Existence of square root 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?
 A: If you are using Completeness Axioms (like existence of least upper bound (LUB) of a set bounded above) then (If I recollect correctly) the "standard" approach is to consider (assuming $r \in (0,1)$)
$$ S = \{ x : x^2 \lt r, \ x \in [0,1]\}$$
and prove that the LUB (or supremum) $l$ of $S$ satisfies $l^2 = r$ by showing
(Hint)

 $l^2 \le r$ and that if $l^2 \lt r$, then there is some $s \gt l$ such that $s^2 \lt r$

(Spoiler)

 Given any $\epsilon \gt 0$, there is an $x \in S$ such that $x \gt l - \epsilon$. 
 Squaring and using the fact that $l \lt 1$ (Why?) leads to $r \gt l^2 - 2 \epsilon$ 
 This implies that $l^2 \le r$ (Why?)  
 If $a = r - l^2 \gt 0$, then we have that by choosing $\epsilon = \frac{a}{3}$ that $\epsilon \lt 1$ (Why?) and $a \gt 2\epsilon + \epsilon^2 \gt 2 \epsilon l + \epsilon^2$ and thus $r \gt (l + \epsilon)^2 = s^2$ with $s \gt l$.  
 We can also show that $s \lt 1$ and thus $s \in S$ contradicting that $l = \sup S$  
 Note that the proof above gives rise to the sequence 
 $x_{n+1} = x_n + \dfrac{r - x_n^2}{3}$ which is quite similar to Henry's sequence.

From what I could see from Google books, the book you are using does use that axiom.
A: (i) Let $s_0 = r$ and $s_{n+1} = s_n + \dfrac{r-s_n^2}{2}$
(ii) If $a \gt 1$, let $r=\dfrac{1}{a}$ and use (i)
