Do the Liouville Numbers form a field?

Solution 1:

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.

Solution 2:

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 $\sqrt2$.