At some point in your beginning algebra class, you were taught that taking the square root is the way to ‘undo’ squaring a number. Although seemingly trivial, this is not exactly true. This happens to be one of the more confusing topics is that frequently glossed-over. Most students just memorize a square root’s two main properties, and move on. Those properties are (i) it is used for finding $x$ when given $x^2$ and (ii) it always outputs a non-negative value (given a non-negative input). But there are some subtle details to be mindful of, and an interesting functional composition behind its existence.

The Problem

Let’s start with squaring things. We know that $5^2 = 25$ and $(-5)^2 = 25$. So, if the square root function ‘undoes’ squaring, then $\sqrt{25}$ should equal both 5 and -5, right? But it doesn’t: $\sqrt{25}=5$. It seems we have lost some information (namely the -5). So why is the $-5$ lost?

The answer comes down to understanding inverse mappings, whether a mapping is a function and the square root function’s role. Let’s review some terminology:

• mapping A fixed relationship between a set of inputs and a set of outputs.

• function A special class of mapping where there is at most a single unique output for each possible set of inputs.

Let’s consider

$$f(x) = x^2.$$

The mapping $f(x)$ is a function (and is quadratic). For every $x$ that is input, there is only one output value. For example, $f(5)=25$ and $f(-5)=25$. However, since both 5 and -5 map to the same value (25), the inverse mapping (the reverse direction), which I will just call $f^{-1}(x)$, must be so that $f^{-1}(25) \in \{5,-5\}$, where $\in$ means that $f^{-1}(25)$ is an ‘element of’ the set $\{5,-5\}$, so can equal either 5 or -5. This is not a function, but is a set-valued mapping.

Note: Once a value is squared, there is no way to determine if the original value was the positive or negative version (hence why $f^{-1}(25)$ should have 2 values).

Functional Composition

Set-valued mappings are complex beasts (especially for beginning algebra students). As with analyzing anything complex, it helps to break it down into simpler parts. So, it is helpful to define a mapping that just outputs one of the two values. We define the mapping $\sqrt{x}$ to output the positive counterpart. Doing so makes $\sqrt{x}$ a nice friendly function, and due to symmetry of $f^{-1}(x)$, we can write $f^{-1}(x)$ as

$$f^{-1}(x) \in \{\sqrt{x},-\sqrt{x}\}, \ \ x \geq 0.$$

Notice our set-valued mapping $f^{-1}(x)$ is composed of another mapping, $\sqrt{x}$, which is a function. So, plugging in 25 yields our solution to the above problem: $f^{-1}(25) \in \{\sqrt{25}, -\sqrt{25}\}= \{5, -5\}$ (we found the -5). This is fantastic because now we can describe our set-valued map with our handy square root function!

Summarizing, the reason that square root function outputs only non-negative numbers is because it was designed that way. This is so that it can play a compositional role as a function used to express the inverse mapping of quadratic functions. By itself, it does not ‘undo’ the operation of squaring a number, but is merely a well-defined stepping stone to get there.