# Proof: |A - B | <

1. Sep 20, 2004

### bullpup

There are many questions i can't get. This is just one. Can anyone give me a hint on what to do? It's probably really simple :(

If A and B are real numbers such that $$|A - B| < \frac{1}{2}$$, show that:

$$|A^2 - B^2| \leq \frac{1}{2} \{2|B|+ \frac{1}{2} \}$$

Last edited: Sep 20, 2004
2. Sep 20, 2004

### arildno

Welcome to PF!
Try to rewrite $$|A^{2}-B^{2}|$$ as a product of two expressions..
(It's a well known identity)

3. Sep 20, 2004

### bullpup

Is this what you meant?

$$|(A+B) (A-B)| \leq |B| + \frac{1}{4}$$

If so, i already had that but didn't know where to go from there...already wasted a lot of paper going no where :grumpy:

Am i supposed to use $$-\frac{1}{2} < A - B < \frac{1}{2}$$ somewhere?

thanks

Last edited: Sep 20, 2004
4. Sep 20, 2004

### arildno

First of all:
You are to PROVE that inequality, not ASSUME it!!
$$|A^{2}-B^{2}|=|A-B||A+B|$$
(Ok so far?)
Now, use the inequality you've been given to derive:
$$|A^{2}-B^{2}|=|A-B||A+B|\leq\frac{1}{2}|A+B|$$
Agreed?
Now you'll need to use your given inequality to provide an estimate of |A+B|!

5. Sep 20, 2004

### arildno

PS:
You'll need to use the triangle inequality as well..