Spivak Calculus chapter 1: absolute values?

[pyrosilver]

Homework Statement

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http://img3.imageshack.us/i/0902091724.jpg/

Homework Equations

The Attempt at a Solution

My problem is that I don't even know where to start on this! My first problem is always forgetting what I can and can't use, because we can only use pretty basic stuff, only the stuff we have already proven. (For anyone not familiar with the Spivak book.) I'm so confused on where to start, so if anyone could help me plot out some sort of plan for a couple of these, I would be so grateful! I'm really struggling with this course because I don't know where to begin. After we do these problems in class, I understand them, but trying to do the homework before is my problem. I don't mean to look like I haven't attempted this and am looking for an easy way out -- I've been thinking about this for quite awhile now and I'm still stuck.

 

[Mark44]

Your scanned image is so fuzzy that it's nearly illegible.

Which problem do you need to do? Pick one (not a whole section) and show us what you've tried. You need to at least have made an attempt.

 

[pyrosilver]

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.

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.

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.

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??

The next question is:

|(|x| - 1)| and again I don't know if I am approaching this the right way. If x is positive, it'd be |x - 1|, I think? But how would I write that? What if x is more than 1 but less than 0? then x - 1 would be a negative number, and from an absolute value, how would that work?

 

[Elucidus]

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

[tex]\left| x \right| = \left\{ \begin{array}{rl} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{array}[/tex]

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.