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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?

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    Looks like contradiction may be useful here – 2017-01-25
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    Downvote: Changing a question after receiving an answer is, in my opinion, but practice. It invalidates the answers. I deleted mine for this reason. Consider asking a new question. – 2017-01-25
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    I realised I needed to generalise my question for full correctness - but the answer is still the same, isn't it @Thomas – 2017-01-25
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    @Alex not a problem. Ask a new question and don't let people who answered your question like fools. – 2017-01-25
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    Technically it could be also bad practise to ask an almost identical question twice @Thomas – 2017-01-25
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    I absolutely do agree. But then _you_ will receive feedback for _your_ actions/sloppyness. This way you make others look like fools. – 2017-01-25
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    @Alex: No, the answer changes. Max is correct for the new version. – 2017-01-25

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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.

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    Thanks, very nice "trick" @Max – 2017-01-25
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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.

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    I realize this is an answer to the version before the edit (the edit made the set of $\pi$-rational numbers and its complement dense). I have a comment on "...are isolated, so there is a closest one." Suppose that $B\subset \mathbb R$ is such that each element of $B$ is an isolated point, and $x\in\mathbb R\setminus B$. It does not follow that there is a closest element of $B$ to $x$. E.g., $B$ is the set of reciprocals of positive integers, and $x=0$., – 2017-01-25
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    For the edited version, this means that we have indeed an open interval $I$ around $\frac{x}{\pi} \neq \frac{m}{n}$ and also an open interval for each $z$ such that $\frac{z}{\pi} = \frac{m}{n}$, right? @JonasMeyer – 2017-01-25
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    @Alex, you are quantifying imprecisely and it is hard to follow. Max has answered what I would guess you are asking. – 2017-01-25