Modules for Prerequisite Mathematics

Subsection 4.1.1 What is Absolute Value

The absolute value of a number \(x\text{,}\) written \(|x|\text{,}\) is a relatively simple thing to define: if \(x\) is positive, then absolute value does nothing; if \(x\) is negative, absolute value makes it positive. In functional notation, we could write this as a piecewise function.

Here are some simple examples.

Absolute value is a useful little tool when we don't care about whether a number is positive or negative, but just want to know how large it is.

Absolute value is natural suited to questions of change. For example, we could wonder which day in a certain place had the greatest temperature change. An increase would be positive and a decrease would be negative, but we only care about the greatest span in temperatures, so we would take the absolute value to get a positive answer.

Absolute value is also very useful for questions of distance. The distance between two number \(a\) and \(b\) (on the number line, say) is given by \(|a-b|\text{.}\) This was, we don't have to know, in advance, whether \(a\) or \(b\) is larger to talk about the distance between them.

Subsection 4.1.2 Solving with Absolute Value

Sometime we have to solve with absolute value in equalities or inequalities. Since absolute value has different behaviours for positive and negative numbers, we usually have to split up the equation or inequality into two pieces. For any absolute value, we have two cases: whether the thing inside the absolute value is negative or positive. In symbols, let \(|\star|\) represent an absolute value, where \(\star\) is some mathematics expression (a number, a variable, a combination of numbers and variables, etc).

• Case 1: when \(\star \geq 0\) we simply drop the absolute value and continue to solve.
• Case 1: when \(\star \lt 0\) we multiply the absolute value term by \((-1)\) and continue to solve.

In each case, when we get a solution, we have to check that the solution fits with the range of each case. Any solutions which are out of the range of the case that produced them are invalid.

Let's demonstrate this by examples.

We have two cases. The first case is when the term inside the asbolute value is positive: \(x + 3 \geq 0\text{.}\) Equiavlently, \(x \geq -3\text{.}\) In this case, we can drop the absolute value, since the thing inside it is always positive.

Then we solve to get \(x=4\text{.}\) When we find a solution like this, we need to check that it part of the case are consiering. The current case is \(x \geq -3\text{,}\) so \(x = 4\) is a fine solution and we can accept it.

Now we procedd with the second case, when the term inside the absolute value is negative. This happens when \(x \lt -3\text{.}\) In this case, when I drop the absolute value, I have to multiply by \((-1)\text{,}\) since multiplying a negative value by a negative makes it positive.

We can solve this by multiplying by \((-1)\text{.}\)

Then we finish by subtracting \(3\) from both sides, giving a solution og \(x = -10\text{.}\) This solution first our case, since \(-10\) is less than \(-3\text{.}\) In total, we conclude that there are two solutions: \(x = 4\) and \(x = -10\text{.}\)

Let's take a very similar example

We solve in cases again. For the first case, we need the expression inside the absolute value to be positive. This happens when \(x \geq 10\text{.}\) In this case, we can drop the absolute value.

Then we can solve and find \(x = 14\text{.}\) Since \(14\) is greater than \(10\text{,}\) this is a valid solution for our first case.

Then we look at the second case, when the term inside the absolute value is negative. This happens when \(x \lt 10\text{.}\) In this case, we multiply by \((-1)\) in place of the absolute value, since multiplying a negative by a negative makes it positive.

Then we can solve by multiplying both sides by \(-1\) and adding 10 to both sides. We get \(x = 6\text{.}\) This value is within the range for this second case, since \(6\) is less than \(10\text{,}\) so this is a valid solution. We conclude we have two solutions: \(x = 6\) and \(x = 14\text{.}\)

This example was very similar to the previous, but it has an important interpretation. The absolute value \(|x-10|\) can be interpreted as the distance between the numbers \(x\) and \(10\text{.}\) So the equation \(|x-10| = 4\) ask which numbers are \(4\) away from \(10\text{;}\) the answer, unsurprisingly, are the two numbers \(6\) and \(14\text{.}\)

To further make this point, here are two more examples.

This is all numbers which are \(12\) away from \(15\text{:}\) these must be \(x=3\) and \(x = 27\text{.}\)

We can rewrite this as a subtraction.

This is all numbers which are \(9\) away from \(-7\text{:}\) these must be \(x=2\) and \(x = -16\text{.}\)

Subsection 4.1.3

We can also have absolute values in inequalities. Consider this example.

