pyrosilver said:
Express each of the following without absolute value signs, treating each case separately when necessary.
|a+b| - |b|
My problem is I don't know how to get started, because I don't know if my starting point is right. It says treating each case separately when necessary, and I don't know if it is necessary but I'll try.
For any problem involving absolute values it is useful to find the "break points" where each argument in an absolute value equals zero. In this case it would be when either b = 0 or b = -a. More on this in a bit.
If a and b are both positive, |a+b| is a+b and |b| is b. So it would be a + b - b which would be a.
I'd buy that.
If a and b are both negative, |a+b| is also a+b and |b| is also b. So it would be a+b - b, this would also be a.
Not buying. If a and b are both negative then a + b is also negative and |a + b| would be (by defintion) -(a + b), while |b| would be -b.
If a is positive and b is negative, |a+b| would be |a-b| and |b| would be b. Here's where I start to get a bit confused. I don't know how to show |a-b| without absolute values the correct way. Sorry if that's a really lame question but I'm kinda bad at this. am I even approaching that the right way??
Unfortunately, it seems there is some fuzziness about how absolute values work. Recall
\left| x \right| = \left\{ \begin{array}{rl} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{array}
In this case, the "break point" equations I mentioned earlier define lines in the
ab-plane. They subdivide the plane into four regions:
(1) where b > 0 and b > -a. (Quadrant I and the upper-right triangular octant of Quadrant IV)
(2) where b < 0 and b > -a. (The upper-right triangular octant of Quadrant II)
(3) where b > 0 and b < -a. (The lower-left triangular octant of Quadrant IV)
(4) where b < 0 and b < -a. (Quadrant III and the lower-left triangular octant of Quadrant II)
If you select values of a and b from these regions you can determine whether the absolute value expressions equal their arguments or their opposites and simplify them from there. Just to get you going in the right direction, assume a and b are from region (1). Since b > -a then a + b > 0 and therefore |a + b| = a + b. Similarly b > 0 implies |b| = b. Hence
|a + b| - |b| = (a + b) - b = a.
The expression simplifies to just a in region (1).
See what happens in the other three regions.
The other question can be handled similarly (although I think you only need a number line).
--Elucidus
EDIT: I forgot to mention, one needs to determine the behavior on the dividing lines between the regions themselves.