I recently learned of so-called 'far' numbers at a talk. In the talk, it was proven that there is a dense subset of the interval $[0,1]$ of far numbers (however, far numbers were only a minor point of the talk, so we quickly went on to bigger and better things).
A number $x$ might be called far if there is a positive constant $c$ s.t. $\displaystyle \left\| \;x - \frac{k}{2^n} \right \| \geq \frac{c}{2^n}$ for all $k, n \in \mathbb{N}$.
It is not so bad to see that for any odd prime $p$, $1/p$ is a far number. For that matter, for most rationals, it's not so hard to see that they are far. It happens to be that the set of far numbers has measure $0$. But I began to wonder:
Is there a 'far' irrational number?
Consider $x = \sum_{j=1}^\infty a_j/4^j$ where each $a_j$ is either $1$ or $2$. Thus the binary expansion of $x$ consists of two-digit blocks which are either $10$ or $01$. Then $x$ is far. But there are uncountably many choices, so all but countably many of them are irrational.