# Hurwitz's theorem for Diophantine approximation

Hurwitz's Theorem in Number Theory states that for every irrational number $$\xi$$, there are infinitely many relatively prime natural numbers $$(p,q)$$ satisfying the equation: $$| \xi−\frac{p}{q}| < \frac{1}{\sqrt{5}q^2}$$

I'm interested in unilateral approximations to $$\xi.$$ Specifically, letting

$$L = \left\{(p,q): \; p \text{ and } q \text{ are relatively prime positive integers such that } \frac{p}{q} < \xi \text{ and } \left|\xi - \frac{p}{q}\right| < \frac{1}{\sqrt{5}q^2} \right\}$$

and

$$U = \left\{(p,q): \; p \text{ and } q \text{ are relatively prime positive integers such that } \frac{p}{q} > \xi \text{ and } \left|\xi - \frac{p}{q}\right| < \frac{1}{\sqrt{5}q^2} \right\},$$

then for each irrational $$\xi$$ we can conclude that AT LEAST ONE of the sets $$L$$ and $$U$$ is infinite. Do we know whether, for each irrational $$\xi,$$ BOTH of the sets are infinite? If not, then do we at least know whether, for each irrational $$\xi,$$ BOTH of the sets are not empty?

• any $p>\xi$, and $q=1$? Nov 26, 2021 at 10:08
• The simple continued fraction gives approximations alternating too big and too small , hence there are in fact inifnite many fractions (and more important : arbitary good fractions both from above and from below) doing the job. Nov 26, 2021 at 10:10
• (I retracted my vote to close because I just understood what op meant; still I think the question should be reformulated to be clearer) Nov 26, 2021 at 10:19
• It might be that if all the even partial quotients (in the continued fraction for $\xi$) are small, while all the odd partial quotients are big, then all the Hurwitz-good approximations will be on the same side of $\xi$. Nov 26, 2021 at 12:20
• @DaveL.Renfro thanks for reformulating. That's indeed what I meant. Nov 27, 2021 at 11:09

For any $$c>1$$ there are uncountably many irrationals $$\xi$$ for which $$0<{a\over b}-\xi<{1\over cb^2}$$ has no rational solutions $$a/b$$. Analogously, for any $$c>1$$ there are uncountably many irrationals $$\xi$$ for which $$0<\xi-(a/b)<1/(cb^2)$$ has no rational solutions.