# Proofs: Absolute Values and Inequalities

1. Feb 8, 2012

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

I am wondering if the general approach to these proofs involving absolute values and inequalities is to do them case-wise? Is that the typical approach (unless pf course you see some 'trick')? For example, I have:

Prove that if

|x-xo| < ε/2 and Prove that if |y-yo| < ε/2,

then

|(x+y)-(xo+yo)| < ε

|(x-y)-(xo-yo)| < ε.

It seems like there are way too many cases, but maybe that's just the way it is. For cases, I see:

x=0, y=0

0 ≤ x, y≤0

oh jeesh .... forget listing them. I am now seeing that there are cases when xo is greater than or less than x (though I have a feeling that xo is supposed to be smaller than x and a positive number, but he (Spivak) did not specify...nor did he specify that ε > 0 though I think that it should be. Seems like these are gearing up for ε-δ proofs.

Anyone have any thoughts on any of these points? This is problem 1.20 from Spivak's Calculus.

2. Feb 8, 2012

### Dick

|(x-x0)+(y-y0)|<=|x-x0|+|y-y0|. It's called the triangle inequality.

Last edited: Feb 8, 2012
3. Feb 8, 2012

Hi Dick I see that it works for one, I need to look at the (-) case a little more to see how it works for that one.

But tell me: How many cases are there if I did not use the triangle inequality? I think I was going overboard in the OP. I need to look at when:

1. x0+y0 < x+y < 0

2. x0+y0 > x+y <0

3. x0+y0 < x+y > 0

4. x0+y0 > x+y > 0

5. x0+y0 > 0 > x+y

6. x0+y0 < 0 < x+y

Am I missing anything? Sorry if it seems like I am beating a dead horse, but I never really got good at these and would like to fix that now.

Thanks again.

4. Feb 8, 2012

### Dick

You are fussing around about nothing. Just try to prove the triangle inequality using cases. Then call it a dead horse. Then you don't need to break every special case of it out into cases.

5. Feb 8, 2012