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I am trying to solve the exercise problems from Spivak Calculus. I am new to proof based approach. Please, go through my reasoning for this problem, and point out whether it is valid or not, or if it is wrong altogether. It's not even close to clean, so if you can provide an elegant solution it would be great.

Problem Statement: Express the following without absolute value signs, treating various cases separately when necessary.

$|a+b|-|b|$

My Solution:
For case 1: $ a,b>0 $

$a+b-b = a$

For case 4: $ a,b<0 $

Then if we consider $c,d>0$ and $a=-c$ and $b=-d$
Then, $|-c-d|-|-d| $
$=|-(c+d)|-d$
$=c+d-d=c$
$c$ is the Positive value of $a$, so case 1 and case 4 yields same result.

For case 2: $ a>0, b<0 $

$Let, b = -c,$ where $c>0$
$|a-c|- |-c| = |a-c|- c $
Let the positive difference $|a-c|=k$
Therefore, $k-c$

For case 3: $ a<0, b>0 $

$Let, a = -d,$ where $c>0$
$|-d+b|- |b|=|b-d|-b$
Let the positive difference $|b-d|=k$ , which is same in value as in case 2.
Therefore, $k-b$. But $b=c$.
Therefore cases 2 and 3 yield same result => $k-b$.

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    I would think it would be desirable to give final answers in terms of $a$ and $b$. – 2017-01-23
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    I am not quite sure how to express $k$. I know |a-b| or |b-a| will yield the same result. Is the solution even correct? Did I get the right solution? – 2017-01-23

2 Answers 2

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Problem Statement: Express the following without absolute value signs, treating various cases separately when necessary.

|a+b|−|b|

Your approach is valid, but I would rather express the final result in terms of the parameters $a,b$. You should start defining the following 4 cases, based on the sign of the expresions within the absolute values:

case 1

$b>0$ and $a+b>0$ ($a + b>0$ is equivalent to $a>-b$)

Then, we get $a+b−b=a$

case 2

$b>0$ and $a<-b$

$|a+b|−|b|=-(a+b)-b=-a-2b$

case 3

$b<0$ and $a>-b$

$|a+b|−|b|=a+b-(-b)=a+2b$

case 4

$b<0$ and $a<-b$

$|a+b|−|b|=-(a+b)-(-b)=-a$

As you noticed, you can group the cases and end just with 2 different formulas. I leave that to you.

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    This is a helpful answer. Thanks. I cannot upvote, otherwise would have. – 2017-01-23
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As noted in a comment, it seems this exercise intends the final answer to use only the variables $a$ and $b$.

For $a,b>0$, you have already done this, and for $a,b>0$, it is easy enough to substitute $-a$ for $c$. For the other two cases, however, you will get hung up on what to do with $k$. For example, if $a>0$ and $b<0$, you don't know whether $a+b$ is positive or negative, so you cannot make a simple substitution for $k$ without using absolute value.

The solution to this problem is to choose better cases. In general, the cases of an absolute value expression are determined not by what variables are used, but by what is inside each of the pairs of absolute value brackets. Specifically, one thing happens when a subexpression inside an absolute value is positive, and a different thing happens when the subexpression is negative.

So separate cases for $b>0$ and $b<0$ still make sense, because $b$ is inside an absolute value by itself, but it makes no sense to do separate cases for $a>0$ and $a<0$, no matter how you combine them with the $b$ cases.