# Prove ||a|-|b|| ≤ |a-b|

1. Dec 26, 2012

### SithsNGiggles

Hey again! I've got another problem I'd like to check for adequacy. I'm pretty sure I've got all the cases covered, but I want to make sure the work here is satisfactory.

1. The problem statement, all variables and given/known data

Prove that for all real numbers $a$ and $b$,
$||a| - |b|| \leq |a - b|$.

2. Relevant equations

The book I'm working with uses the following definition for absolute value:

$|x| = \begin{cases} x, & \mbox{if } x \geq 0 \\ -x, & \mbox{if } x < 0 \end{cases}$

And for later use:
$|a| = \begin{cases} a, & \mbox{if } a \geq 0 \\ -a, & \mbox{if } a < 0 \end{cases}$
$|b| = \begin{cases} b, & \mbox{if } b \geq 0 \\ -b, & \mbox{if } b < 0 \end{cases}$
$|a-b| = \begin{cases} a-b, & \mbox{if } a-b \geq 0 \\ b-a, & \mbox{if } a-b < 0 \end{cases}$

3. The attempt at a solution
By definition,
$||a|-|b|| = \begin{cases} |a|-|b|, & \mbox{if } |a|-|b| \geq 0, \mbox{ i.e. } |a|\geq|b| \\ -(|a|-|b|) = |b|-|a|, & \mbox{if } |a|-|b|< 0, \mbox{ i.e. } |a|<|b| \end{cases}$

Suppose $|a| = |b|$. (I initially let a = b, but using |a|=|b| is more fitting, right?)
Then, $||a|-|b|| = ||a|-|a|| = |0| = 0$. So, $0 \leq |a-b|$.
If $a=b$, then $a-b=0$ and we have $0 \leq 0$, which is true. If $a \not= b$, then $|a-b|>0$ and we have $0 \leq |a-b|$, which is true.

For this reason, I modify the definition of $||a|-|b||$ to be
$||a|-|b|| = \begin{cases} |a|-|b|, & \mbox{if } |a|>|b| \\ |b|-|a|, & \mbox{if } |a|<|b| \end{cases}$
(I hope there aren't any problems with this step. I thought it would simplify the later sub-cases.)

Case 1: Suppose $|a|>|b|$.

Case 1(a): Suppose $a,b \geq 0$. Then $|a|=a$ and $|b|=b$.
We thus have $||a|-|b|| = |a-b| \leq |a-b|$, which is true.

Case 1(b): Suppose $a,b<0$. Then $|a|=-a$ and $|b|=-b$.
$(|a|>|b|) \; \Rightarrow \; (-a>-b) \; \Rightarrow \; (a-b<0) \; \Rightarrow \; (|a-b|=b-a)$.
We have $||a|-|b|| = |a|-|b| = -a-(-b) = b-a$, and so we have $b-a \leq |a-b| = b-a$, which is true.

Case 1(c): Suppose $a\geq 0, b<0$. Then $|a|=a$ and $|b|=-b$.
$(a\geq 0, b<0) \; \Rightarrow \; (a>b) \; \Rightarrow \; (a-b>0) \; \Rightarrow \; (|a-b|= a-b = a+(-b) = |a|+|b|)$.
We have $||a|-|b|| = |a-(-b)| = |a+b|$, which means we have
$|a+b| \leq |a|+|b|$, which is true (by the Triangle Inequality Theorem).

Case 1(d): Suppose $a<0, b\geq 0$. Then $|a|=-a$ and $|b|=b$.
$(a<0, b\geq 0) \; \Rightarrow \; (b>a) \; \Rightarrow \; (a-b<0) \; \Rightarrow \; (|a-b|= b-a = b+(-a) = |a|+|b|)$.
We have $||a|-|b|| = |-a-b| = |-(a+b)| = |a+b|$, and so we have
$|a+b| \leq |a|+|b|$, as in Case 1(c).

Case 2: Suppose $|a|<|b|$.

Case 2(a): Suppose $a,b \geq 0$. This case is identical to Case 1(a).

Case 2(b): Suppose $a,b<0$. Then $|a|=-a$ and $|b|=-b$.
$(|a|<|b|) \; \Rightarrow \; (-a<-b) \; \Rightarrow \; (a>b) \; \Rightarrow \; (a-b>0) \; \Rightarrow \; (|a-b|=a-b)$.
We also have $||a|-|b|| = |b|-|a| = -b-(-a) = a-b$, and so we have
$a-b \leq |a-b| = a-b$, which is true.

Case 2(c): Suppose $a\geq 0, b<0$. This case is identical to Case 1(c).

Case 2(d): Suppose $a<0, b \geq 0$. This case is identical to Case 1(d).​

Thus, for all real numbers $a$ and $b$,
$||a| - |b|| \leq |a - b|$.

(I say the sub-cases under case 2 are identical to those in case 1 since they don't utilize the modified definition.)

If I'm missing anything important, please do tell! Thanks

2. Dec 26, 2012

### MarneMath

Just a quick read, it looks right, but tedious. There is a much much much easier way to do this. If you're interested, just tell me and I'll help you out.

3. Dec 26, 2012

### SithsNGiggles

I had a feeling there was, but I wanted to be thorough. What can I omit? I would think some of the algebra isn't all that necessary.

4. Dec 26, 2012

### MarneMath

If you can prove lx|-|y| less than or equal to |x - y| then it's pretty simple if you think about it the right way.

5. Dec 26, 2012

### SithsNGiggles

It looks like I partially proved that in cases 1(b) and 2(b).
Would it have anything to do with the fact that they're the only cases that actually consider the assumptions made in "Case 1:..." and "Case 2:..."?

6. Dec 26, 2012

### MarneMath

More along the line that if you make a 'clever' change in variables, then the equality just appears. (By the way, if you're really not interested in learning an alternative proof, this convo isn't important. I just like short simple proofs. I reread your proof and I haven't found a mistake, besides some minor math grammar things, but other than that it looks good. It's the way I would've done it when I was learning how to write proofs.)

7. Dec 26, 2012

### SithsNGiggles

Ah, well, at least I'm on the right track. Thanks for the input, though, much obliged!