Question related to Diophantine approximations and Roth's theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T20:39:11Z http://mathoverflow.net/feeds/question/58569 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58569/question-related-to-diophantine-approximations-and-roths-theorem Question related to Diophantine approximations and Roth's theorem Ramin 2011-03-15T20:17:10Z 2012-02-09T00:22:19Z <p>The following question came up in my arithmetic geometry course yesterday. Suppose $\alpha$ is an irrational real algebraic integer, and suppose $\epsilon >0$ is given. Then by Roth's theorem there are at most finitely many rational numbers $\frac{h}{q}$ with $\gcd(h,q)=1$, $q>1$, such that $$\left| \alpha - \frac{h}{q}\right| &lt; \frac{1}{q^{2+\epsilon}}.$$ Are there any results on how large such $q$ can be? Thanks. </p> http://mathoverflow.net/questions/58569/question-related-to-diophantine-approximations-and-roths-theorem/58571#58571 Answer by Felipe Voloch for Question related to Diophantine approximations and Roth's theorem Felipe Voloch 2011-03-15T20:25:42Z 2011-03-15T20:25:42Z <p>Of course, there may be none or there may be a very small one (if $|\alpha|&lt;1$ then $q=1,h=0$ works). If there is a small very good approximation, the others must be very large. There are lots of papers on effective diophantine approximation dealing with this. I recall there was a paper of Davenport and Roth which first established such a bound.</p> http://mathoverflow.net/questions/58569/question-related-to-diophantine-approximations-and-roths-theorem/58723#58723 Answer by Felipe Voloch for Question related to Diophantine approximations and Roth's theorem Felipe Voloch 2011-03-17T05:41:12Z 2011-03-17T05:41:12Z <p>My other answer was for the first version of this question. The question has now been changed completely. As Antoine mentioned in his comment, an effective Roth's theorem is not known in general. Finding such a result is the probably the main open problem in Diophantine approximation. There some instances in which a non-trivial effective result can be proved. Most notable is the Baker-Feldman theorem which provides such a result with an exponent of the form $\deg \alpha - \epsilon$ (instead of $2+\epsilon$) for a suitable small positive $\epsilon$.</p>