Dirichlet's approximation theorem with even or odd denominators

Question

It follows from Dirichlet's approximation theorem that for any irrational $\alpha,\ 0<\left\lvert \frac{p}{q} - \alpha \right\rvert < \frac{1}{q^2} $ for infinitely many pairs of integers $(p,q).$ In fact, Hurwitz's theorem is a nice result which says that the best we can do with the above is replace $\ \frac{1}{q^2}\ $ with $\ \frac{1}{\sqrt{5}q^2},$ and that the result does not hold if we replace $\sqrt{5}$ with any number $A>\sqrt{5},$ so this $\ \sqrt{5}$ is an (least) upper bound.

My question is the following. Is it true that, for any irrational $\alpha,\ 0<\left\lvert \frac{p}{q} - \alpha \right\rvert < \frac{1}{q^2} $ for infinitely many pairs of integers $(p,q)$ with $q$ even?

And similar question for $q$ odd?

Or can we construct an irrational number such that the above inequality is only true for finitely many even (similarly: odd) $q$?

Answer 1

Every irrational $\alpha$ has infinitely many such rational approximations with $q$ odd. Indeed, it is known that all of the convergents (in the sense of continued fractions) to $\alpha$ satisfy the inequality in Dirichlet's approximation theorem. If the continued fraction of $\alpha$ equals $[a_0;a_1,a_2,\dots]$, then the denominators of the convergents of $\alpha$ satisfy $k_0 = 1$, $k_1=a_1$, and $k_n=a_nk_{k-1}+k_{n-2}$ for $n\ge2$. In particular, if $k_{n-2}$ is odd and $k_{n-1}$ is even, then $k_n$ is automatically odd. In other words, it's impossible for the sequence of denominators to have two consecutive even terms, and therefore there are infinitely many convergents with odd denominator.

There are certainly irrational numbers all of whose convergents have odd denominators; the simplest one is $\frac1{\sqrt2} = [0;1,2,2,2,\dots]$. However, this only happens when the partial quotients $a_j$ are all even, which forces the improved inequality $|\alpha-\frac{h_n}{k_n}| < \frac1{a_{n+1}k_n^2} \le \frac1{2k_n^2}$. I haven't worked out the details, but I believe it's possible to show that when $k_{n-1}$ and $k_n$ are both odd, this forces the mediant $\frac{h_{n-1}+h_n}{k_{n-1}+k_n}$ (the next "secondary convergent"), which has even denominator, to satisfy $|\alpha-\frac{h_{n-1}+h_n}{k_{n-1}+k_n}| < \frac1{(k_{n-1}+k_n)^2}$. In other words, I believe the answer is still "yes" for infinitely many even denominators.