I have a problem in computing and understanding in description of multicones with Farey intervals. Let $M^{*}$ be the monoid On the generators $F_{+}$ , $F_{-}$ operating on words in $A,B$ by the substitutions \begin{align} F_{+}&: A\mapsto A &&B\mapsto AB \\ F_{-}&:A\mapsto BA &&B\mapsto B \end{align} We identify $M^*$ with $\mathbb{Q}\cap(0,1)$ via the canonical bijection j: for $F\in M^*$, $j(F)=p/q$ if $F(AB)$ has length $q$ and contains $p$ times the letter B. We have $j(id_{M^*})=1/2$.
For $F\in M^*$, with $j(F)=p/q$, denote by $O(p/q)$ the set of words of length $q$ deduced from $F(AB)$ by cyclic permutation. This set can also be described in the following way. Consider the map $$ R_{p/q}:[0,1)\longrightarrow [0,1) \text{ defined by }x\mapsto x+p/q \bmod 1 $$ and set $$ \theta(x)=\begin{cases} A&\text{if $x \in [0,1-p/q)$}\\ B&\text{if $x \in [1-p/q,1)$} \end{cases} $$ Then setting $\Theta(x)=(\theta({R^i}_{p/q}(x)))_{0\leq i < q}$, the image of $\Theta$ is $O(p/q)$.
Let $F\in M^*$ , with $j(F)=p/q$ ; let $[p_{0}/q_{0}, p_{1}/q_{1}]$ be the Farey interval with center $p/q$. Recall that $$ p_{0}+p_{1}=p, \,\,\,\, q_{0}+q_{1}=q, \,\,\, p_{1}q_{0}-p_{0}q_{1}=1 \,\,\,\, (**) $$ Then $O(p_{0}/q_{0})$ is the set of words deduced from $F(A)$ by cyclic permutation, and $O(p_{1}/q_{1})$ is similarly the set of words deduced from $F(B)$ by cyclic permutation. Here we extend the definition of $O(p/q)$ setting $O(0/1)=\{A\}$ and $O(1/1)=\{B\}$. In the main contex said that from (**) it is obvious that $${R^{q_{1}}}_{p/q}(0)={R^{-q_{0}}}_{p/q}(0)=1-1/q$$ but I couldn't conclusion it from $(**)$ from any computation. Could you help me to reach the above line from $ (**)$?
Thanks in advance.