a) Reverse opeartions $A$ and $B$ to get only one way to get $\frac{20}{23}$ $1 \to 2 \to \frac{2}{3} \to \frac{20}{3} \to \frac{20}{23}$ b) Obvious, that we can get only positive rationals. Let $r$ is rational that can not be got by this operations and $r=\frac{a}{b}, a>0,b>0, (a,b)=1$ and sum $(a+b)$ is minimal Then $r<1$ because if $r>1$ then $r-1=\frac{a-b}{b}$ has minimal sum. So $a<b$ But then number $\frac{a}{b-a}$ has less sum, so it is possible to get this number, and then using operation $B$ it is possible to get $\frac{\frac{a}{b-a}}{\frac{a}{b-a}+1}=\frac{a}{b}=r$ So every positive $r$ can be presented Other way to prove is using reverse operations $A':\frac{a}{b} \to \frac{a-b}{b}$ for $a>b$ and $B': \frac{a}{b} \to \frac{a}{b-a}$ for $b>a$ Not hard to show that these operations similar to Euclidian algorithm, and so after some operations from $r$ we will get number $\frac{c}{1}$ and then $1$ or number $\frac{1}{c}$ and then using $B'$ to get $1$