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Suppose $Q \in (0,1)$. Then clearly, $\exists$ $x \in \mathbb{R}$, $\ni$ $ \sin(x) = Q$. We define a set

\begin{equation*} P = \left\{ x \textrm{ } | \textrm{ } Q \leq \sin(x) < 1 \right\} \end{equation*}

This set will be non-empty. My question is (with Archana Puran Singh accent), is this set $P \cap \mathbb{N}$ non-empty ? Or is there some $Q$ for which the set $P \cap \mathbb{N}$ is empty ? If we take $Q$ near 1, will it ensure that $P \cap \mathbb{N} = \varnothing$ ?

My MO is that I need to find some $x,y \in S$, $0<x<y$ and $y-x > 1$. If I can prove this, then $P \cap \mathbb{N}$ being non-empty will easily follow.

rtybase
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Minto P
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1 Answers1

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Use the fact that

$\{\sin{n} \mid n \in \mathbb{N}\}$ is dense in $[-1,1]$

proved here. As a result $\forall x,y \in \mathbb{R}$ with $Q<x<y<1$, $\exists n \in \mathbb{N}$ such that $Q<x<\sin{n}<y<1$, which means $n \in P$ and $P\cap \mathbb{N}\ne \varnothing$.

rtybase
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  • But is there any way to prove this problem without using the denseness property of $\left{\sin n \textrm{ } | \textrm{ } n \in \mathbb{N} \right}$ ? – Minto P Aug 20 '18 at 07:16
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    It will be a tedious proof of course, but you can look at continued fractions and convergents of $\frac{\pi}{2}$. – rtybase Aug 20 '18 at 08:36