Question

# Deduce from the Completeness Axiom that there exists a square root of a real number a if and only if a ≥ 0

Deduce from the Completeness Axiom that there exists a square root of a real number a if and only if $$a \geq 0$$

2021-02-25

Suppose that sqrta exists. Then $$\displaystyle{a}={\left(\sqrt{{a}}\right)}^{{2}}$$, and we know that a square of a real number is alwais nonnegative, so $$\displaystyle{a}\Rightarrow{0}$$.
Suppose that $$\displaystyle{a}\Rightarrow{0}$$. Define a set $$\displaystyle{S}={\left\lbrace{x}∈{R}{\mid}{x}^{{2}}\le{a}\right\rbrace}$$
Then $$\displaystyle{0}∈{S}$$, so $$S =0$$. Furthermore, S has an upper bound, fi $$\displaystyle{a}\le{1},{1}$$ can be an upper bound, and if $$a>1$$, a can be an upper bound (because $$\displaystyle{a}^{{2}}{>}{a}$$ when a>1).
By the Completeness Axiom, there exists a supremum of S: $$\displaystyle{s}=\supset{S}$$
Now we can see that we must have $$\displaystyle{s}^{{2}}={a}$$ (since it is the least upper bound of S, otherwise, if $$\displaystyle{s}^{{2}}{>}{a}$$, we could find a smaller upper bound, which is a contradiction). Thus, $$\displaystyle{s}=\sqrt{{a}}$$
as required.