CMO 2007 Problem 4

For two real numbers $a$ , $b$ , with $ab\neq 1$ , define the $\ast$ operation by $$a\ast b=\frac{a+b-2ab}{1-ab}.$$ Start with a list of $n\geq 2$ real numbers whose entries $x$ all satisfy $0<x<1$ . Select any two numbers $a$ and $b$ in the list; remove them and put the number $a\ast b$ at the end of the list, thereby reducing its length by one. Repeat this procedure until a single number remains.

$a.$ Prove that this single number is the same regardless of the choice of pair at each stage. $b.$ Suppose that the condition on the numbers $x$ is weakened to $0<x\leq 1$ . What happens if the list contains exactly one $1$ ?