# 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.

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