end{equation} \]

Which sign obtains depends on the parity of \( n \). To see this, work through the example of \( n = 2 \). Multiplying the original equation by \( 2! \) results in:

\[ a\left(2 + 2 + 1 + {1 \over 3} + {1 \over {3 \cdot 4}} + {1 \over {3 \cdot 4 \cdot 5}} + \cdots\right) + b \left(2 - 2 + 1 - {1 \over 3} + {1 \over {3 \cdot 4}} - {1 \over {3 \cdot 4 \cdot 5}} + \cdots\right) = c. \]

\[ \therefore \; a\left({1 \over 3} + {1 \over {3 \cdot 4}} + {1 \over {3 \cdot 4 \cdot 5}} + \cdots\right) + b \left(-{1 \over 3} + {1 \over {3 \cdot 4}} - {1 \over {3 \cdot 4 \cdot 5}} + \cdots\right) = c - 5a - b = \mu, \]

\[ \text{so } \; {a \over 3} \left(1 + {1 \over 4} + {1 \over {4 \cdot 5}} + \cdots\right) + {b \over 3} \left(-1 + {1 \over 4} - {1 \over {4 \cdot 5}} + \cdots\right) = \mu. \]

\[ \text{That is: } \; a \sigma_2 - b \tau_2 = \mu. \]

The minus sign appears similarly whenever \( n \) is even, and in like way a positive sign is in order when \( n \) is odd. The upshot is that (1) can be written more specifically as:

\[ \begin{equation}{a \sigma_n + (-1)^{n-1} b \tau_n = \mu.}\tag{2} \end{equation} \]

\( \sigma_n \) is obviously a finite value greater than \( 0 \), considering that all its terms are positive and \( \sigma_n = n! \cdot e - \text{ some integer} \). In fact, the third term in the sum for \( \sigma_n \)is less than \( 1 / (n+2)^2 \) and in general the \( k^{\text{th}} \) term is less than \( 1 / (n+2)^{k-1} \), so \( \sigma_n \) is less than the infinite geometric sum with ratio \( 1 / (n+2) \):

\[ \begin{align*}
\sigma_n &\leq {1 \over {n+1}} \sum_{k=0}^\infty \left({1 \over {n+2}}\right)^k\\
&= {1 \over {n+1}}\left({1 \over {1-{1 \over{n+2}}}}\right)\\
&= {1 \over {n+1}}\left({{n+2} \over {(n+2)-1}}\right)\\
&= {{n+2} \over {(n+1)^2}} \rightarrow 0.
\end{align*} \]

The sum in \( \tau_n \) is an alternating series with initial term equal to one and subsequent terms decreasing monotonically in absolute value, so that sum converges and \( \tau_n \rightarrow 0 \) like \( \sigma_n \) does. In fact, \( \tau_n \lt \sigma_n \). The \( \tau_n \) are positive too. To see this, note that the alternating nature of the sum in \( \tau_n \) implies that:

\[ \begin{align*}
\tau_n &\gt {1 \over {n+1}}\left(1 - {1 \over {n+2}}\right)\\
&= {1 \over {n+1}}\left({{(n+2)-1} \over {n+2}}\right)\\
&= {1 \over {n+1}}\left({{n+1} \over {n+2}}\right)\\
&= {1 \over {n+2}}
\end{align*} \]

Putting this together:

\[ {1 \over {n+2}} \lt \tau_n \lt \sigma_n \lt {{n+2} \over {(n+1)^2}}, \]

and the chart here shows that the four values are close (note that the ratio of the lower and upper limits is one). Recall that \( a \) and \( b \) are fixed integers, coefficients of the original quadratic equation. Like Liouville says, we can assume \( a \gt 0. \) We want to show that the left side of (2) can be constrained to be strictly between \( 0 \) and \( 1 \), which would be a contradiction, considering that the right side of (2), \( \mu \), is an integer. If \( b \) is positive, taking increasingly large odd values of \( n \) results in everything on the left side of (2) being positive and as small as desired — less than one, in particular. Contrarily if \( b \) is negative, take large positive values of \( n \) to reverse the sign of \( \tau_n \), again driving the value of the left side of (2) down to a small positive value. Either way there is a contradiction — there can be not quadratic equation with integer coefficients satisfied by \( e \). QED.

Mike Bertrand

February 11, 2018

^ 1. Sur l'irrationalité du nombre e = 2.718..., by J. Liouville, Journal de Mathématiques Pures et Appliquées 5, 1840, p. 192.

^ 2. Joseph Liouville 1809–1882: Master of Pure and Applied Mathematics, by Jesper Lützen (Springer-Verlag, 1990), ISBN 978-1-4612-6973-1. Liouville regarded it as such: p. 79.

^ 3. 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.

^ 4. Ueber die Transcendenz der Zahlen e und π, by David Hilbert, Mathematische Annalen, Vol. 43 (2-3), June 1893, pp. 216-219.