Question 168
Let $a \in [0 ; 1]$. Define the sequence $\left(x_{n}\right)$ by
$x_{0} = a$ and $x_{n+1} = 1 - \left|1 - 2 x_{n}\right|$, for all $n \geq 0$.
Prove that the sequence $\left(x_{n}\right)$ is eventually periodic if and only if $a$ is a rational number.
(Note: we say that $\left(x_{n}\right)$ is eventually periodic if there exist integers $T \gt 0$ and $n \geqslant 0$ such that $x_{k+T} = x_{k}$ for all $k \geqslant n$.)
$x_{0} = a$ and $x_{n+1} = 1 - \left|1 - 2 x_{n}\right|$, for all $n \geq 0$.
Prove that the sequence $\left(x_{n}\right)$ is eventually periodic if and only if $a$ is a rational number.
(Note: we say that $\left(x_{n}\right)$ is eventually periodic if there exist integers $T \gt 0$ and $n \geqslant 0$ such that $x_{k+T} = x_{k}$ for all $k \geqslant n$.)