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Inequality with Absolute Value: What Does |x| Mean?

Date: 06/14/99 at 20:20:39
From: Paul A. Schroeter
Subject: Algebraic inequality |x| > a  

In all the texbooks I've seen, the solution of the absolute value 
inequality  |x| > a  is usually written as  x > a AND x < (-a).

Is it improper for me to combine the solutions and write this as   
(-a) > x > a ?

I find that using this notation makes finding the solutions of 
absolute value inequalities easier, as I only have to:

   Remove the absolute value symbols
   Copy the inquality symbol to the far left of the variable quantity
   Place (-a) to the left of this

The usual way this is taught makes it seem more complicated to me.

Thank you,

Date: 06/15/99 at 16:50:47
From: Doctor Peterson
Subject: Re: Algebraic inequality |x| > a  

Hi, Paul.

What you've suggested isn't correct. This is a tricky area, and you 
have to be very careful in solving these problems when they get beyond 
the basics.

Part of the problem here is that you've misstated what |x| > a means. 
It's not

   x > a  AND  x < (-a)


   x > a  OR  x < (-a)

That is, the inequality holds if EITHER x is greater than a, OR x is 
less than -a. Assuming a is positive, these can't both be true at the 
same time. It gets confusing, because this OR statement can be 
rewritten using the word AND: the set

    {x : x > a  OR  x < -a}

is the union of the sets

    {x : x > a} AND {x : x < -a}

so that there are two cases, one where x > a AND another where x < -a. 
But it's still really an OR in disguise: either one thing is true OR 
the other. And that's very important in solving this sort of problem.

If you write

   -a > x > a

you are saying that both inequalities are true at once for the same x; 
that is, it implies an AND. As a result, the transitivity of 
inequality implies that

   -a > a

which is not true unless a < 0. An OR statement has to be written as 
two separate inequalities; an AND can be combined as long as we remain 
aware that it means "AND."

On the other hand, it is valid to write

   -a < x < a

because this means the same as

   -a < x  AND x < a

which makes sense as long as a > 0; x is between -a and a.

Your shortcut to solving inequalities of this sort may work in many 
simple cases because you're not doing anything that depends on the two 
parts being true at the same time. But sooner or later you will run 
across a problem where you will forget that you are really combining 
two distinct cases into one line, and you will get in trouble. It's 
best to maintain the habit of writing the two cases separately, 
because |x| > a is in fact more complicated than |x| < a, and 
disguising it doesn't change that fact.

Let me add one more thing. Your translation of |x| > a into two 
inequalities is actually a little wrong, because you also have to take 
into account whether x is positive or negative. The more careful way 
to state it is this:

   x >= 0 AND x > a


   x < 0 AND x < -a

If a is positive, these simplify into the two cases we've talked about 
before. But in many problems you can't skip over this. Take this 
problem, for example:

   |x - 2| > 2x

We have to break this into two cases, depending on whether x - 2 is 
positive or negative:

   x >= 2 AND x - 2 > 2x


   x < 2 AND -(x - 2) > 2x

The first case becomes

   x >= 2 AND x < -2  which is never true (think about it).

and the second case becomes

   x < 2 AND 3x < 2  which is true when x is less than both 2 and 2/3.

Combining the cases, we find that the inequality is true for x < 2/3. 
If I had not watched the sign of x - 2, I would have thought that the 
x < -2 case also worked. If I had tried your method, it might have 
gone like this:

    |x - 2| > 2x

    -2x > x - 2 > 2x

    -3x > -2 > x

    x < 2/3 and x < -2

so you'd probably say the answer is x < 2/3 and be right; but there's 
a lot of detail in there you would have missed, which in a somewhat 
bigger problem would have been essential.

I hope this illustrates why the complexity of the solution is 
warranted. I tried to find a problem where you couldn't just stumble 
onto the right answer, and I couldn't do it without getting much more 
complicated. You might want to go to our search page and look for 
"absolute value inequality." You'll find a couple more examples.

- Doctor Peterson, The Math Forum   
Associated Topics:
High School Basic Algebra

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