Some conjectures on the irrationality measure

Conjecture 2 is true, while conjectures 1, 2.1 and 3 are false. The key is that by considering continued fractions, it is actually quite easy to control the irrationality measure of the number. Specifically, we have the following formula: for an irrational $\alpha$ with continued fraction $[a_0;a_1,a_2,\dots]$ and convergents $p_n/q_n$, we have $$\mu(\alpha)=2+\limsup\frac{\ln a_{n+1}}{\ln q_n}.$$ Since $q_n$ only depends on $a_0,\dots,a_n$, we can freely choose $a_{n+1}$ and make this $\limsup$ anything we want. For instance, picking any $c\geq 0$ and letting $a_{n+1}=\lfloor\exp(c\ln q_n)\rfloor$, we get $\mu(\alpha)=2+c$. This immediately disproves conjecture 3, taking $c$ rational. Note also since the formula above is asymptotic, by letting $a_0,\dots,a_k$ be arbitrary up to some $k$ and only then requiring this recurrence to hold, we in fact get that that numbers with irrationality measure $2+c$ are dense.

The other proofs and disproofs depend on a similar interplay, in which initial segments of the continued fractions approximately determine the value, and we can adjust the limiting behavior accordingly. For instance, here is a sketch of how to show fixed points of $\mu$ are dense, proving 2 and disproving 1. First pick some rational $p/q=[a_0;a_1,\dots,a_k]>2$. Then $p/q$ will be a convergent of any irrational $\alpha$ with continued fraction beginning this way, so $|\alpha-p/q|<1/q^2$. Choose the following coefficients arbitrarily, until we get a convergent denominator $q_m$ large enough that, for any $q_n\geq q_m$, we can find some $a_{n+1}$ satisfying $$\left|2+\frac{\ln a_{n+1}}{\ln q_n}-\frac{p}{q}\right|<\frac{1}{q^2}$$ and demand that the future coefficients satisfy this inequality. This will guarantee that we get $$\left|\mu(\alpha)-\frac{p}{q}\right|\leq\frac{1}{q^2}$$ and thus also $|\mu(\alpha)-\alpha|<2/q^2$. Now we can repeat this construction with $p_m/q_m$ in place of $p/q$, and impose conditions which force $|\mu(\alpha)-\alpha|<2/q_m^2$, and so on. The resulting conditions will give a fixed point $\alpha=\mu(\alpha)$ satisfying $|\alpha-p/q|<1/q^2$, thus giving the desired density.

Together with the disproof of conjecture 3, this gives a disproof of 1 as stated. Indeed, let $\beta=\mu(\beta)$ and let $\alpha\neq\beta$ satisfy $\mu(\alpha)=\beta$, then the iteration of $\mu$ does not converge to $1$. There are more interesting ways for this to fail as well though - by adapting the construction above, it is possible to construct numbers $\alpha\neq\beta$ such that $\mu(\alpha)=\beta,\mu(\beta)=\alpha$, and in fact such pairs will be dense in the product $(2,\infty)^2$. Thus iteration of $\mu$ needn't converge at all.

More generally, for any $k$, we can have length $k$ loops $\alpha_1,\dots,\alpha_k$ which are in fact dense in $(2,\infty)^k$. Lastly, this iteration need not fall into a loop at all, and can behave quite arbitrarily - through a lot of bookkeeping, one can adapt the proof to show that for any choice of intervals $(a_k,b_k)\subseteq(2,\infty)$ there is some irrational $\alpha$ such that $\mu^k(\alpha)\in(a_k,b_k)$ for every $k$.

(This answer was written at the same time as Wojowu's, so it may overlap to some extent.)

Here's an attempt at proving constructively that for any open intervals $I \subseteq \mathbb{R}$ and $J \,\subseteq\, \mathopen]2,+\infty\mathclose[$ and for any continuous order-preserving $f\colon I\to J$ there exists a $\xi \in I$ (and therefore infinitely many) such that $\mu(\xi) = f(\xi)$.

Recall that two rationals $r = p/q$ and $r' = p'/q'$ (all my rationals are implicitly assumed to be in irreducible form with positive denominator) are adjacent when $pq' - p'q = \pm 1$ (that is, $r - r' = \pm\frac{1}{qq'}$).

Lemma 1: Given $r = p/q$ and $\varepsilon \in \{-1,1\}$, for any $q_{\text{min}}, q_{\text{max}}$ with $q_{\text{max}} - q_{\text{min}} \geq q$, one can find $r' = p'/q'$ adjacent to $r$ with $pq' - p'q = \varepsilon$ and $q_{\text{min}} < q' \leq q_{\text{max}}$.

Proof. Consider $\varepsilon p^{-1}$ mod $q$ (where $p^{-1}$ denotes the modular inverse, which makes sense since $p/q$ is irreducible). Since $q_{\text{max}} - q_{\text{min}} \geq q$ there is a representative $q'$ of this class mod $q$ which satisfies $q_{\text{min}} < q' \leq q_{\text{max}}$. So now $pq' \equiv \varepsilon \pmod{q}$, so there is $p'$ such that $pq' - pq' = \varepsilon$, as claimed. ∎

Lemma 2: If $(r_n)$ with is a sequence of rationals with $r_n = p_n / q_n$ and $p_n q_{n+1} - p_{n+1} q_n = (-1)^{n+1}$, and $(\alpha_n)$ is a sequence of reals with $\alpha_{n+1} \geq \alpha_n > 2$, having limit $\alpha$, and if $q_n^{\alpha_n - 1} < q_{n+1} \leq C \cdot q_n^{\alpha_n - 1}$ holds for each $n$, for some constant $C>1$ independent of $n$, then $(r_n)$ has a limit $\xi$ having irrationality measure $\alpha$.

