Roth's theorem, Lang's conjecture and beyond - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T03:28:29Z http://mathoverflow.net/feeds/question/59485 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59485/roths-theorem-langs-conjecture-and-beyond Roth's theorem, Lang's conjecture and beyond David Feldman 2011-03-24T21:13:25Z 2011-03-24T21:13:25Z <p>Lang conjectured that for an irrational algebraic number $\alpha$ and $\epsilon > 0$, there exist only finitely may rationals $p/q$ such that $$\left| \alpha - \frac{p}{q} \right| &lt;\frac{1}{q^2(\ln q)^{1+\epsilon}} \ .$$ Presumably the heuristic here rests on the observation that the set of all reals that violate this condition has measure zero (which follows because a certain sum of characteristic functions has a finite integral). But this applies equally to $$\left| \alpha - \frac{p}{q} \right| &lt;\frac{1}{q^2\ln q (\ln \ln q)^{1+\epsilon}}$$ and $$\left| \alpha - \frac{p}{q} \right| &lt;\frac{1}{q^2\ln q \ln \ln q (\ln \ln \ln q )^{1+\epsilon}},$$ etc.</p> <p>So, my questions:</p> <p>1) Do these strengthenings of Roth's theorem/Lang's conjecture appear in the literature? If not, is it because</p> <p>a) there's no pragmatic point in considering them until someone settles Lang's conjecture or</p> <p>b) there's some theoretical or known reason actually to doubt their truth.</p> <p>2) The heuristic above for Lang's conjecture fails if we take $\epsilon=0$, but I don't know where one looks for counter-examples - meaning algebraic numbers (with deg>2) with infinitely many rational approximations that satisfy for some $c$ $$\left| \alpha - \frac{p}{q} \right| &lt;\frac{c}{q^2\ln q}$$ or if such do exist, then analogous statements with more iterated logarithms in the denominator on the right.</p>