What is a Square Root?

We opened our exploration in real analysis with the notion of $\sqrt 2$ . We could technically define it in the following manner:

$\sqrt{2} := \sup\{x \in \mathbb Q : x^2 < 2\}.$

However, it would take a ton of work to show that we do get the property $(\sqrt 2)^2 = 2$ , and isn’t easily generalisable to other kinds of roots. Instead, we will take advantage of the intermediate value theorem, and define roots as inverses. The intuition is that

$y = \sqrt[n]{x} \quad \iff \quad y^n = x.$

Setting $n = 2$ and $x = 2$ gives us a definition for $\sqrt 2$ , where $\sqrt[2]{\cdot} \equiv \sqrt{\cdot}$ for brevity. But first, a warm-up exercise.

Theorem 1. For any $n \in \mathbb N$ , the function $f_n : \mathbb R \to \mathbb R$ defined by $f_n(x) = x^n$ is continuous.

Proof. It is obvious that $f_1$ is continuous. Since products of continuous functions are continuous, $f_n = f_1^n$ is continuous by induction.

How do we know these functions have inverses? When $n$ is odd, life is nice.

Lemma 1. Fix $i \in \mathbb N$ . Consider the sequence $\{f_i(n)\}$ . Then $f_i(n) \to \infty$ as $n \to \infty$ . Furthermore, for odd $i$ , $f_i(-n) \to -\infty$ as $n \to \infty$ . Finally, for even $i$ , $f_i(-n) \to \infty$ as $n \to \infty$ .

Proof. By definition, $f_i(n) = n^i$ . Fix $M > 0$ . Use the Archimedean property to find $N \in \mathbb N$ such that $N > \max\{1, M\}$ . Then

$f_i(N) = N^i > \max\{1, M^i\} \geq M.$

Hence, for $n > N$ , $f_i(n) > f_i(N) > M$ .

Furthermore, if $i$ is odd, then $f_i(-n) = (-n)^i = -n^i = -f_i(n)$ . It suffices to prove a more general result: if $x_n \to \infty$ , then $-x_n \to -\infty$ .

To that end, assume $x_n \to \infty$ . Fix $M > 0$ . Since $x_n \to \infty$ , find $N \in \mathbb N$ such that $n > N$ implies $x_n > M \iff -x_n < -M$ , which proves $-x_n \to -\infty$ as required.

Finally, for even $i$ , $f_i(-n) = (-n)^i = n^i = f_i(n) \to \infty$ .

Theorem 2. Let $n \in \mathbb N$ be odd. Then $f_n : \mathbb R \to \mathbb R$ is bijective and $f_n^{-1} := \sqrt[n]{\cdot}$ exists.

Proof. We first prove injectivity. Fix $x,y \in \mathbb R$ with $x < y$ .

If $x \geq 0$ , then $f_n(x) = x^n < y^n = f_n(y)$ .
If $x < 0$ and $y > 0$ , then $f_n(x) = x^n < 0 < y^n = f_n(y)$ .
If $x < y < 0$ , then $-x > -y > 0$ implies $-x^n = (-x)^n > (-y)^n = -y^n$ , which yields $f_n(x) = x^n < y^n = f_n(y)$ .

Thus $f_n$ is strictly increasing, and thus injective. For surjectivity, fix $y \in \mathbb R$ . Since $f_n(\pm k) \to \pm \infty$ as $k \to \infty$ , find $M \in \mathbb N$ such that

$f_n(-M) < y < f_n(M).$

By the intermediate value theorem, there exists $c \in (-M,M)$ such that $y = f_n(c) = c^n$ .

A similar argument shows that for even $n$ , if we restrict the domain of $f_n$ to $\mathbb R_{\geq 0}$ , we still get inverses.

Theorem 3. Let $n \in \mathbb N$ be even. Then $f_n : \mathbb R_{\geq 0} \to \mathbb R$ is bijective and $f_n^{-1} := \sqrt[n]{\cdot}$ exists.

This will help us define rational exponents in a future post.

—Joel Kindiak, 19 Dec 24, 2138H

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What is a Square Root?

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