# 2 Proofs: Does this work?

1. Feb 14, 2012

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

Prove the following:

(i) $|x|-|y| \le |x-y|$

and

(ii) $|(|x|-|y|)| \le |x-y|\qquad$ (Why does this immediately follow from (i) ?)

2. Relevant equations

$|z| = \sqrt{z^2}$

3. The attempt at a solution

(i) $(|x|-|y|)^2 = |x|^2 - 2|x||y| + |y|^2 = x^2 - 2|x||y| + y^2 \le x^2 - 2xy + y^2= (x-y)^2 \implies \boxed{|x|-|y| \le |x-y|.}$

(ii) For this part, I looked at the question "Why does this immediately follow from (i)" for inspiration and saw that if I could show that $|(|x|-|y|)| \le |x-y|$ then the proof is complete by transitivity.

Is it as simple as:

$|(|x|-|y|)| = \sqrt{(|(|x|-|y|)|)^2} = \sqrt{(|x|-|y|)^2} = |x|-|y|?$

I think that it is, but it is getting late

2. Feb 14, 2012

### HallsofIvy

Staff Emeritus
If |x|> |y|, (ii) is identical to (i). If |y|< |x|, swap the two.

3. Feb 14, 2012

### LCKurtz

No, it isn't. $\sqrt{(|x|-|y|)^2} = |x|-|y|$ only when $|x|\ge |y|$.

4. Feb 14, 2012

### LCKurtz

Remember that $\sqrt{a^2}=|a|$ and if $0\le a\le b$ then $\sqrt a\le \sqrt b$. So how can you change the above conclusion?

5. Feb 14, 2012

### mtayab1994

Isn't this what is called the kaushy swartz?