Let $S$ be a subset of the irrationals. Also, lets assume that $S$ has infinitely many elements. My very general question is, under what non-trivial conditions does there exist an element $x\in S$ such that $x^2$ is rational. For example, if $S$ were the entire set of irrationals, then the usual proof that $\sqrt{2}$ is irrational would find such an $x$. On the other hand, the same methods would not necessarily be sufficient if $S$ is a strict subset of the irrationals. Is there some natural set of actions we can throw into $S$ to find such an $x$? For example, we could assume $S$ is closed under addition and multiplication. Is there some general theory about any of this?
I would appreciate examples of such non-trivial sets $S$.
The exact condition is that $S$ intersects $\{\pm\sqrt{r}: 0 \le r \in {\mathbb Q}\}$, where $\mathbb Q$ is the rationals. It's a dense set, but countable, and therefore very "easy to miss" by anything that doesn't have to contain an open interval.