Proof of Existence and Uniqueness of the Nonnegative Square Root

Let

Then $0\in S$, so $S$ is nonempty. Also, if $x\in S$, then $x^2\le a$, hence $x\le a+1$; for if $x>a+1$, then $x>1$ and so $x^2>x>a\ge 0$, a contradiction. Thus $S$ is bounded above. By the least upper bound property, the set $S$ has a supremum $r=\sup S$ in $\mathbb{R}$.

We prove that $r^2=a$. First, suppose for contradiction that $r^2<a$. Since $a-r^2>0$, choose a real number $h>0$ such that

Then

Also $r+h\ge 0$, so $r+h\in S$. But $r+h>r$, contradicting that $r$ is an upper bound for $S$.

Next, suppose for contradiction that $r^2>a$. Since $r^2-a>0$, choose a real number $h>0$ such that

If $x\in S$ and $x\ge r-h$, then

Because $x\in S$, one has $x^2\le a$. Also $0\le r-x\le h<1$ and $x\le r$, so

Hence

a contradiction. Therefore every $x\in S$ satisfies $x<r-h$, so $r-h$ is still an upper bound for $S$. This contradicts the definition of $r=\sup S$.

Thus $r^2=a$. By construction $r\ge 0$, so existence is proved.

For uniqueness, let $s\in\mathbb{R}$ also satisfy $0\le s$ and $s^2=a$. Without loss of generality suppose $r\le s$. Then

Since $s+r\ge 0$ and in fact $s+r=0$ would force $r=s=0$, it follows that $s-r=0$. Hence $s=r$. Therefore the nonnegative square root is unique.