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The set of Liouville numbers is defined as all irrational $x$ such that for each natural $n$ there exists integers $p$ and $q > 1$ such that $|x - p/q| < 1 / (q^n)$.

QUESTION: Is a Liouville number plus a rational number again a Liouville number? How do you prove this?

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up vote 2 down vote accepted

Yes. Let $x$ be a Liouville number, and let $a/b$ be a rational number (of course the irrationality of $x+a/b$ is easy). For each $m$, let $p_m,q_m$ be integers such that $|x-p_m/q_m|<1/q_m^m$. Note that $q_m\to\infty$ as $m\to\infty$.

The obvious candidate for rational approximations to $x+a/b$ are $p_n/q_n+a/b = r_n/(q_nb)$, where $r_m = ap_n+bq_n$. Given $n\ge1$, choose $m>n$ so large that $q_m>b^n$. Then $$ \bigg| \bigg( x+\frac ab\bigg) - \bigg( \frac{r_m}{q_mb} \bigg) \bigg| = \bigg| x-\frac{p_m}{q_m} \bigg| < \frac1{q_m^m} \le \frac1{q_m^{n+1}} < \frac1{q_m^n b^n} = \frac1{(q_mb)^n}. $$

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