7
$\begingroup$

Given $x \in \mathbb{R}$ we will write $\{x\}$ for the fractional part of $x$ and $\|x\|$ for the distance of $x$ from the nearest integer, in such a way that $\{x\} = x - \lfloor x \rfloor$ and $\|x\| = \min(\{x\}, 1 - \{x\})$, where $\lfloor x \rfloor$ is, as usual, the greatest integer $\le x$.

Let $x$ be a fixed irrational number and $\mu$ its irrationality measure, namely the infimum of the set of all positive exponents $s$ such that

$$0 < \left|x - \frac{m}{n}\right| < \frac{1}{n^s}$$ for finitely many pairs $(m,n) \in \mathbb{Z} \times \mathbb{N}^+$. Assume that $\mu$ is finite (added later: see N. Elkies' answer and the comments below). Recall that $\mu \ge 2$ by (a corollary of) Dirichlet's (approximation) theorem. Then, given $s \in \{2\} \cup [2,\mu[$, define $G_s$ as the set

$$\{n \in \mathbb{N}^+: \|nx\| < n^{1-s}\},$$ and let $G_s^{-} := \big\{n \in G_s: \|nx\| = \{nx\}\big\}$ and $G_s^+ := G_s \setminus G_s^{-}$.

We have that $G_s$ is infinite (by Dirichlet's theorem and the definition of $\mu$), so at least one of $G_s^-$ or $G_s^+$ is infinite too, and I'm tempted to claim that each of them must be infinite. I don't have a serious argument in support of this: My only point is that it would be weird, I believe, to observe a similar "asymmetry" in the diophantine approximations of $x$ (yet, I would be happy to hear that my expectation is completely wrong). This leads to the following:

Q1. Is it true that $G_s^-$ is infinite for each $s \in \{2\} \cup [2,\mu[$? Q2. If Q1 is well-established, would you kindly provide me with a reference?

Some remarks: (i) If the answer to Q1 is yes, then it is as well true that $G_s^+$ is infinite for each $s \in \{2\} \cup [2,\mu[$. (ii) I can prove that the answer to Q1 is positive if $s = 2$ (this follows from Dirichlet's theorem and some elementary properties of the simple continued fraction expansion of $x$). (iii) By (ii) and Khintchine's theorem (which yields that the irrationality measure of most real numbers is equal to $2$), the answer to Q1 is yes for almost all $x$.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ What if the partial quotients in the continued fraction for $x$ alternate between big and small? Would that cause an asymmetry? $\endgroup$
    – Gerry Myerson
    Commented Nov 1, 2013 at 21:42

1 Answer 1

Reset to default
5
$\begingroup$

This can't be right in general. For example $$ x = 0.101000001000000000000000001\ldots = \sum_{k=0}^\infty 10^{-3^k} $$ has plenty of lower approximants with exponent just under 3 (the partial sums) but no upper ones.

Improve this answer
$\endgroup$
4
  • $\begingroup$ [I wanted to extend the decimal expansion to 81 digits but that made it too wide.] $\endgroup$
    – Noam D. Elkies
    Commented Nov 1, 2013 at 21:50
  • $\begingroup$ Right. And the irrationality measure of your $x$ is $3$ (and not $\infty$ as I had previously commented in a hurry). $\endgroup$
    – Salvo Tringali
    Commented Nov 1, 2013 at 22:32
  • $\begingroup$ Do you have a proof that there are no upper ones? I think defining $x$ through its continued fraction may be more tractable, say $x=[0;10,1,100,1,1000,1,\dots]$. $\endgroup$
    – Kevin O'Bryant
    Commented Dec 13, 2013 at 23:03
  • $\begingroup$ The lower approximants with $\mu \rightarrow 3$ have denominators that are spaced close enough that there can't be any upper approximants with $\mu$ much larger than $2$ (by the usual argument with $|p/q - p'/q'| \geq 1/qq'$). $\endgroup$
    – Noam D. Elkies
    Commented Dec 15, 2013 at 4:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .