# Absolute Value

The answers to the True-False quiz at the end of the last post are all false. This brings us to absolute value, another topic I want to concentrate on for my upcoming Algebra 1 class. Absolute value becomes a concern in calculus too which I will discuss as the last example below.

There a several “definitions” of absolute value that I’ve seen over the years which I mostly do not like

• The number without a sign – awful: all numbers except zero have a sign
• The distance from zero on the number line – true, but not too useful especially with variables
• The larger of a number and its opposite – true, but not to useful with variables

So I propose to give them an algorithm: If the number is positive, then the absolute value is the same number; if the number is negative, then its absolute value is its opposite. Of course, this is really the definition.

So I’ll soon express this in symbols

$\left| a \right|=\left\{ \begin{matrix} a & \text{if }a\ge 0 \\ -a & \text{if }a<0 \\ \end{matrix} \right.$

Now interestingly this is probably the first piecewise defined function an Algebra 1 student may see, or at least the first one that’s not artificial.  So this is a good place to start talking about piecewise defined function and the importance of talking about the domain. And of course, we’ll have to take a look at the graph.

Sometimes we will have to start solving equations and inequalities with absolute values. So here is the next thing I understand but do not like and will try to avoid. Solve the equation: $\left| x \right|=3$ , Answer including work: $x=\pm 3$. But I think a longer way around is also better:

If $x<0$ then $\left| x \right|=-x=3$ so $x=-3$ or if $x>0$, then $\left| x \right|=x=3$ Solution: $x=3$ or $x=-3$.

Longer? Sure. I hope that by making the students write that a few times that when they get to solving  $\left| x \right|>3$ that it will be natural to say

If  $x<0$ then $\left| x \right|=-x>3$  so $x<-3$ or if $x>0$, then $\left| x \right|=x>3$ Solution:  $x<-3$ or $x>3$

The last case may take a little more discussion. Solve $\left| x \right|<3$. Starting the same way

If $x<0$ then $\left| x \right|=-x<3$ so $x>-3$ which really means $-3

if $x>0$, then $\left| x \right|=x<3$ which really means $0\le x<3$ . Then the union of these two sets looks like an intersection. The solution is $-3

Quite often the equation and the two types of inequalities are treated as separate problems: with = you go with $\pm$ on the other side, with > you have a union pointing away from the origin and with < you have somehow an intersection.  Who needs to remember all that when this idea works all the time?

Example:  Solve $\left| 4x-10 \right|<8$

If $4x-10<0$, then $x<\tfrac{5}{2}$ and $\left| 4x-10 \right|=-\left( 4x-10 \right)=-4x+10<8$, so $-4x<-2$ and $x>\tfrac{1}{2}$ or more precisely  $\tfrac{1}{2} If$latex 4x-10\ge 0\$, then $x\ge \tfrac{5}{2}$  and $\left| 4x-10 \right|=4x-10<8$, so $4x<18$ and $x<\tfrac{9}{2}$ or more precisely $\tfrac{5}{2}. The union again becomes an intersection and the answer is $\tfrac{1}{2}

Finally an example from calculus. On the 2008 AB exam, question 5 asked student to find the particular solution of a differential equation with the initial condition $f\left( 2 \right)=0$. After separating the variables, integrating, including the “+C” and substituting the initial condition students arrived at this equation which they now need to solve for y:

$\displaystyle \left| y-1 \right|={{e}^{\tfrac{1}{2}-\tfrac{1}{x}}}$

How can you lose the absolute value sign? Simple, near the initial condition where y = 0,  $\left( y-1 \right)<0$ so replace $\left| y-1 \right|$ with $-\left( y-1 \right)$ and then go ahead and solve for y

$\displaystyle -\left( y-1 \right)={{e}^{\tfrac{1}{2}-\tfrac{1}{x}}}$

$\displaystyle y=1-{{e}^{\tfrac{1}{2}-\tfrac{1}{x}}}$

I don’t think I’ll try this one in Algebra 1, but maybe it will come in handy when they get to calculus.

# Absolutely

##### Absolute Value

The majority of students learn about absolute value long before high school. That is, they learn a lot of wrong things about absolute value.

• They learn that “the absolute value of a number is the number without its sign” or some such nonsense. All numbers, except zero have a sign!  This sort of works with numbers, but becomes a problem when variables appear. True or false | x | = x? True or false | –x | = x? Most kids will say they are both true; in fact, as you know, they are both false.
• They also learn that “the absolute value of a number is its distance from zero on the number line.” True and works for numbers, but what about variables?
• They learn that “the absolute value of a number is the larger of the number and its opposite.” True again. How do you use it with variables?
• They learn $\left| x \right|=\sqrt{{{x}^{2}}}$ which is correct, useful for order-of-operation practice, and useful in other ways later, But they still compute $\sqrt{{{\left( -3 \right)}^{2}}}=-3$ and  $\sqrt{{{x}^{2}}}=x$ since the square and square “cancel each other out.”

So here is a good vertical team topic. Get to those teachers in elementary and middle school and be sure they are not doing any of the above. They should start with the correct definition in words:

• The absolute value of a negative number is its opposite.
• The absolute value of a positive number (or zero) number is the same number.

This works all the time and will continue to work all the time. Teaching anything else will eventually require unlearning what they are using, and unlearning is far more difficult than learning.

When they start using variables and reading symbols translated into English, then the definition becomes their first piecewise define function:

• $\text{ If }x\ge 0,\text{ then }\ \left| x \right|=x;$  and if $x<0,\text{ then }\left| x \right|=-x$
• $\left| x \right|=\left\{ \begin{matrix} x & \text{ if }x\ge 0 \\ -x & \text{ if }x<0 \\ \end{matrix} \right.$

When reading this definition be sure to say “the opposite of the number” not “negative x” which in this case is probably a positive number.

Give variations of the two True-False questions above on every quiz and test until everyone gets it right!

When you see absolute value bars and want to be rid of them the first question to ask is, “Is the argument positive or negative? “Any time there is an absolute value situation, this is the way to proceed.

And yes, this does show up on the AP Calculus exams. Consider $\int_{0}^{1}{\left| x-1 \right|dx}$ which appeared as a multiple-choice question a few years ago. Give it a try before reading on.

On the interval of integration, [0,1], $\left( x-1 \right)\le 0$ so $\left| x-1 \right|=-\left( x-1 \right)$

$\displaystyle \int_{0}^{1}{\left| x-1 \right|dx}=\int_{0}^{1}{-\left( x-1 \right)}dx=\left. -\tfrac{1}{2}{{x}^{2}}+x \right|_{0}^{1}=-\tfrac{1}{2}+1-0=\tfrac{1}{2}$

Now try $\displaystyle \int_{0}^{1}{\sqrt{{{x}^{2}}-2x+1}\,dx}$, or did we do this one already?