Existence of Square Roots of Positive Real Number

Theorem
Let $r \in \R_{\ge 0}$ be a positive real number.

Then:
 * $\exists y_1 \in \R_{\ge 0}: {y_1}^2 = r$
 * $\exists y_2 \in \R_{\le 0}: {y_2}^2 = r$

Proof
Let $S = \set {x \in \R: x^2 < r}$.

As $0 \in S$, it follows that $S$ is non-empty.

To show that $S$ is bounded above, we note that $r + 1$ is an upper bound:
 * $y > r + 1 \implies y^2 > r^2 + 2 r + 1 > r$

and so $y \notin S$.

Thus $x \in S \implies x < r + 1$.

By the Completeness Axiom, $S$ has a supremum, say:
 * $u = \sup S$

We already have that $u \ge 0$, as $0 \in S$ as seen.

It remains to demonstrate that $u^2 = r$.

$u^2 \ne r$.

Then either $u^2 > r$ or $u^2 < r$.

Suppose that $u^2 > r$.

Then:
 * $\dfrac {u^2 - r} {2 u} > 0$

So there exists $n \in \N$ such that:
 * $0 < \dfrac 1 n < \dfrac {u^2 - r} {2 u}$

Then:

Hence:

which contradicts the leastness of $u$.

Suppose instead that $u^2 < r$.

Then $\exists n \in \N$ such that:
 * $0 < \dfrac 1 n \le \dfrac {r - u^2} {4 u}$

and:
 * $\dfrac 1 n < 2 u$

Then:

Hence:
 * $u + \dfrac 1 n \in S$

which contradicts the fact that $u$ is an upper bound of $S$.

Hence by Proof by Contradiction it follows that $u^2 = r$.

Hence let:
 * $y_1=u$

and:
 * $y_2=-u$