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**ThePerfectHacker** Let $\displaystyle r\in \mathbb{R}$ with $\displaystyle h(r) > 0$ WLOG. Take $\displaystyle \epsilon > 0$ so that $\displaystyle h(r) - \epsilon > 0$. Then it means there is $\displaystyle \delta > 0$ such that if $\displaystyle x\in (r-\delta,r+\delta) \implies h(x) > h(r) - \epsilon > 0$. The problem is that on any interval we can find a number having the form $\displaystyle \tfrac{m}{2^n}$. And $\displaystyle h( \tfrac{m}{2^n} ) = 0 \not > 0$. This is a contradiction.