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 $u^2 = r$.
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
- 1937: Eric Temple Bell: Men of Mathematics ... (previous) ... (next): Chapter $\text{IX}$: Analysis Incarnate
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): square root