Proving the Triangle Inequality for Absolute Values

[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.

Homework Statement

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

Homework 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}

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

 

[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.

 

[SithsNGiggles]

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.

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.

 

[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.

 

[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:..."?

 

[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.)

 

[SithsNGiggles]

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