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I know that every irrational number $\xi$ can be approximated by rational numbers in such a way that $$ \left| \xi - \frac{p}{q} \right| \leq \frac{c}{q^2} $$ for infinitely many choices of $p$ and $q$. Here $c>0$ is some constant that depends on $\xi$.

My question is: for what (irrational) numbers can we have a precise asymptotic estimate, namely $$ \frac{c_1}{q^2} \leq \left| \xi - \frac{p}{q} \right| \leq \frac{c_2}{q^2} $$ as $q \to +\infty$?

Edit

My question comes from the following problem:

Does there exist a cluster point for the sequence $n \mapsto 2n \{n\xi \}-n$? Here $\xi$ is an irrational number and $\{.\}$ denotes the fractional part of a real number.

If I assume that $x \in \mathbb{R}$ is a cluster point, in particular for every $a<x$ and every $b>x$ there are infinitely many integers $p$ and $q$ such that $$ \frac{2a}{q^2} < \xi - \frac{p}{q} < \frac{2b}{q^2}. $$ At this point I am kind of stuck. I can't understand if the answer is positive or negative, although I suspect that it is more probably negative.

I know that every irrational number $\xi$ can be approximated by rational numbers in such a way that $$ \left| \xi - \frac{p}{q} \right| \leq \frac{c}{q^2} $$ for infinitely many choices of $p$ and $q$. Here $c>0$ is some constant that depends on $\xi$.

My question is: for what (irrational) numbers can we have a precise asymptotic estimate, namely $$ \frac{c_1}{q^2} \leq \left| \xi - \frac{p}{q} \right| \leq \frac{c_2}{q^2} $$ as $q \to +\infty$?

Edit

My question comes from the following problem:

Does there exist a cluster point for the sequence $n \mapsto 2n \{n\xi \}-n$? Here $\xi$ is an irrational number and $\{.\}$ denotes the fractional part of a real number.

If I assume that $x \in \mathbb{R}$ is a cluster point, in particular for every $a<x$ and every $b>x$ there are infinitely many integers $p$ and $q$ such that $$ \frac{2a}{q^2} < \xi - \frac{p}{q} < \frac{2b}{q^2}. $$ Furthermore, $p = [q\xi]$, the integer part of $q\xi$.

At this point I am kind of stuck. I can't understand if the answer is positive or negative, although I suspect that it is more probably negative.

I know that every irrational number $\xi$ can be approximated by rational numbers in such a way that $$ \left| \xi - \frac{p}{q} \right| \leq \frac{c}{q^2} $$ for infinitely many choices of $p$ and $q$. Here $c>0$ is some constant that depends on $\xi$.

My question is: for what (irrational) numbers can we have a precise asymptotic estimate, namely $$ \frac{c_1}{q^2} \leq \left| \xi - \frac{p}{q} \right| \leq \frac{c_2}{q^2} $$ as $q \to +\infty$?

I know that every irrational number $\xi$ can be approximated by rational numbers in such a way that $$ \left| \xi - \frac{p}{q} \right| \leq \frac{c}{q^2} $$ for infinitely many choices of $p$ and $q$. Here $c>0$ is some constant that depends on $\xi$.

My question is: for what (irrational) numbers can we have a precise asymptotic estimate, namely $$ \frac{c_1}{q^2} \leq \left| \xi - \frac{p}{q} \right| \leq \frac{c_2}{q^2} $$ as $q \to +\infty$?

Edit

My question comes from the following problem:

Does there exist a cluster point for the sequence $n \mapsto 2n \{n\xi \}-n$? Here $\xi$ is an irrational number and $\{.\}$ denotes the fractional part of a real number.

If I assume that $x \in \mathbb{R}$ is a cluster point, in particular for every $a<x$ and every $b>x$ there are infinitely many integers $p$ and $q$ such that $$ \frac{2a}{q^2} < \xi - \frac{p}{q} < \frac{2b}{q^2}. $$ At this point I am kind of stuck. I can't understand if the answer is positive or negative, although I suspect that it is more probably negative.