# Are there lower bounds on the quality of a rational approximation?

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.

• Doesn't the article on Liouville numbers answer your question? (I.e., for a Liouville number, such an A cannot possible exist?) – David Cohen Aug 25 '12 at 20:56
• David: Perhaps I should attempt to clarify. The question is not whether or not such an $A$ exists for each irrational number, but rather, given a <i>particular</i> irrational number, is there such an $A$? In the case of the Liouville numbers, the answer is no. In the case of irrational roots of quadratics, the answer is yes. – Xander Henderson Aug 25 '12 at 23:06
• @Xander: Could you make your question more precise? Do you have a specific number in mind? Do you want an if and only if statement that is easier than the approximation condition (which almost certainly doesn't exist)? Have you looked at Khintchine's theorem, which says that the set of numbers with such an A has full measure? – Felipe Voloch Aug 26 '12 at 0:01
• This is a duplicate of mathoverflow.net/questions/43381/…. – Douglas Zare Aug 26 '12 at 0:05
• The existence of such an $A$ is equivalent (for irrational $\xi$) to the boundedness of the partial quotients of the continued fraction of $\xi$. Almost all numbers have unbounded partial quotients, so there is no such $A$. Except for quadratic irrationals and irrationals defined through their cf, I don't think any number is known to have bounded partial quotients. There are a very few, such as $e$, that are known to have unbounded partial quotients. – Kevin O'Bryant Aug 26 '12 at 1:31

1. 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.
2. 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.
3. There are continuum many $\xi$'s such that $\left|\frac{p}{q}-\xi\right|\geq\frac{1}{3q^2}$ for any rational number $\frac{p}{q}$.