Is there an open interval about a number that is $\textbf{not}$ $\pi$-rational?

Question

Suppose I have an $x$ such that $x$ is not $\pi$-rational, i.e. $\frac{x}{\pi} \neq \frac{n}{m}$ where $n$ and $m$ are some integers. Does this mean that there is some open interval $I = (x - \epsilon, x+ \epsilon)$ about $x$ such that each $y \in I$ is also not $\pi$-rational?

Answer 1

Note: This was answering a previous version of the question where the $\pi-$ rational numbers were $\frac n{\pi}$, integers divided by $\pi$ instead of rationals divided by $\pi$.

Yes. It is true as long as $x$ is not $\pi-$rational. All the $\pi-$rational numbers are isolated, so there is a closest one. Take $\epsilon$ small enough.

Answer 2

There isn't any such interval. Indeed let $x$ be any real number, and let $r_n$ be any sequence of rationals converging to $x$. Then let $q_n$ be any sequence of rationals converging to $\frac{1}{\pi}$. Then $r_n q_n \pi$ converges to $x$ and is a sequence of $\pi$-rational numbers, so any open interval containing $x$ will contain $\pi$-rational numbers, whether $x$ is $\pi$-rational or not.