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More precisely, what real numbers $r$ have the following property: for any $\epsilon > 0$ there exist infinitely many pairs $(p, q)$ of integers such that

$$\left| \frac{p}{q} - r \right| < \frac{\epsilon}{q^2}.$$

I think that this is impossible if $r$ is a quadratic irrational. On the other hand, it's certainly possible for any number with irrationality measure strictly greater than $2$.

What I really want to know is if the real numbers which don't have the property above have measure zero. If that's true, it would answer the last part of this math.SE question.

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up vote 13 down vote accepted

This is a well-studied question in diophantine approximation. You can look up Markov or Lagrange spectrum for a "description" of the numbers for which you cannot take $\epsilon$ arbitrarily small. For the answer to your last question, look up Khinchin's theorem (the answer is no, they have full measure).

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Thanks, Felipe! I can't seem to find a statement of Khinchin's theorem (I assume you're not referring to the theorem about Khinchin's constant) online. Do you have a reference? – Qiaochu Yuan Oct 24 '10 at 14:47
2 – Felipe Voloch Oct 24 '10 at 14:56
If I read it correctly, the answer is yes, almost all real numbers can be approximated pretty well by rationals. – Douglas Zare Oct 25 '10 at 20:14
For intuition, large coefficients in the simple continued fraction expansion produce good rational approximations. There is a limiting distribution for the coefficients, the Gauss-Kuzmin distribution, which is supported on all positive integers, so you expect that with some coarse independence that you get infinitely many coefficients greater than any fixed size except on a set of measure $0$. – Douglas Zare Oct 25 '10 at 20:32
@Douglas: Now I am confused, too many double negations. Most numbers can be well approximated. – Felipe Voloch Oct 25 '10 at 20:40

Hardy and Wright devoted a chapter (chapter 9) to these questions. One interesting theorem related to your question is theorem 196.

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