Proof. First note that the assumptions imply $q_{n+1} > q_n$ for all $n$. The adjacency relations $p_{n-1} q_n - p_n q_{n-1} = (-1)^n$ and $p_n q_{n+1} - p_{n+1} q_n = (-1)^{n+1}$ along with $q_{n+1} > q_{n-1}$ ensure that $r_{n+1}$ is always between $r_{n-1}$ and $r_n$. The sequence $(r_n)$ converges: indeed, we have constructed its limit $\xi$ by the convergents $p_n/q_n$ of its continued fraction expansion. We have $r_0 < r_2 < r_4 < \cdots < \xi < \cdots < r_5 < r_3 < r_1$.

Let $\xi$ be the limit of $(r_n)$. I claim that $\xi$ has irrationality measure $\alpha$.

First, let us show that it has irrationality measure at least $\alpha$. For all $n$ we have $|r_{n+1} - r_n| = \frac{1}{q_n\,q_{n+1}} < \frac{1}{q_n^{\alpha_n}}$, and since $\xi$ is between $r_n$ and $r_{n+1}$ we have $|\xi - r_n| < \frac{1}{q_n^{\alpha_n}}$. In particular, if $n\geq k$, we have $|\xi - r_n| < \frac{1}{q_n^{\alpha_k}}$. Since the $(r_n)$ for $n\geq k$ are infinitely many, this shows that $\xi$ is approximated at least to order $\alpha_k$ (for any $k$), and since this is true for all $k$ we conclude that $\mu(\xi) \geq \alpha$.

On the other hand, if $r = p/q$ is any rational with denominator $q_{n-1} \leq q < q_n$ (which implies $q_n < C \cdot q_{n-1}^{\alpha_{(n-1)}-1} \leq C \cdot q^{\alpha_n - 1}$), we have $|r - r_n| \geq \frac{1}{q q_n}$ but (as we have seen in the previous paragraph) $|\xi - r_n| < \frac{1}{q_n^{\alpha_n}}$, giving $$ \begin{aligned} |\xi - r| &> \frac{1}{q q_n} - \frac{1}{q_n^{\alpha_n}}\\ & = \frac{1}{q q_n}\left(1 - \frac{q}{q_n^{(\alpha_n)-1}}\right)\\ & \geq \frac{1}{q q_n}\left(1 - \frac{1}{q^{(\alpha_n)-2}}\right)\\ & \geq \frac{1}{q q_n}\left(1 - \frac{1}{q^{(\alpha_0)-2}}\right)\\ & \geq \frac{1}{2 q q_n}\quad\text{for $q$ large enough}\\ & \geq \frac{1}{2C \cdot q^{\alpha_n}}\\ \end{aligned} $$ So if $\beta > \alpha$ we have $|\xi - r| > \frac{1}{q^\beta}$ for all but finitely many $r$, and this shows that $\xi$ cannot be approximated to order $\beta$.

Therefore the irrationality measure of $\xi$ is exactly $\alpha$. ∎

Proposition: For any open intervals $I \subseteq \mathbb{R}$ and $J \,\subseteq\, \mathopen]2,+\infty\mathclose[$ and for any continuous order-preserving $f\colon I\to J$ there exists a $\xi \in I$ such that $\mu(\xi) = f(\xi)$.

Proof. Let $r_0 = p_0 / q_0$ and $r_1 = p_1 / q_1$ be adjacent rationals in $I$ with $2 < q_0 < q_1$ and $r_0 < r_1$. Using lemma 1, we construct by induction a sequence $r_n = p_n / q_n$ satisfying $p_n q_{n+1} - p_{n+1} q_n = (-1)^{n+1}$ and $q_n^{\alpha_n-1} < q_{n+1} \leq 2 q_n^{\alpha_n-1}$ where $\alpha_n = f(r_m)$ for $m := 2\lfloor n/2\rfloor$ the largest even integer $\leq n$. This is possible because $q_n^{\alpha_n-1} \geq q_n$. Now (since $r_0 < r_2 < r_4 < \cdots$ so $\alpha_0 \leq \alpha_1 \leq \alpha_2 \leq \cdots$) we are in the conditions where lemma 2 applies so $r_n$ has a limit $\xi$ having irrationality measure the limit of the $\alpha$, which is none other than $f(\xi)$. ∎

(By applying this to smaller intervals $I$ we get infinitely many such $\xi$.)

In particular, applying this to $f$ being the identity function and the constant function, we get:

Corollary: The real numbers such that $\mu(\xi) = \xi$ are dense in $]2,+\infty\mathclose[$, and for any $d>2$ the real numbers such that $\mu(\xi) = d$ are dense in $\mathbb{R}$ [note: the latter statement is also true for $d=2$, but the above proposition doesn't apply].

Note that, even though I didn't always phrase everything in a constructively ḥalāl way, these proofs are easily modified to be valid in any reasonably form of constructive math, and give effective algorithms, e.g., given any pair of rationals $2<r<s$ we can effectively compute a sequence $p_n/q_n$ of rationals converging (in an effective sense) to $\xi$ in $\mathopen]r,s\mathclose[$ with $\mu(\xi) = \xi$.