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  • 2
    $\begingroup$ It shouldn't be too difficult to show, even constructively, that there exists a solution to $μ(α)=α$ in any given interval above $2$: you start with rationals $α_0 = p_0/q_0$ and $α_1 = p_1/q_1$ in the given interval with $p_0 q_1 - p_1 q_0 = -1$, then you construct $p_{n+1},q_{n+1}$ such that $q_n^{α_n-1} < q_{n+1} < 2 q_n^{α_n-1}$ and $p_n q_{n+1} - p_{n+1} q_n = -1$ (by taking $q_{n+1}$ congruent to $-p_n^{-1}$ mod $q_n$), so the $p_n/q_n$ approximate $α$ to order exactly $α$, and since they will be the best we can do no better. $\endgroup$ Commented 2 days ago
  • 2
    $\begingroup$ [contd.] I mean, basically, take a constructive proof of the fact that for any $d>2$ there is an irrational with irrationality measure exactly $d$ (it will proceed by constructing a sequence $p_n / q_n$ converging at the desired rate) and you just replace $d$ at each step by the approximation $p_n / q_n$ constructed at the previous step. If you want I can try to write it all down, but I don't think there's any substantial difficulty. (Did you even try?) $\endgroup$ Commented 2 days ago
  • 1
    $\begingroup$ PS: One normally calls “conjecture” something which one strongly believes. Your “conjectures” contradict each other, so obviously you don't strongly believe all of them. (It's not that I care about gatekeeping the use of terminology, but it's confusing, because I wondered whether I had misunderstood them.) $\endgroup$ Commented 2 days ago
  • $\begingroup$ @Gro-Tsen : It seems to work even though I’m not fully convinced. Maybe post it as an answer. Also, some conjectures may be « too easy », I apologize for that, I’m far from being an expert in this field. And for your remark on the conjectures, I don’t see any « obsvious » contradictions but yes I do not strongly believe each statement. I wanted to make this in questions but I started in conjectures so I had to state something not to ask something. $\endgroup$ Commented 2 days ago
  • 3
    $\begingroup$ Using the relation between irrationality measure and continued fraction expansion (see Wikipedia) it is easy to show every number $\geq 2$ is an irrationality measure of some irrational. This immediately disproves 3, and by picking $\alpha$ with irrationality measure equal to a Liouville number, or some number such in conjecture 2, disproves 1. Same result together with Gro-Tsen's idea should let you show that fixed points of $\mu$ are dense in $(2,\infty)$. $\endgroup$ Commented 2 days ago