What can be said about this "modified" irrationality measure?

Question

Given $x \in \mathbb{R}$, we define its irrationality measure $\mu=\mu(x)$ by $$\mu = \inf\left\{\lambda : 0<\left|x-\frac{p}{q}\right|<\frac{1}{q^{\lambda}} \ \text{admits at most finitely many integral solutions} \ (p,q), \ q \neq 0 \ \right\}$$

This is the usual definition. Note of course that $\mu=+\infty$ is possible.

I propose a variant of this.

Given $x \in \mathbb{R}$, we define its modified irrationality measure $\mu'=\mu'(x)$ by $$\mu' = \inf\left\{\lambda : 0<\left|x-\frac{p}{q}\right|<\frac{1}{q^{\lambda}} \ \text{admits no integral solutions} \ (p,q), \ q \neq 0 \ \right\}$$

Clearly, $\mu \leq \mu'$, for all $x \in \mathbb{R}$. Is there more that can be said? For instance, can we say that $\mu = \mu'$ for all $x$? This does not seem obvious to me.

If it's not true that $\mu=\mu'$ for all $x$, then I suppose I would like to know why the original definition of $\mu$ is more useful/common as the definition for irrationality measure than $\mu'$? Also, in this case, an explicit example where $\mu \neq \mu'$ would be nice!

Answer 1

The problem with saying no integer solutions is that for every $x\notin\mathbb Z$, every $\lambda$ has at least one solution: $$ 0 < \left|x - \frac{\lfloor x \rceil}{1}\right| < \frac{1}{1^\lambda}, $$ where $\lfloor x \rceil$ is the nearest integer to $x$.

OK, with that out of the way, the problem persists even if we modify the definition of $\mu'$ to not allow $q=1$. I think it might be more obvious why by looking at the irrationality measure a different way. An equivalent definition of the measure is \begin{eqnarray} \mu(x) = \limsup_{q\rightarrow\infty}\left( -\frac{\ln|x-p/q|}{\ln q}\right) \end{eqnarray} where $p$ is always chosen to minimize the difference $|x-p/q|$. It's not hard to show your measure is (if we require $q\ge 2$) $$ \mu'(x) = \sup_{q\ge 2}\left( -\frac{\ln|x-p/q|}{\ln q}\right). $$ The problem is that for almost all numbers, this supremum occurs at small values of $q$ because $\ln q$ is small there. We're more interested in the typical behavior of this measure over all $q$, so the $\limsup$ is more useful.

Now, the question of which numbers have $\mu(x) = \mu'(x)$ is somewhat interesting. The trivial cases are the integers ($\mu(x) = \mu'(x) = 1$) and the Liouville numbers ($\mu(x) = \mu'(x) = \infty$). However, I'm not sure if there are any numbers with $1 < \mu(x) = \mu'(x) < \infty$.

Answer 2

I think I have an answer to my own question. We want the irrationality measure to somehow measure how well an irrational number can be approximated by rationals. The original definition $\mu$ accomplishes this, but my modified definition $\mu'$ fails to.

Consider, for instance, an irrational number whose decimal expansion is of the $\alpha=1.1000000000000000\beta_1\beta_2\beta_3..$ for some (permissible) sequence of 'tail' digits $\{\beta_i\}_{i \geq 1}$. By permissible, we mean that they don't eventually become constant, or start repeating (we want the number to be irrational). Notice that $|\alpha-\frac{11}{10}|< \frac{1}{10^6}\leq \frac{1}{10^{\lambda}}$ for each $\lambda \leq 6$. It follows that $\mu'(\alpha) \geq 6$. This should suggest that $\alpha$ is well-approximable by rationals. But we have no information on the $\beta_i$'s! Hence, we don't actually know for sure that $\alpha$ can be well-approximated by rationals to arbitrary precision. We only that $\alpha$ happens to be (unusually) close to $\frac{11}{10}$, but being unusually close to one particular rational does not imply well-approximability by rationals in general. For that, it's crucial to have infinitely many denominators $q$.