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Section 27.5 Going Deeper: The Absolute Value Function

Consider the following question: Is -1,000,000 a bigger number than 1? The answer you give depends on how you interpret the notion of size. For example, we can say that \(1 \gt -1,000,000\) by just thinking about the number line, and if we think that "bigger" is the same as "greater than" we can conclude that -1,000,000 is not bigger than 1. On the other hand, if you think about "bigger" as meaning the "size" of a number, then we might conclude that -1,000,000 is a pretty big number relative to 1. The notion of the "size" of a number is not captured by the inequalities we used earlier, and so we need to introduce a new mathematical object.

The absolute value of a number has several different representations, and the choice of representation depends on the details of the situation. We will start with the geometric interpretation.

Definition 27.5.1. Absolute Value.

The absolute value of a number \(x\) is denoted by \(|x|\) and is the distance between 0 and \(x\) on the number line.

Because of how the number line is constructed, it's easy to see that when \(x \gt 0\text{,}\) we have \(|x| = x\text{.}\) Here is an example of that:

When we have a negative number, takes the value of when we ignore the negative sign.

This framework for the absolute value should remind you of some of the ideas we used when we were working with subtraction. In fact, we can use this to define the distance between two points. The distance between the numbers \(a\) and \(b\) is \(|b - a|\text{.}\) Notice that this has the same value as \(|a - b|\text{.}\) You will see both definitions used, but the first definition draws a better connection between distance and displacement. In fact, sometimes distance is simply defined to be the absolute value of the displacement, and you're just expected to know how to compute the displacement.

The algebraic definition is slightly more complicated because it requires a special type of notation:

Definition 27.5.2. Absolute Value.

The absolute value of a number \(x\) is denoted by \(|x|\) and is given by the following formula:

\begin{equation*} |x| = \begin{cases} x \amp \text{if $x \geq 0$} \\ -x \amp \text{if $x \lt 0$} \end{cases} \end{equation*}

We first need to describe this notation. This is the notation that is used for a piecewise defined function, which is a function that uses different formulas depending on the value that it has been given. You can see that this definition is broken into two parts. The first formula is applied when \(x \geq 0\) and the second formula is applied when \(x \lt 0\text{.}\) This means that you have to decide which category your number is in before you apply the formulas. If you take a precalculus course, you will probably run into this idea there and get a lot more experience. For now, it is enough to understand that this is a special way of building a function using multiple different pieces.

From this definition, we can see that this is the same as we had with the geometric definition. When the value of \(x\) is positive, then we have \(|x| = x\text{.}\) This is also true when \(x = 0\text{.}\) When \(x\) is negative, the geometric definition led us to ignoring the negative sign. Unfortunately, that concept doesn't have a good algebraic equivalent. However, it's computationally true that the negative of a negative number is a positive number, and so that's how we communicate it.

Working with equations involving absolute value can feel very different depending on whether you're thinking about it from the geometric perspective or the algebraic perspective. For example, let's consider the equation \(|x| = 4\text{.}\)

Geometrically, this is asking us to find the numbers that are a distance 4 from the number 0. And we can immediately determine this by looking at the number line and moving 4 spaces in either direction starting at the origin.

This tells us that the solutions are \(x = -4\) and \(x = 4\text{.}\) We can express this as \(x = \pm 4\text{,}\) where the symbol \(\pm\) is a shorthand for two different possibilities.

In order to solve this from the algebraic perspective, we need to think about the two different situations. When \(x \geq 0\text{,}\) we have \(|x| = x\text{,}\) so that the equation becomes \(x = 4\text{.}\) On the other hand, if \(x \lt 0\text{,}\) then \(|x| = -x\text{,}\) so that our equation becomes \(-x = 4\text{,}\) which we can solve to get \(x = -4\text{.}\) We still arrived at \(x = \pm 4\) as the solution, but it required us to work with two different cases. Here is one way of presenting that work:

\begin{equation*} \begin{array}{c} |x| = 4 \\ \\ \begin{array}{c} \text{If $x \geq 0$:} \\ \begin{aligned} x \amp = 4 \\ \\ \end{aligned} \end{array} \qquad \begin{array}{c} \text{If $x \lt 0$:} \\ \begin{aligned} -x \amp = 4 \\ x \amp = -4 \end{aligned} \end{array} \end{array} \end{equation*}