As with equalities, we work in cases, depending on whether the expression inside the absolute value is positive or negative. The first cases is \(x \geq 4\text{,}\) where we can just drop the absolute value.

Then we can solve normally. Adding \(4\) to both sides of the equation give \(x \gt 10\text{.}\) Therefore, in the case \(x \geq 4\text{,}\) the inequality describes all number larger than or equal to 10. All those numbers are also larger than 4, so they all fit the case.

In the second case, when \(x \lt 4\text{,}\) the term in the absolute value is negative, so we multiply by \((-1)\) to make it positive.

Now we have to solve. We first multiply both sides by \((-1)\text{,}\) which reverses the inequality.

Then adding \(4\) to both sides of the equation gives \(x \lt -2\text{.}\) The solution consists of all numbers stritcly less than \(-2\text{.}\) Our case is \(x \lt 4\text{,}\) and all number less than \(-2\) are also less than \(4\text{,}\) so all these numbers fit the case.

We conclude that the original inequality describes two sets of numbers: all number strictly larger than \(10\) and all number strictly less than \(-2\text{.}\) The solution consists of both collections of numbers.

The left side of the original inequality can be interpreted as the distance between \(x\) and \(4\text{.}\) Therefore, this is describe all numbers which are at least \(6\) away from \(4\) on the number line. Above \(6\text{,}\) that is all numbers larger than \(10\text{,}\) and below \(6\text{,}\) that is all numbers less than \(-2\text{,}\) as we found. In this interpretation, it makes sense that there should be two ranges, one above and one below.

Inequalities with absolute value can be more complicated, of course. Here is another example which mostly illustrates how messy inequalities and absolute values can get. At several points, we'll have to carefully think through what the various inequalities mean and how to proceed reasonably.

First we'll subtract \(9\) from both sides.

When we square root both sides. This preserves the inequality because both sides are positive. Since the absolute value is positive, we don't need to worry about the possible negative square root here.

Now that we've worked our way down to the absolute value term, we can now work in two cases. In the first case, the term inside the absolute value is positive. This happens when \(x^2 \geq 8\text{.}\) This case covers \(x \geq \sqrt{8}\text{,}\) but also \(x \leq -\sqrt{8}\text{.}\) In this case, we can just drop the absolute value.

Then we can solve. First we add \(8\) to both sides.

Then we take the square root. We get \(x \lt \sqrt{12}\text{.}\) However, as discussed in Subsection 3.1.4 we also need \(x \gt -\sqrt{12}\text{.}\) Finally, we have to compare this to the numbers that this case works for. This case is all number \(x \geq \sqrt{8}\) and \(x \leq \sqrt{8}\text{.}\) Matching this solution to the case gives us two collections of numbers:

In interval notation (as described in Chapter 10):

That's the first case. For the second case, we consider the situation where the term inside the inequality is negative. This happens if \(x^2 \lt 8\text{.}\) In the square root, this is \(x \lt \sqrt{8}\text{,}\) but also \(x \gt -\sqrt{8}\text{,}\) so that \(x\) is also not a large negative number. We can express this case as one range: \(-\sqrt{8} \lt x \lt \sqrt{8}\text{.}\) In this case, the absolute value become multiplication by \(-1\text{.}\)

Then we can solve. First we multiply by \((-1)\text{,}\) which reverses the inequality.

Then we add \(8\) to both sides.

Then we square root to get \(x \gt 2\text{.}\) However, as discussed in Subsection 3.1.4 we also need \(x \lt -2\text{.}\) So we get two ranges of numbers, those above \(2\) and those below \(-2\text{.}\) However, our case was restricted to those numbers between \(-\sqrt{8}\) and \(\sqrt{8}\text{.}\) But that together, this case gives two groups of numbers:

In interval notation (as described in Chapter 10):

Now we have to put it all together. The two cases together describe four collections of numbesr, but those collections of number can be combined. The first case gave us negative number from \(-\sqrt{12}\) to \(-\sqrt{8}\) and then the second case gave us negative number from \(\sqrt{8}\) to \(-2\text{.}\) Together we get the range from \(-\sqrt{12}\) to \(-2\text{.}\) The same is true for the positive ranges; they fit together on either side of \(\sqrt{8}\text{.}\) All in all, the solution to this inequality is the two collection of numbers described by

In interval notation (as described in Chapter 10):

That is the solution to the inequality. You can see that the organizating and bookkeeping, with multiple cases and multiple intervals for each case, get a bit hectic.