Fix an irrational number $\alpha \in (0,1)$. Let $\{x\}$ denote the fractional part of the real number $x$. Consider the sequence $$\{{\{\alpha}n\}:n=1,2,\cdots \}.$$ This sequence is uniformly distributed in the unit interval. However, is it the case that $${\{\alpha}n\}\ge {\frac{1}{n}}$$ for sufficiently large positive integers $n$?
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The answer is NO! I found a theorem in Ivan Niven's book Diophantine Approximation (Theorem 3.1, page 23). Not sure I should post it as an answer since I didn't solve it! – student May 06 '20 at 04:45
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The statement of the theorem is as follows: Given an irrational number $\alpha$ there is an increasing sequence of positive integers : ${k_1}<{k_2}<\cdots$ such that for $i\geq1$, $${{{\alpha}k_i}} < {\frac{1}{k_{i+1}}}$$. The proof may be found on pp 23-24 of the book by Ivan Niven Diophantine Approximation. – student May 06 '20 at 04:53