# Do the Liouville Numbers form a field?

The Liouville numbers are those which are better-than-polynomially approximated by rationals. More precisely, we say $x\in\mathbb{R}$ is Liouville when for all $n\in\mathbb{N}$ there is a $\tfrac pq\in\mathbb{Q}$ with $$\left|x-\frac{p}{q}\right|<\frac{1}{q^n}.$$ For the purposes of this question, we will take rational numbers to be Liouville.

Do the Liouville numbers form a field? It seems to me that if $x\simeq \tfrac pq$ and $x'\simeq \tfrac {p'}{q'}$ then $x+x',x-x',xx'$ and $x/x'$ are approximated by $\tfrac pq+\tfrac {p'}{q'},\tfrac pq-\tfrac {p'}{q'},\tfrac pq\tfrac {p'}{q'}$ and $\tfrac pq/\tfrac {p'}{q'}$, with the approximation only getting quadratically worse in each case.

But I've never seen it mentioned anywhere that the Liouville numbers form a field, and you would think that if they were the wikipedia page would at least mention it.

So are they or not?

• for the multiplication it works, but not for the addition : $|x-p/q| < \frac1{q^n}$ and $|y-r/s| < \frac1{s^n} \implies |x+y - \frac{ps+rq}{qs}| < \frac1{q^n}+\frac1{s^n} = \frac{s^n+q^n}{(qs)^n}$ Jan 31, 2016 at 14:56

Not even an additive group. One of the celebrated results by Paul Erdős is that for every real number $t$ there exists Liouville numbers $x$, $y$, $u$, $v$ such that

$$t=x+y=uv$$

The reference is

Paul Erdős. Representations of Real Numbers as Sums and Products of Liouville Numbers. Michigan Math. Journal 9, pp.59--60, 1962.

• That is quite astonishing. Can you provide a reference? Mar 29, 2015 at 14:10
• @JackD'Aurizio A quick Google suggests renyi.hu/~p_erdos/1962-18.pdf . Perhaps more remarkably, it's not coauthored with anyone! Mar 29, 2015 at 14:13

This is more of a comment, but it's a bit too long for one. As marwalix mentions in their answer, every real number can be written as the sum of two Liouville numbers. (And every real number can be written as the product of two Liouville numbers too. At least it's true that the reciprocal of a Liouville number is again a Liouville number, with the obvious rational approximations.) I thought it would be instructive to demonstrate the proof with a specific example.

With \begin{multline*} \sqrt2= 1.414213562373095048801688724209698078569671875376948073176679\\ 7379907324784621070388503875343276415727350138462309122970249248\dots, \end{multline*} define \begin{multline*} a= 1.010000562373095048801688000000000000000000000000000000000000\\ 0000000000000000000000000000000000000000000000000000000000009248\dots \end{multline*} and \begin{multline*} b= 0.404213000000000000000000724209698078569671875376948073176679\\ 7379907324784621070388503875343276415727350138462309122970240000\dots. \end{multline*} (For $b$, the run of $0$s at the end of the line starts at the $(5!+1)$st decimal place and goes to the $6!$th place; $a$ will then have a run of $0$s starting at the $(6!+1)$st decimal place and going to the $7!$th place; then $b$ will have another run of $0$s, and so on.)

The usual proof shows that $a$ and $b$ are Liouville numbers, and they clearly sum to $\sqrt2$.

• I came to know about this problem in a different context (not knowing that this was a result of Erdős), and had been trying several (perhaps overcomplicated) approaches for a long time to no avail, only to finally give in and look for some online inspiration. I was afraid of directly finding the solution without a hint or a push in the right direction. Your comment was EXACTLY what I needed to see. :) Sep 12, 2020 at 6:01