The question is: Let $h:\mathbb{R}\rightarrow\mathbb{R}$ be continuous on $\mathbb{R}$ satisfying $h(m/2^n)=0$ for all $m\in \mathbb{Z},n\in \mathbb{N}$. Show that $h(x)=0$ for all $x\in \mathbb{R}$.
My attempt: Suppose that there exists $x$ such that $h(x)\neq0$. Then, since $h$ is continuous, given $\epsilon>0$ there exists $\delta>0$ such that if $\vert{x-c}\vert<\delta$, then $\vert{h(x)-h(c)}\vert<\epsilon$.
I have no clue of what to do next. Can somebody give me a continued solution or any other solution to this question? Thank you.
Given any $\epsilon > 0$ and $x \in \mathbb{R}$ find $m$ and $n$ such that $|x - \frac{m}{2^n}| < \delta$ where $\delta$ corresponds to $\epsilon$ is the definition of continuity of $f$. Thus $|f(x) - f(\frac{m}{2^n})| = |f(x)| < \epsilon$ and it follows $f(x) = 0$. You should argue why such an $m$ and $n$ exist.