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Chapter 27 Negatives on the Number Line

Section 27.1 Through the Looking Glass

When we introduced the number line, we showed that the negative numbers were the numbers to the left of zero. But we have not yet worked with negative numbers in our arithmetic.

Activity 27.1. Addition with Negative Numbers.

It turns out that if we add to a positive number or subtract from a positive number, there are no conceptual changes to how we work with the number line. Here is an example for addition. Notice that the process has not changed.
Try it!
Calculate \(-8 + 4\) using a number line.
Solution.

Activity 27.2. Subtraction with Negative Numbers.

Here are two subtraction calculations. Notice that it does not matter whether we start from a negative value or simply cross over from positive values to negative values. In both situations, the process remains exactly the same.
Try it!
Calculate \(2 - 7\) using a number line.
Solution.
At this point, it is important that the number we are adding or subtracting is positive, and that all we are doing is allowing negatives as the starting position. We will look at the other possibilities in another section. But as long as that restriction is met, we can expand our ideas to allow us to think about addition and subtraction with larger numbers.

Activity 27.3. Number Line Addition with Negative Numbers.

Let’s consider the calculation \(-75 + 27\text{.}\) When we try to draw this on a number line, there are two observations we need to make. The first is that we are starting at a negative value, so we must begin our movement to the left of zero. The second observation is that the movement to the right is smaller than the distance to zero. This means that we will stay to the left of zero when we are done.
From here, it doesn’t matter how you get to your answer. If you need to do the calculation in two steps, then do it in two steps. If you can perform the calculation mentally, that’s also fine. But avoid using a calculator. It is worth the time to practice your mental arithmetic skills, as this feeds directly into your general mathematical confidence. The important thing is to think about the relative locations of the numbers.
This is mostly about the practice of thinking through the diagram to understand the underlying logic, not about following rules. It would be possible to write out a set of 5 or 6 rules for how to do all of these problems, but if you simply think about the picture, many of those rules will work out naturally and intuitively.
Try it!
Calculate \(-53 + 38\) using a number line.
Solution.

Activity 27.4. Number Line Subtraction with Negative Numbers.

Sometimes, the calculation takes us past zero. In this case, we just need to look at the picture and think about what that means. Consider \(22 - 54\text{.}\) If we start at 22 and move 54 spaces to the left, we have definitely gone past zero.
Conceptually, it’s easiest to think of this as simply returning to zero and then taking the remaining number of steps to finish the movement.
If you need to move a total of steps to the left, and you’ve already moved steps, how many steps are left? You should hopefully be able to think that through and figure out that there are steps remaining. And that leads us to the final result.
Try it!
Calculate \(34 - 78\) using a number line.
Solution.

Activity 27.5. Arithmetic Practice with the Number Line.

You do not want to think about these calculations as rules. You just want to think about them as following the natural logic of the problem. Draw the diagram and let the diagram guide your thinking.
Try it!
Calculate \(-15 - 32\) using a number line.
Solution.

Section 27.2 Worksheets

PDF Version of these Worksheets
 1 
external/worksheets/27-Worksheets.pdf

Worksheet Worksheet 1

1.
Calculate \(-3 + 5\) using a number line. Draw all the points and count out the steps.
2.
Calculate \(3 - 8\) using a number line. Draw all the points and count out the steps.
3.
Calculate \(-3 - 4\) using a number line. Draw all the points and count out the steps.
4.
Without performing the calculation, explain why \(387 - 749\) will result in a negative number.
5.
Without performing the calculation, explain why \(-178 - 455\) will result in a negative number.

Worksheet Worksheet 2

1.
Calculate \(19 - 45\) using a number line.
2.
Calculate \(-28 - 46\) using a number line.
3.
Calculate \(-15 + 73\) using a number line.
4.
Practice your mental arithmetic by performing the following calculations without drawing a number line (though you may certainly visualize one).
\begin{equation*} \begin{aligned} 33 - 57 \amp = \\ -38 + 19 \amp = \\ -32 - 29 \amp = \\ -42 + 70 \amp = \\ -43 + 37 \amp = \\ 72 - 34 \amp = \\ 38 - 14 \amp = \\ -52 + 31 \amp = \\ -22 + 49 \amp = \\ -18 - 35 \amp = \\ 28 - 63 \amp = \\ -24 + 15 \amp = \end{aligned} \end{equation*}

