Let $0<a<b<1$ be reals numbers and \[g(x)=\left\{\begin{array}{cc}x+1-a,&\mbox{ if } 0<x<b\\b-a, & \mbox{ if } x=a \\x-a, & \mbox{ if } a<x<b\\1-a ,&\mbox{ if } x=b \\ x-a ,&\mbox{ if } b<x<1 \end{array}\right.\] Give that there exist $n+1$ reals numbers $0<x_0<x_1<...<x_n<1$, for which $g^{[n]}(x_i)=x_i \ (0 \leq i \leq n)$. Prove that there exists a positive integer $N$, such that $g^{[N]}(x)=x$ for all $0<x<1$. ($g^{[n]}(x)= \underbrace{g(g(....(g(x))....))}_{\text{n times}}$) Official solution here