Existence of Square Roots of Positive Real Number

From ProofWiki
Jump to navigation Jump to search

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 $u^2 = r$.


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:

$0 < \dfrac 1 n < \dfrac {u^2 - r} {2 u}$


We note that:

\(\ds u - \dfrac {u^2 - 2 r} {2 u}\) \(=\) \(\ds \dfrac {u^2 + 2 r} {2 u}\) algebraic manipulation
\(\ds \) \(>\) \(\ds 0\) as $u > 0$ and $r > 0$
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 n\) \(<\) \(\ds u\)
\(\ds \leadsto \ \ \) \(\ds u - \dfrac 1 n\) \(>\) \(\ds 0\)


Then:

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

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 $u^2 = r$.

Hence let:

$y_1=u$

and:

$y_2=-u$

$\blacksquare$


Examples

Example: $\sqrt 2$

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


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Existence_of_Square_Roots_of_Positive_Real_Number&oldid=677438"