, 2006

Since $x$ is rational, its digits repeat periodically starting at some point. We wish to show that this is also true for the digits of $y$, implying that $y$ is rational.

Let $d$ be the length of the period of $x$ and let $d = 2^{u} \cdot v$, where $v$ is odd. There is a positive integer $w$ such that
$$2^{w} \equiv 1 \quad (\bmod v) .$$
(For instance, one can choose $w$ to be $\varphi(v)$, the value of Euler's function at $v$.) Therefore
$$2^{n+w} = 2^{n} \cdot 2^{w} \equiv 2^{n} \quad (\bmod v)$$
for each $n$. Also, for $n \geq u$ we have
$$2^{n+w} \equiv 2^{n} \equiv 0 \quad \left(\bmod 2^{u}\right)$$
It follows that, for all $n \geq u$, the relation
$$2^{n+w} \equiv 2^{n} \quad (\bmod d)$$
holds. Thus, for $n$ sufficiently large, the $2^{n+w}$th digit of $x$ is in the same spot in the cycle of $x$ as its $2^{n}$th digit, and so these digits are equal. Hence the $(n+w)$th digit of $y$ is equal to its $n$th digit. This means that the digits of $y$ repeat periodically with period $w$ from some point on, as required.