Roth's theorem states that for every real algebraic $\alpha$ and $\epsilon>0$, there is a $c>0$ such that $$|\alpha -\frac{p}{q}| > \frac{c}{q^{2+\epsilon}}.$$

Lang's conjecture strengthened this to $$|\alpha -\frac{p}{q}| > \frac{c}{q^2 (\log q)^{1+\epsilon}}.$$

A naive further strengthening would be to ask for $$|\alpha -\frac{p}{q}| > \frac{c}{q^2},$$ and this is satisfied by all $\alpha$ with bounded partial quotients $a_n$ (Khinchin Thm 23).

Life would be nice and simple if it were true that all algebraic $\alpha$ has bounded $a_n$. This seems too much to be asking for but I am ignorant of any counterexample. Does anyone knows a counter example or even some heuristic reasoning why this is unlikely to be true ? Thanks in advance.

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    $\begingroup$ The question of boundedness of partial quotients of algebraic numbers is addressed in several article by Enrico Bombieri and Alf van der Poorten, and there are too computational articles by Serge Lang (included in the last edition of his Diophantine Approximations, as far as I remember). It is likely that the continued fractions of algebraic numbers are "random" (behave as generic continued fractions following the Gauss-Kuzmin statistics), so there partial quotients are unbounded. See the answer of Kevin to mathoverflow.net/questions/23111. $\endgroup$ Commented Jun 16, 2010 at 5:22
  • $\begingroup$ @Wadim, you are of course referring to algebraic numbers of degree at least three. @OP, "Khinchin Thm 23" isn't a useful way to give a reference. I suspect you'll find that it's an "if and only if" statement, so that if algebraic numbers of degree at least three all have unbounded partial quotients (which, as others have mentioned, is where the smart money is), then there is no such $\alpha$ for which such a $c>0$ exists. $\endgroup$ Commented Jun 16, 2010 at 5:34
  • $\begingroup$ @Gerry: I call algebraic numbers of degree 2 quadratic irrationalities. :-) Sorry, for confusion. $\endgroup$ Commented Jun 16, 2010 at 5:46
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    $\begingroup$ @Wadim, you'll also need a name for algebraic numbers of degree 1. $\endgroup$ Commented Jun 16, 2010 at 7:02
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    $\begingroup$ @Gerry, hmmmm. Let call them rationalities! :-) $\endgroup$ Commented Jun 16, 2010 at 7:22

1 Answer 1


There is no example of an algebraic number of degree $> 2$ for which the boundedness or not of the entries of the continued fraction has been determined. In a 1976 Annals of Math paper, Baum & Sweet treat the analogous problem with the ring of rational integers replaced by $\mathbb{F}_ 2[x]$ (and its infinite place). They found algebraic quantities over $\mathbb{F}_ 2(x)$ contained in $\mathbb{F}_ 2((x))$ of degree $> 2$ whose "continued fraction" entries have bounded degree, and others with unbounded degree. So it all looks a bit puzzling.

  • $\begingroup$ @Boyarsky. Thanks for your answer and the reference $\endgroup$
    – Chua KS
    Commented Jun 16, 2010 at 5:03

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