A Liouville number is an irrational number $x$ with the property that, for every positive integer $n$, there exist integers $p$ and $q$ with $q > 1$ and such that $0 < \mid x - \frac{p}{q} \mid < \frac{1}{q^n} $.

I'm looking for either hints or a complete proof for the fact that $e$ is not a Liouville number. I can prove that $e$ is irrational and even that it is transcendental, but I'm a bit stuck here.

Here's my research:

The wikipedia article about Liouville numbers states:

[...] not every transcendental number is a Liouville number. The terms in the continued fraction expansion of every Liouville number are unbounded; using a counting argument, one can then show that there must be uncountably many transcendental numbers which are not Liouville. Using the explicit continued fraction expansion of $e$, one can show that e is an example of a transcendental number that is not Liouville.

However, theres clearly more to the argument then the boundedness of the continued fraction of $e$, because the terms of $e$'s continued fraction expansion are unbounded and yet it is not a Liouville number. Also, if possible, i would like to avoid using continued fractions at all.

This book has the following as an exercise:

Prove that $e$ is not a Liouville number. (Hint: Follow the irrationality proof of $e^n$ given in the supplements to Chapter 1.)

Unfortunately, the supplements to Chapter 1 are not publically available in the sample and I do not want to buy that book.

This book states:

Given any $\varepsilon > 0$, there exists a constant $c(e,\varepsilon) > 0$ such that for all $p/q$ there holds $\frac{c(e,\varepsilon)}{q^{2+\varepsilon}} < \mid e - \frac{p}{q} \mid$. [...] Using [this] inequality, show that $e$ is not a Liouville number.

Which, given the inequality, I managed to do. But I do not have any idea of how one would go about proving that inequality.

I greatly appreciate any help!

  • 1
    $\begingroup$ 1) Should "Using [this] inequality, show that $e$ is a Liouville number" be "is not a Liouville number?" 2) The desired inequality by definition means that $e$ has irrationality measure $\mu(e)=2$. Wikipedia notes this in passing, and Mathworld's page on irrationality measure cites Borwein & Borwein to that effect; alas, there's no preview of that book on Google books. $\endgroup$ Sep 29, 2016 at 14:24
  • $\begingroup$ Thanks, I corrected that. Thanks for pointing out, that the equality in question essentially means that $e$ has irrationality measure of 2. However this seem to be hard to prove and from a quick search I could not come up with a freely available proof. $\endgroup$
    – cdwe
    Sep 29, 2016 at 14:47

2 Answers 2


Using Gauss continued fraction for $\tanh$, it is not difficult to show that the continued fraction of $e$ has the following structure: $$ e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,\ldots]\tag{1}$$ then by studying the sequence of convergents $\left\{\frac{p_n}{q_n}\right\}_{n\geq 1}$ through $$\left|\frac{p_n}{q_n}-\frac{p_{n+1}}{q_{n+1}}\right|=\frac{1}{q_n q_{n+1}}=\frac{1}{q_n(\alpha_{n+1}q_n+q_{n-1})}\tag{2}$$ and $$ \left|e-\frac{p_n}{q_n}\right| = \left|\sum_{k\geq n}\frac{(-1)^k}{q_k q_{k+1}}\right| \tag{3} $$ we may easily get that there is no rational approximation such that $$ \left|e-\frac{p_n}{q_n}\right|\leq \frac{1}{q_n^4}\tag{4} $$ hence $e$ is not a Liouville number. It is not difficult to use $(1)$ to prove the stronger statement

The irrationality measure of $e$ is $2$.

  • $\begingroup$ Thank you, that works for me. $\endgroup$
    – cdwe
    Sep 29, 2016 at 17:30

In continued fractions for $\alpha=[a_0,a_1,a_2,\dots,a_n,\dots]$ we have $$\left|\frac{P_n}{Q_n} -\alpha\right|>\frac{1}{2}\left|\frac{P_n}{Q_n}-\frac{P_{n+1}}{Q_{n+1}}\right|=\frac{1}{2Q_nQ_{n+1}}\geq\frac{1}{2Q_n^2(a_n+1)}$$.

So the fact that $Q_n\geq F_n$ (with $F_n$ the $n$th Fibonacci number) and $a_n=O(n)$ when $\alpha=e$ means that $2Q_n^2(a_n+1)$ can't get bigger than $Q_n^3$ for large $n$.

This provides at least one necessary condition for $\alpha$ being Liouville - that there are for any $k$ infinitely man $n$ with $a_n\geq Q_n^k/2$.


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