I am self-studying real analysis for fun and came across the following question (from A First Course in Real Analysis by Berberian).
(i) Prove the following: If $r$ is a real number such that $0 \leq r \leq 1$, then there exists a unique real number $s$ such that $0 \leq s \leq 1$ and $s^2 = r$.
(ii) Generalize (i) to the following: If $a \in \mathbb{R}$, $a \geq 0$, then there exists a unique $b \in \mathbb{R}$, $b \geq 0$ such that $b^2 = a$.
Put $y = 1-r$ and $x = 1-s$. Then we want to find a real number $x \in [0,1]$ such that $(1-x)^2 = 1-y$. I am trying to find an increasing sequence $(x_n)$ such that $0 \leq x_n \leq 1$ (defined recursively).
For (ii) how could I use (i) to get the result?