Posted by Jason Polak on 04. November 2017 · 2 comments · Categories: elementary · Tags: ,

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?

Posted by Jason Polak on 09. October 2012 · Write a comment · Categories: number-theory · Tags: , ,

(Or would "The Number 1, Part e" be more interesting?!)

Let's talk about the number $ e$, my favourite number. Of course, to talk about it we need a definition, so we define $ e$ as

$ e = 1/0! + 1/1! + 1/2! + 1/3! + \cdots.$

Exercise: check that this converges! In Part 2, we shall see Hermite's argument that $ e$ is transcendental. In order to warm up, let us prove that $ e$ is not an integer first, and then let us prove that $ e$ irrational.
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