# What is a Liouville number?

An irrational number $r$ is called a Liouville number if for every positive integer $n$ there exists integers $p$ and $q \gt 1$ such that
$$| r – p/q | \lt 1/q^n.$$
The requirement that $q \gt 1$ is crucial because otherwise, all irrational numbers would satisfy this definition. Liouville numbers are transcendental. In an intuitive way you could think of Liouville numbers as having pretty rapidly converging rational approximations. Here is an example of a Liouville number:
$$r = 0.1001000000000000000000000010\dots$$
The pattern might not be apparent but it is the number between $0$ and $1$ whose decimal expansion is all zeroes except ones in the $1^1,2^2,3^3,\dots$ places. More succintly, we can write
$$r = \sum_{k=1}^\infty 10^{-k^k}.$$
Let us prove that it is a Liouville number. Fix an $n$ and let
$$r_n = \sum_{k=1}^n 10^{-k^k}$$
Then $r_n$ is a rational number and can be written as $p/q$ with $q = 10^{-n^n}$. Then:
$$|r – r_n| = |r – p/q| \lt 1.1\times 10^{-(n+1)^{n+1}} \le 10^{-n^{n+1}} = 1/q^n$$
Therefore, $r$ is a Liouville number. This proof worked so well because I made the sequences of consecutive zeros increase very rapidly in length. Here's another question:

Is the irrational number $M = \sum_{k=1}^\infty 10^{-k^2}$ a Liouville number?

If you tried the proof I just gave for this number $M$ (so-called because it is the first letter of mystery), it won't work. But that doesn't prove that $M$ is not a Liouville number. Maybe we just weren't judicious enough in our choice of $p$ and $q$. Can you show that $M$ is or is not a Liouville number? I'll leave this to the dear reader.

There are many of Liouville numbers in the sense of cardinality: the preceding proof works if you replace the $1$s with any nonzero numbers (you can replace some of them by zero too, but you can't just replace all of them by zero). So we see the Liouville numbers have cardinality of the continuum. It's also pretty easy to see that the set of Liouville numbers is dense.

On the other hand, Liouville numbers form a set of measure zero. Therefore, there are plenty of transcendentals that are not Liouville numbers. Can you write down a specific transcendental number that is not Liouville?

### 2 Comments

• Philippe says:

Hey thanks for your blog posts in general.
Just commenting on this one so that you can change Louiville everywhere to Liouville :-)
Arguably an unusual spelling even for a Frenchman like myself.
PN

• Jason Polak says:

Goodness you are absolutely right! It is a rather peculiar spelling. Thanks for the comment and for the correction :)