Demystifying the Square Root

Many rookie students will solve the equation $x^2 = 4$ by taking square roots, then concluding that $x = \pm 2$ . This is correct. However, they will then make the conclusion that $\sqrt 4 = \pm 2$ , in that there are two possible square roots for $4$ .

While there is a place for this notion when studying complex numbers, as far as real numbers are concerned, the notion of two square roots is rather useless. Therefore, we will only say $\sqrt 4 = 2$ . But what do we mean by useless?

Formally, we want to answer the following question: Is there a square-root function, denoted $\sqrt{\cdot}$ , such that we can un-do squaring? That is, for any $x \in \mathbb R$ and for any $y \in \mathbb R_{\geq 0}$ ,

$\sqrt{x^2} = x\quad \text{and}\quad (\sqrt y)^2 = y$ ?

If such a function exists, we say that $\sqrt{\cdot}$ is the inverse of the squaring function $f(x) = x^2$ . But when does an inverse function exist? In fact, we have discussed such an idea before.

Theorem 1. Let $f : K \to L$ be a function. Then $f^{-1}$ exists as a function if and only if $f$ is bijective.

We have proven this in the aforementioned post.

More generally, the function $f^{-1}$ acts as an un-do button in the following sense.

Theorem 2. Let $f : K \to L$ be a function. Then $f^{-1}$ exists as a function if and only if

$f^{-1} \circ f = \mathrm{id}_K,\quad f \circ f^{-1} = \mathrm{id}_L$

as relations. Here, we recall that $\mathrm{id}_K$ is the identity relation defined by

$\mathrm{id}_K := \{(x,x) \subseteq K \times K : x \in K\}.$

In fact, $\mathrm{id}_K$ is a function.

Proof. Let’s first prove $(\Rightarrow)$ . By a previous result, we have shown that $f^{-1} \circ f = \mathrm{id}_K$ if and only if $f$ is injective. Since $f$ is bijective, $f^{-1}$ is a bijection. Thus, both $f,f^{-1}$ are injective, yielding the desired results.

To prove $(\Leftarrow)$ , we need to prove that $f$ is bijective. Injectivity follows from the observation that $f^{-1} \circ f \subseteq \mathrm{id}_K$ . Surjectivity follows from the observation that $\mathrm{id}_L \subseteq f \circ f^{-1}$ .

This result formalises our intuition that the functions $f$ and $f^{-1}$ cancel each other out.

We can finally answer the question that motivated this discussion to begin with.

Theorem 3. Let $f : \mathbb Z \to \mathbb Z$ be defined by $f(n) = n^2$ whenever $n \in K$ . Then $g : \mathbb N \to f(\mathbb N)$ defined by $g(n) = f(n)$ is bijective, but $f$ is not.

Thus, only $g$ has an inverse, denoted by

$g^{-1} : f(\mathbb N) \to \mathbb N,\quad x=f^{-1}(y) \quad \iff \quad y = f(x) = x^2.$

We define $\sqrt{\cdot} := g^{-1}$ . In fact, it is possible to define the function $f : \mathbb R \to \mathbb R_{\geq 0}$ by $f(x) = x^2$ , so that $g : \mathbb R_{\geq 0} \to \mathbb R_{\geq 0}$ is bijective. This result, however, requires more effort to thoroughly establish.

The point is this: for any $x \in \mathbb R_{\geq 0}$ , $\sqrt{x}$ is the one-and-only positive square root of $x$ , since $f$ defined in this manner can be shown to bijective. Therefore, in the context of real numbers, $\sqrt{4} = 2$ unambiguously.