TITLE: Are there lower bounds on the quality of a rational approximation?
QUESTION [5 upvotes]: For which irrational numbers $\xi$ does there exist a constant $A$ such that $\left|\frac{p}{q}-\xi\right|<\frac{1}{Aq^2}$ (where $p/q$ is a rational number) has only finitely many solutions?

Background
I apologize if this is a terribly elementary question—my background is in analysis, not number theory.  If there is some well-known source that answers this question, I would appreciate a link or citation.
As far as I can tell, there are definitely some irrational $\xi$ for which such an $A$ exists.  For instance, if $\xi = \frac{1+\sqrt{5}}{2}$, then for any $A>\sqrt{5}$ the inequality $\left|\frac{p}{q}-\xi\right|<\frac{1}{Aq^2}$ has only finitely many solutions (this seems to be a very elementary result, which appears in a number of texts on elementary number theory).  Additionally, if $\xi$ is an irrational root of a quadratic polynomial, then there is some $A$ for which $\left|\frac{p}{q}-\xi\right|<\frac{A}{q^2}$ has no solutions (see the Wikipedia article on Liouville numbers).  Thus for any irrational root of a quadratic polynomial, there is some constant $A$ of the kind desired.
According to MathWorld, there are some results in this direction:  the article on Hurwitz's Irrational Number Theorem suggests that the answer to this question is be related to the existence or nonexistence of a Lagrange number associated to a particular irrational number.  However, there are only countably many Lagrange numbers, and each Lagrange number is associated to only a countable number of irrationals, so it seems that the existence/nonexistence of a Lagrange number associated to a particular irrational is only a part of the answer.
Again, I apologize if this is a really basic question, and appreciate any help or advice.

REPLY [5 votes]: Here are some standard facts that you can find in many textbooks, e.g. in Cassels: An introduction to diophantine approximation.

There exists an $A>0$ such that $\left|\frac{p}{q}-\xi\right|<\frac{1}{Aq^2}$ has only finitely many rational solutions $\frac{p}{q}$ if and only if the continued fraction expansion of $\xi$ consists of bounded digits. In particular, quadratic irrationals have this property because they have a periodic continued fraction expansion.
There exists an $A<3$ such that $\left|\frac{p}{q}-\xi\right|<\frac{1}{Aq^2}$ has only finitely many rational solutions $\frac{p}{q}$ if and only if $\xi$ lies in an $\mathrm{SL}_2(\mathbb{Z})$-orbit associated with some Markov number (see here). In particular, only countably many $\xi$'s belong to this case.
There are continuum many $\xi$'s such that $\left|\frac{p}{q}-\xi\right|\geq\frac{1}{3q^2}$ for any rational number $\frac{p}{q}$.