Express with at least one less pair of absolute value signs

[wifi]

We have $||a+b|-|a|-|b||$. The only way I can think to eliminate one pair of absolute value signs is to consider the cases separately and determine which pair(s) can be removed without affecting the final answer. I'm trying to get better at doing these types of problems, so feel free to scrutinize this analysis.

Case 1: $a>b>0$,

we have $|a+b-a-b| = 0$.

Case 2a: $a>0>b$ with $b>-a \Rightarrow a+b>0$,

we have $|a+b-a+b| = |2b| = |2| \cdot |b| = -2b$.

Case 2b: $a>0>b$ with $b<-a \Rightarrow a+b<0$,

we have $|-a-b-a+b| = |-2a| = |-2| \cdot |a| = 2a$.

Case 3: $0>a>b$,

we have $|-a-b+a+b| = 0$.

After some inspection, one can see that the outermost pair of absolute value signs can be removed without affecting the final answer as long as a negative sign is added in its place. This is possible since the outermost absolute value bar serves only to change the sign in cases 2a and 2b.

So what we have is $-(|a+b|-|a|-|b|) = |a|+|b|-|a+b|$.

As a check, let's make sure this yields the same answers as the original expression.

Case 1: $a>b>0$,

we have $a+b-a-b = 0$.

Case 2a: $a>0>b$ with $b>-a \Rightarrow a+b>0$,

we have $a-b-a-b= -2b$.

Case 2b: $a>0>b$ with $b<-a \Rightarrow a+b<0$,

we have $a-b+a+b = 2a$.

Case 3: $0>a>b$,

we have $-a-b+a+b = 0$.

Same as above!

EDIT: Now that I'm thinking about it more, there are several cases I hadn't included:

If $a=b$ we have $||a+a|-|a|-|a|| = ||2a|-|2a||= 0$.

If $a=0, b>0$ we have $|b-b|=0$.

If $a=0, b<0$ we have $|-b+b|=0$.

If $a>0, b=0$ we have $|a-a|=0$.

If $a<0, b=0$ we have $|-a+a|=0$.

Last but not least, for the trivial case $a=0, b=0$ we just have 0.

So we can see that in every case the outer bars may be removed without affecting the final answer. Thus, the answer is the same as described above. Sorry! Hopefully it's correct now.

 

[Mandelbroth]

We have $||a+b|-|a|-|b||$. The only way I can think to eliminate one pair of absolute value signs is to consider the cases separately and determine which pair(s) can be removed without affecting the final answer. I'm trying to get better at doing these types of problems, so feel free to scrutinize this analysis.

Can't that be written as $||a+b|-(|a|+|b|)|$? Are you familiar with the triangle inequality?

The triangle inequality applies to the real number line as a vector space by using the absolute value function as the norm. The triangle inequality states that $|a+b|\leq|a|+|b|$. What does that imply here?

 

[wifi]

Can't that be written as $||a+b|-(|a|+|b|)|$? Are you familiar with the triangle inequality?

The triangle inequality applies to the real number line as a vector space by using the absolute value function as the norm. The triangle inequality states that $|a+b|\leq|a|+|b|$. What does that imply here?

Yes, I am familiar with that inequality.

This implies that $|a+b|-(|a|+|b|)\leq 0$. Thus, we can include the negative sign and remove the outermost absolute values bars as I did. Now I feel like what I did was a bit excessive.

I feel as though I discover only the least clever method for solving problems.

 

[Mandelbroth]

Yes, I am familiar with that inequality.

This implies that $|a+b|-(|a|+|b|)\leq 0$. Thus, we can include the negative sign and remove the outermost absolute values bars as I did. Now I feel like what I did was a bit excessive.

Proof by exhaustion is perfectly acceptable. Though, my suggestion for you would be to wait before you decide to separate it into every single case and solve. That will only work when there is a finite number of cases. I like this depiction:

 

[wifi]

Proof by exhaustion is perfectly acceptable. Though, my suggestion for you would be to wait before you decide to separate it into every single case and solve. That will only work when there is a finite number of cases. I like this depiction:

Ha! I definitely learned my lesson.