The derived set of $S=\{ m+ \sqrt{2}n: m,n \in Z \} $ is?

Question

The derived set of \begin{equation} S= \{ m+ \sqrt{2}n: m,n \in Z \} \end{equation} is, In the same set like if m and n belong to natural numbers it is clear that the derived set is empty as all points are isolated points. and if m and n belong to rational numbers then maybe I'm not correct I still convinced myself that 'Real numbers' is the derived set as if we take n to be zero still we have a rational number set and that have whole real numbers as the limit points but I'm don't know how to tackle this if m and n as given in the problem. one more doubt for the same set can we say this set is dense in real numbers or not? Any help would be appreciated

Answer 1

Hint: Note that $S$ is such that $\alpha \in S \implies m \alpha \in S$ for all $m \in \Bbb Z$. With that said, it sufficient to show that for all $n \in \Bbb Z$, there exists an $\alpha \in S$ with $0 < \alpha < 1/n$. In order to prove this, consider the fractional part of $k\sqrt{2}$ for all $k \in \Bbb Z$ and apply the pigeonhole principle.