# 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 $S = \set {x \in \Q: x^2 < r}$.

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

To show that $S$ is bounded above, we note that $n + 1$ is an upper bound:

$y > n + 1 \implies y^2 > n^2 + 2 n + 1 > n$

and so $y \notin S$.

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

By the Completeness Axiom, $S$ has a supremum, say:

$u = \sup S$

We already have that $u \ge 1$, as $0 \in S$ as seen.

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

Aiming for a contradiction, suppose $u^2 \ne r$.

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:

 $\ds \paren {u - \dfrac 1 n}^2$ $=$ $\ds u^2 - \dfrac {2 u} n + \dfrac 1 {n^2}$ $\ds$ $>$ $\ds u^2 - \dfrac {2 u} n$ $\ds$ $>$ $\ds u^2 - \paren {u^2 - r}$ $\ds$ $=$ $\ds r$

Hence:

 $\ds x$ $\in$ $\ds S$ $\ds \leadsto \ \$ $\ds x^2$ $<$ $\ds r$ $\ds$ $<$ $\ds \paren {u - \dfrac 1 n}^2$ $\ds \leadsto \ \$ $\ds x$ $<$ $\ds u - \dfrac 1 n$

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:

 $\ds \paren {u + \dfrac 1 n}^2$ $=$ $\ds u^2 + \dfrac {2 u} n + \dfrac 1 {n^2}$ $\ds$ $<$ $\ds u^2 + \dfrac {2 u} n + \dfrac {2 u} n$ as $\dfrac 1 n < 2 u$ $\ds$ $\le$ $\ds u^2 + r - u^2$ as $r - u^2 \ge \dfrac {4 u} n$ $\ds$ $=$ $\ds r$

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 $u^2 = r$.

$\blacksquare$

## Examples

### Example: $\sqrt 2$

There exists $u \in \R$ such that $u^2 = 2$.