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By definition, $$0<\left\lvert \frac{p}{q}-x\right\rvert<\frac1{q^{\mu(x)-\epsilon}}$$ has infinitely many solutions $(p,q)$ for every $\epsilon>0$.

However, to prove a theorem in a current project, I would need the following inequality to have infinitely many solutions: $$ 0<\frac{p}{q}-x<\frac1{q^{\mu(x)-\epsilon}}$$ In other words, I need infinitely many good over-estimated rational approximations for $x$.

Since $x$ in my project is transcendental, I used $<0$ instead of $\le0$.

My questions are

  1. Does the definition of irrationality measure directly implies the second inequality has infinitely many solutions?
  2. If no, then what conditions can we add on $x$ such that the second inequality has infinitely many solutions? What are some examples of $x$ that satisfies the new conditions?
  3. Or, indeed, is it impossible for the second inequality to have infinitely many solutions?

This question may lack context; I apologize for being unable to provide more.

Thanks in advance.

Szeto
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  • @Hurkyl Yes, sorry about that. I will add it. – Szeto Jun 23 '18 at 08:51
  • Perhaps you shoud add a definition of $\mu(x)$ or refer to your question https://math.stackexchange.com/q/2828095. I would not rule out the possibilty that there are two irrationality measures $\mu_\pm(x)$ - one overestimating and one underestimating. To show that $\mu_+(x) = \mu_-(x)$ is certainly non-trivial (if it is true). – Paul Frost Jun 23 '18 at 10:04

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