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How to express any arbitrary real number $\alpha$ as sum of two Liouville numbers?

Recalling a Liouville number is a real number such that $ 0<| \alpha - \frac{p}{q}|< \frac{1}{q^n}$ has infinitely many solutions in $\frac{p}{q}$ for all $n \geq 0.$

Explicitely $\sum_{n=1}^{\infty} \frac{a_n}{b^{n!}}$ such that $a_n=0,...b-1$ and $b\geq 2$ are Liouville numbers.

Given any arbitrary real number $\alpha$ and after considering its decimal expansion how do we construct two Liouville numbers that will add up to $\alpha$?

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    $\begingroup$ This is a result of Erdos, here. In that paper he also shows that ever real is the product of two Liouville numbers. $\endgroup$ – lulu Sep 16 at 13:12
  • $\begingroup$ @lulu Very interesting. Has such a result also been established for the sum ? $\endgroup$ – Peter Sep 18 at 10:28
  • $\begingroup$ @Peter What do you mean? That paper handles both sum and product. $\endgroup$ – lulu Sep 18 at 11:21

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