Bijection Principle

Answer:A,B
Notice that every fraction of the form $\frac{a}{b}$ that lies in $S$ satisfies $ 0< \frac{a}{b} < 1. $ Since gcd$(a,b)= $ gcd$(b-a,b)$. For every fraction $\frac{a}{b},$ such that $0<\frac{a}{b}<\frac{1}{2}$, there is another fraction $\frac{b-a}{b}$ such that $ \frac{1}{2}< 1-\frac{a}{b}< 1. $ Notice that gcd$(a,b)= $ gcd$(b-a,b) =1.$ Therefore, there is a bijection from the elements of set $S$ lying in the interval $ (0,\frac{1}{2})$ to those elements of $S$ lying in the set $(\frac{1}{2},1).$ Similarly, for every fraction $\frac{a}{b},$ such that $0<\frac{a}{b}<\frac{1}{3}$, there is another fraction $\frac{b-a}{b}$ such that $ \frac{2}{3}< 1-\frac{a}{b}< 1. $ Therefore, there is a bijection from elements of set $S$ lying in the interval $ (0,\frac{1}{3})$ to the elements of the set $S$ lying in the interval $(\frac{2}{3},1).$ Therefore, options $(A)$ and $(B)$ are correct.