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Let $a>1$, and $b$ be irrational numbers, show that: There exist $n\in N^{+}$, such that $$\{nb\}>\dfrac{1}{a}$$ where $\{x\}=x-[x]$

My idea: since $$\{nb\}=nb-[nb]$$

we only prove $$nb-[nb]>\dfrac{1}{a}$$ then $$nb-\dfrac{1}{a}>[nb]$$ then I can't, thank you very much

math110
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  • see this http://math.stackexchange.com/questions/19565/proving-that-the-terms-of-the-sequence-nx-nx-is-dense-in-0-1 – r9m Mar 07 '14 at 12:42
  • Hello,Then? Thank you – math110 Mar 07 '14 at 12:45
  • Once you have ${{nb}:n\in \mathbb {Z}}$ is dense in $(0,1)$, it means for any $\epsilon \in (0,1)$, you get ${n'b} > \epsilon$ for some $n' \in \mathbb{Z}$ – r9m Mar 07 '14 at 12:48

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