The same ideas can be employed for more complicated equations. When you have mathematical expressions inside of the absolute value sign, you need to treat that as a single object.

\begin{equation*} \begin{array}{c} |x + 7| = 3 \\ \\ \begin{array}{c} \text{If $x + 7 \geq 0$:} \\ \begin{aligned} x + 7 \amp = 3 \\ x \amp = -4 \\ \\ \end{aligned} \end{array} \qquad \begin{array}{c} \text{If $x + 7 \lt 0$:} \\ \begin{aligned} -(x + 7) \amp = 3 \\ x + 7 \amp = -3 \\ x \amp = -10 \end{aligned} \end{array} \end{array} \end{equation*}

Most textbooks do this calculation in a much more condensed presentation:

\begin{equation*} \begin{aligned} | x + 7 | \amp = 3 \\ x + 7 \amp = \pm 3 \\ x \amp = -7 \pm 3 \\ x \amp = -4 \text{ or } -10 \end{aligned} \end{equation*}

This is ultimately the same process, but it starts to mask some of the ideas and turns it more into an exercise of executing rote calculations rather than providing insight into how the absolute value function works. Since our focus is on the ideas, we've left it in the slightly longer form. Ultimately, you can do this however you choose.

There's a catch to the algebraic approach, which is that it is possible to go through the algebra and get an answer, but find that the answers don't satisfy the original equation. For example, let's look at \(|x| = -4\text{.}\) If we simply pushed ahead with the algebra, it would look like this:

\begin{equation*} \begin{array}{c} |x| = -4 \\ \\ \begin{array}{c} \text{If $x \geq 0$:} \\ \begin{aligned} x \amp = -4 \\ \\ \end{aligned} \end{array} \qquad \begin{array}{c} \text{If $x \lt 0$:} \\ \begin{aligned} -x \amp = -4 \\ x \amp = 4 \end{aligned} \end{array} \end{array} \end{equation*}

If we only looked at the last lines, we would say that \(x = \pm 4\text{,}\) just as before. But if we tried to plug those values in, we would find that they aren't actually solutions.

The problem is that the conclusion in each column is in contradiction with the assumption. On the left, we were supposing that is a nonnegative number, but we concluded that \(x = -4\text{.}\) On the right, we have a similar problem. These are sometimes called apparent solutions (or sometimes phantom solutions) because they seem like they should be solutions even though they aren't.

There are a number of methods that you can employ to identify these situations. The first is to simply check that your solutions really work in the original equation. This is the most common approach that math textbooks teach. The reason this is often suggested is that it's straight-forward to perform all the calculations, and you don't really have to think very much about what's going on. You can simply grind out the calculations from beginning to end without really thinking about things. And from a practical perspective, there's nothing wrong with this approach. It will get the job done.

However, It can be helpful to practice being more thoughtful. By thinking ahead, you can sometimes save yourself the extra work by simply asking whether your absolute value equation is equal to a positive or a negative value. For the equation \(|x| = -4\text{,}\) we can immediately conclude that there are no solutions because the absolute value function will never give a negative value.

Ultimately, neither approach is inherently better or worse. For simpler equations, thinking ahead can save some work. But there are more complicated examples (such as when the variable is both inside and outside the absolute value) where it can be more work to think through the possibilities than to simply grind out the algebra and check to see if they work. Regardless of the approach, this is a feature that you will need to keep in mind when solving equations involving the absolute value signs.

(There is another set of complications that arise when working with absolute value inequalities. You end up needing to work with compound intervals and paying close attention to the exact form of your equation. These skills are sometimes taught in courses, but sometimes not. It usually ends up being another set of memorized algebraic rules, and it has limited value to students at this level. So we're just not going to go there.)