Worksheet Worksheet 3

1.
Calculate \(159 - 217\) using a number line.
2.
Calculate \(-228 - 146\) using a number line.
3.
Calculate \(-215 + 273\) using a number line.
4.
Practice your mental arithmetic by performing the following calculations without drawing a number line (though you may certainly visualize one).
\begin{equation*} \begin{aligned} -211 - 157 \amp = \\ -118 + 205 \amp = \\ -312 + 129 \amp = \\ -138 - 124 \amp = \\ -152 + 361 \amp = \\ -212 + 188 \amp = \\ -242 + 170 \amp = \\ -143 + 137 \amp = \\ 238 - 314 \amp = \end{aligned} \end{equation*}

Worksheet Worksheet 4

1.
Calculate \(-53 + 27\) using a number line.
2.
Earlier in the section, there was a warning about adding and subtracting in columns when working with negative numbers. We are going to explore the challenges that arise in this setting in order to more fully understand the challenges that arise from working in columns.
Below a possible first step (ones column) of the calculation \(-53 + 27\) when performed using columns:
Identify the error that has already taken place in this calculation.
3.
There is a "rule" that can be used for doing this calculation in columns. To calculate \(-a + b\) (where and are positive numbers).
  • Step 1: Identify the larger number.
  • Step 2: Perform the calculation "larger minus smaller."
  • Step 3: Give your result the same sign as the larger number in the original problem.
Apply this "rule" to the calculation \(-53 + 27\text{.}\) Explain how this "rule" is a more complicated expression of the number line calculation.

Worksheet Worksheet 5

1.
In the previous worksheet, we saw that calculations in columns can be problematic and lead to errors. The "rule" that was provided is the common way that this is taught. But this is not the only way to think about doing this calculation in columns. We’re going to explore this in a different way.
Rather than working with digits, we will work with values. We will look at the calculation \(-53 + 27\) again, but rewriting the calculations using expanded form. From here, it is much easier to perform the calculation while avoiding errors.
Using the expanded form version of writing the calculation, calculation \(42 - 76\text{.}\)
2.
What this is showing is that the calculation can be performed if we focus on individual place values. Upon a deeper investigation, this would also reveal that the real issue comes down to the steps of "carrying the one" or "borrowing." The digit manipulations that one might normally do are incompatible with the algorithms for addition or subtraction in columns. Here are two of the most reasonable attempts at performing this calculation using the traditional algorithms.
As best as you can, try to explain the conceptual errors of each attempt.

Section 27.3 Deliberate Practice: Practice with Negatives

Algebra is a skill, which means it requires practice to become proficient. But it will take more than rote repetition to get there. Deliberate practice is the thoughtful repetition of a task. For each of these sections, you will be given a list of specific skills or ideas to focus on as you practice thinking through the problems.
Focus on these skills:
  • Do the problem mentally before drawing any part of the number line diagram.
  • You are free to draw as few or as many steps in your number line calculation as you find necessary.
  • Present your work legibly.

Worksheet Worksheet

Instructions: Perform the given calculation mentally, then draw out a number line diagram to perform the calculation.
1.
Calculate \(-34 + 18\) mentally and on number line.
2.
Calculate \(23 - 49\) mentally and on number line.
3.
Calculate \(55 - 27\) mentally and on number line.
4.
Calculate \(-48 + 32\) mentally and on number line.
5.
Calculate \(-32 + 58\) mentally and on number line.
6.
Calculate \(35 - 71\) mentally and on number line.
7.
Calculate \(45 - 18\) mentally and on number line.
8.
Calculate \(-328 + 455\) mentally and on number line.
9.
Calculate \(243 - 345\) mentally and on number line.
10.
Calculate \(-314 + 146\) mentally and on number line.

Section 27.4 Closing Ideas

In this section, we took a concept that is traditionally very confusing for students and presented it as a single image that most people find simple and intuitive. We are once again leveraging both your intuition and experience with numbers to create a framework that is easy for you to understand and use. But this really isn’t about the calculations. It’s about the idea of the calculation. We are in the process of exploring just how far we an go with addition and subtraction using the number line. What we have shown in this section is that the number line is a more powerful idea than the addition and subtraction algorithms.
And that is the general situation in mathematics. Very often, a good idea will take you further than just applying formulas or prescribed algorithms. Developing a depth of mathematical thinking can help you solve problems that you’ve never seen before by relating it back to things that you have. These calculations with negative numbers are extremely accessible and understandable without needing to do a lot of extra work because we were able to relate it back to the number line, which we have had plenty of experience with over the last few sections. This is far better than simply having you memorize a complex set of rules for adding and subtracting in columns.
Whenever possible, you should try to focus on ideas rather than rules. You will be able to go so much further if you do.

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 27.1. 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 27.2. 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.)