# Absolute Value Difficulties

1. Feb 12, 2010

### Bruce Tonkin

ABSOLUTE VALUE: FUNCTIONAL OR VISUAL?

One of the more troubling aspects of teaching a concept is the disconnect between understanding and the ability to do problem-solving. Perhaps the most elementary such concept is that of absolute value.
Consider the meaning of the equation |x – 3| = 5. The usual “book” explanation is “all points that are a distance of five from three”. It takes a little mental gymnastics to parse the equation into the meaning, but once the idea has been grasped few students have trouble with it.
The more formal definition (given in nearly all texts) is that of a piecewise function:
If x ≥ 0, |x| = x.
If x < 0, |x| = -x.
It takes but a little effort to verify that the two definitions have the same meaning. But we already have a problem: students will prefer to use the first definition (the “visual” one) to solve their exercises. Further, their textbooks often cater to this desire. And, for a time, all is well.
The texts, often enough, continue with an example like |x – 3| = 5. It is easy enough to say that
x – 3 = 5 or x – 3 = -5. It even seems that the second definition is being followed. But it isn’t (the negative sign is on the wrong side; this is a function definition), and there are consequences for a non-standard definition of a function. This becomes especially clear for inequalities.
After all, if a function is defined as f(x) = 2x – 5, then what is f(x + 1)? It is 2(x+ 1) – 5 = 2x – 3. The value is placed into the x part of the function and the function “engine” performs according to the definition: the value is doubled and then 5 is subtracted.
If we had called the absolute value function by a different name, say S(x), and used the same definition given earlier, how would we then solve S(x – 3) = 5?
We would say there are two possibilities: that x – 3 is not negative, and then S(x – 3) = x – 3. In that case, x – 3 = 5. Then x = 8.
The second possibility is that x – 3 is negative. Then S(x – 3) = -(x – 3) = 5, and –x + 3 = 5. In that case, x = -2. That’s how we would handle matters for a function named S with the same definition as the absolute value function.
Too often, that’s not how things are done. When we get to absolute value inequalities, this creates problems in the book and with our students. Consider |x – 3| < 5
The usual textbook will explain things this way: “This is the collection of points that are a distance of less than 5 from 3”. The equation will be transformed into –5 < x – 3 < 5. This is also clearly correct and consistent with both the formal and visual definitions.
But students have more trouble with a compound inequality such as this, so most texts break the inequality into two parts. One part is obvious: x – 3 < 5. The second part requires that x – 3 > -5. Compared to the previous equation, the right hand side has been multiplied by –1 and the direction of the inequality has changed. So the author of the text must explain this, and the usual explanation is the introduction of a rule, which applies only to inequalities involving absolute values.
The students are often confused by this. They want to put the minus sign on the right hand side, as was done for the equality. They often cannot understand why this should not work.
It would be perhaps better to take the compound inequality and break it into two parts, thus eliminating the need for a special rule. This is seldom done in the texts I have seen.
Let us backtrack to the functional definition to see how this can be handled more precisely. We begin with the case of an absolute value equality.
For the problem |x – 3| = 5, either (x – 3) = 5, or –(x – 3) = 5. The solutions match the usual texts exactly.
Now consider |x – 3| < 5. This must mean (x – 3) < 5 or –(x – 3) < 5. There is no difference between solving the equality and the inequality, no special rule, no complication, and no conflict with the definition of a function. It’s also consistent with the compound inequality that nearly all texts show.
But we can do much more with the more formal functional definition.
Problems such as |x – 3| < |2x – 1| become nightmares without using the functional definition. Many texts avoid such problems entirely, and it’s certainly not clear how the compound inequality method would help solve it.
Let’s analyze it using the function definition. There are four possibilities:
a. (x – 3) < (2x – 1), for which each side must be non-negative.
b. -(x – 3) < (2x – 1), for which x – 3 must be negative and 2x – 1 non-negative.
c. (x – 3) < -(2x – 1), for which x – 3 must be non-negative and 2x – 1 negative.
d. -(x – 3) < -(2x – 1), for which both x – 3 and 2x – 1 must be negative.
For (a) we get –2 < x. By the function definition, we require x – 3 ≥ 0, so x ≥ 3. We also require 2x – 1 ≥ 0, so x ≥ ½. All these must be true, so x ≥ 3 is necessary.
For (b) we get 4/3 < x. x – 3 < 0, so x < 3. 2x – 1 ≥ 0, so x ≥ ½. All together, 4/3 < x < 3.
For (c) we get x < 4/3. This cannot give a valid solution, because x – 3 ≥ 0, so x ≥ 3.
For (d) we get –x + 3 < -2x + 1, so x < -2. Also, x – 3 < 0, so x < 3. 2x – 1 < 0, so x < ½. Since all must hold, x < -2.
Now combine all the solutions. Note that each is combined with the others using an OR operation, since (a) or (b) or (c) or (d) is true. We have x ≥ 3, 4/3 < x < 3, x < -2. This gives the solution set
{(-∞, -2) ∪ (4/3, ∞)}. This solution set satisfies the original absolute value inequality.
The functional definition is easily extended to quite complex absolute value problems. In each problem, the key to the solution lies in following the definition precisely. There is no need to create special rules for each kind of problem.
Since most courses that teach absolute value also teach the definition of a function at about the same time, a solution algorithm for absolute value problems that does not follow the way a function is defined is not only confusing but either unnecessarily complicated or just plain wrong.
A “visual” definition of absolute value can be useful, but it should not be the basis of solving more complex problems.