Question 14
Let $x_{0}, x_{1}, x_{2}, \ldots$ be a sequence of rational numbers defined by recurrence as follows: $x_{0}$ is any rational number, and, for $n \geq 0$, $$
x_{n+1}= \begin{cases}\left|\frac{x_{n}}{2}-1\right| & \text { se il numeratore di } x_{n} \text { è pari }, \\ \left|\frac{1}{x_{n}}-1\right| & \text { se il numeratore di } x_{n} \text { è dispari },\end{cases}
$$ where for numerator of a rational number means that of the fraction reduced to a minimum. Prove that for each value of $x_{0}$:
(a) the sequence contains only a finite number of distinct terms;
(b) The sequence contains exactly one of the numbers 0 and $2 / 3$ (i.and.: either there is an index $k$ such that $x_{k}=0$, or there is an index $m$ such that $x_{m}=2 / 3$, but they do not exist both).
x_{n+1}= \begin{cases}\left|\frac{x_{n}}{2}-1\right| & \text { se il numeratore di } x_{n} \text { è pari }, \\ \left|\frac{1}{x_{n}}-1\right| & \text { se il numeratore di } x_{n} \text { è dispari },\end{cases}
$$ where for numerator of a rational number means that of the fraction reduced to a minimum. Prove that for each value of $x_{0}$:
(a) the sequence contains only a finite number of distinct terms;
(b) The sequence contains exactly one of the numbers 0 and $2 / 3$ (i.and.: either there is an index $k$ such that $x_{k}=0$, or there is an index $m$ such that $x_{m}=2 / 3$, but they do not exist both).