This is a follow-up to Question 146635, namely Expected symmetry in the diophantine approximations of an irrational number, which I will refer to for notation and terminology used here without explanation.

Let $x$ be an irrational number and $\mu$ its irrationality measure. For each $s \in \{2\} \cup [2,\mu[$ set

$$G_s := \{n \in \mathbb{N}^+: \|nx\| < n^{1-s}\},$$ and then define $G_s^- := \big\{n \in \mathbb{N}^+: \|nx\| = \{nx\}\big\}$ and $G_s^+ := G_s \setminus G_s^-$ (recall that $\mu \ge 2$). It was already remarked in the other thread that $G_s$ is infinite (by Dirichlet's approximation theorem and the definition itself of $\mu$), so at least one of $G_s^-$ or $G_s^+$ is infinite too.

However, while $G_2^-$ and $G_2^+$ are both infinite, Noam Elkies' example proves that it can well happen that one of $G_s^-$ or $G_s^+$ is finite for all $s \in \; ]2,\mu[$. Moreover, the same example can be generalized to produce a countably infinite family of such examples, by considering Mahler numbers of the form $\sum_{n=0}^\infty a^{-b^n}$, where $a$ and $b$ are integers with $a \ge 2$ and $b \ge 3$ (each of these numbers has irrationality measure equal to $b$; see here for a discussion on this point). As for the case $\mu = \infty$, we can consider Liouville numbers of the form $\sum_{n=0}^\infty 2^{-(n!)^k}$ for $k = 1, 2, \ldots$

Having the above in mind, let us say that $x$ is a balanced number if both $G_s^-$ and $G_s^+$ are infinite for all $s \in \;]2, \mu[$, and unbalanced otherwise. As mentioned in the other thread, almost all numbers are balanced (just because almost all numbers have an irrationality measure equal to $2$). It seems then natural to ask the following:

Q1. Are there uncountably many unbalanced numbers? Q2. More specifically, are there uncountably many [un]balanced numbers of any prescribed irrationality measure (either finite or not)?

I'm pretty sure that the question of deciding whether or not a number is balanced has been stated and studied before, possibly in a more general form. If this is actually the case, then:

Q3. Would you kindly provide me with a reference?

Some further remarks: When $2 \le \mu < \infty$, a rough idea could be to consider a number $x$ of the form $2 \cdot \sum_{n = v}^\infty 3^{-\lfloor \lambda \mu^n\rfloor}$, where $\lambda \in \mathbb{R}^+$ and $v$ is a sufficiently large integer (depending on $\lambda$ and $\mu$). It is known after Y. Bugeaud's Diophantine approximations and Cantor sets (Math. Ann. 341, No. 3 (2008), 677-684) that the irrationality measure of $x$ is $\mu$. However, it seems a little difficult to make Noam's reasoning work in this case too (but I would be rather happy to learn that I'm wrong). Furthermore, this would not cover the case when $\mu$ is infinite.


Yes, there are uncountably many examples (maybe I should have mentioned this variation with my previous answer). For example, $\sum_{k=0}^\infty b_k 10^{-3^k}$ with each $b_k=1$ or $2$. There are uncountably many choices (cardinality of the continuum, even), and for each there are infinitely many lower approximants with exponent approaching 3 (the partial sums, as before), and no comparably good upper approximants.

  • $\begingroup$ Cool! This answers Q2 (and hence Q1) in the unbalanced case ($\mu = \infty$ can now be treated in a similar way). But what about the balanced case? $\endgroup$ – Salvo Tringali Nov 3 '13 at 0:39
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    $\begingroup$ Sure: just let $b_k$ be $1$ or $-1$ instead. (If you want to make sure they're exactly balanced then require also that $b_{2k+2} = -b_{2k+1}$.) $\endgroup$ – Noam D. Elkies Nov 3 '13 at 0:44

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