Problem 1769

［Solution］(1) From the given, we have
$$x_{n+1}=\left\{\begin{array}{l} 2 x_{n}, \text { if } 0 \leqslant x_{n}<\frac{1}{2} \\ 2-2 x_{n}, \text { if } \frac{1}{2} \leqslant x_{n} \leqslant 1 \end{array}\right.$$

If $x_{1}$ is a rational number, we can set $x_{1}=\frac{p}{q},(p, q)=1, p \in N, q \in N$. Note that for all $n$, it is clear that $0 \leqslant x_{n} \leqslant 1$, and $x_{n}=\frac{p_{n}}{q}, p_{n} \in\{0,1, \cdots, q\}$. Therefore, there must exist $n_{1}, n_{2}$ and $n_{1}<n_{2}$, such that
$$p_{n_{1}}=p_{n_{2}}$$

Thus, we have $\quad x_{n_{1}}=x_{n_{2}}$.
Therefore, from (1), we can see that $\left\{x_{n}\right\}$ is periodic after the $n_{1}$-th term.
Conversely, if the sequence is periodic after the $n_{1}$-th term, and the period is $T$. We represent $x_{n_{1}}$ in binary, denoted as
$$x_{n_{1}}=\sum_{k=1}^{\infty} a_{k} \cdot 2^{-k} \text {, where } a_{k} \in\{0,1\} \text {. }$$

We also denote
$$\bar{a}_{k}=1-a_{k}, k \in N$$

Thus, from (1), we get
$$\begin{array}{l} x_{n_{1}+1}=\left\{\begin{array}{l} \sum_{k=1}^{\infty} a_{k+1} \cdot 2^{-k}, \text { if } a_{1}=0 ； \\ \sum_{k=1}^{\infty} \overline{a_{k+1}} \cdot 2^{-k}, \text { if } a_{1}=1 . \end{array}\right. \\ x_{n_{1}+2}=\left\{\begin{array}{l} \sum_{k=1}^{\infty} a_{k+2} \cdot 2^{-k}, \text { if } a_{1}+a_{2} \equiv 0(\bmod 2) ； \\ \sum_{k=1}^{\infty}-a_{k+2} \cdot 2^{-k}, \text { if } a_{1}+a_{2} \equiv 1(\bmod 2) ， \end{array}\right. \end{array}$$

By mathematical induction, it is easy to get
$$x_{n_{1}+T}=\left\{\begin{array}{l} \sum_{k=1}^{\infty} a_{k+T} \cdot 2^{-k}, \text { if } a_{1}+a_{2}+\cdots+a_{T} \equiv 0(\bmod 2) ; \\ \sum_{k=1}^{\infty} \bar{a}_{k+T} \cdot 2^{-k}, \text { if } a_{1}+a_{2}+\cdots+a_{T} \equiv 1(\bmod 2) \end{array}\right.$$

Since $\quad x_{n_{1}+T}=x_{n_{1}}$,
Therefore, when $a_{1}+a_{2}+\cdots+a_{T} \equiv 0(\bmod 2)$, we immediately get $a_{k}=a_{k+T}$, $k \in N$,
Thus, it shows that $x_{n_{1}}$ is a binary repeating decimal, hence it is a rational number.
When $a_{1}+a_{2}+\cdots+a_{T} \equiv 1(\bmod 2)$, from
$$x_{n_{1}+T}=x_{n_{1}}$$

We get $\left\{\begin{array}{ll}a_{k}=\vec{a}_{k+T}=1-a_{k+T}, & k \in N, \\ a_{k+T}=\vec{a}_{k+2 T}=1-a_{k+2 T}, & k \in N .\end{array}\right.$
Thus, we have
$$a_{k}=1-a_{k+T}=1-\left(1-a_{k+2 T}\right)=a_{k+2 T}, k \in N$$

Therefore, $x_{n_{1}}$ is also a rational number.
From (1), we know that $x_{n_{1}}$ is derived from $x_{1}$ through $n_{1}-1$ rational operations, so $x_{1}$ must also be a rational number.
(2) If we take
$$x_{1}=(0 . \dot{1} 10 \dot{0})_{2}, x_{1}=(0 . \underbrace{\dot{1} \cdots 10}_{m-1 \uparrow})_{2} \text {, }$$

Then, the corresponding $\left\{x_{n}\right\}$ are periodic with $T=2$ and $T=m, m \geqslant 3$, respectively.
If $x_{1}$ is equal to $\frac{1}{2^{k}}, k \in N$ of the values of $x_{1}$ in (2), then the period $T$ of the corresponding sequence $\left\{x_{n}\right\}$ after a certain term remains unchanged. Therefore, for each $T=2,3, \cdots$, there are infinitely many $x_{1}$ such that the sequence $\left\{x_{n}\right\}$ is periodic with $T$ after a certain term.