# Math Help - Absolute Value + Inequality Proofs

1. ## Absolute Value + Inequality Proofs

Got a couple proofs I'm stuck on.....

1. Prove |x| > a iff x > a or x < -a

I know it has a couple different cases...but from there...I'm lost! yikes!

2. Prove |x| - |y| < |x-y|

Again, I know it has to do with the triangle inequality...but everything I try doesn't seem to get me anywhere....

2. Originally Posted by mikijo
Got a couple proofs I'm stuck on.....

1. Prove |x| > a iff x > a or x < -a

I know it has a couple different cases...but from there...I'm lost! yikes!

2. Prove |x| - |y| < |x-y|

Again, I know it has to do with the triangle inequality...but everything I try doesn't seem to get me anywhere....
Triangle inequality

$| \,a + b\, | \le |\,a\,| + |\,b\,|$ so $| \,a + b\, | - |\,b\,| \le |\,a\,|$ then let $a = x - y,\; b = y$.

3. Originally Posted by mikijo
Got a couple proofs I'm stuck on.....

1. Prove |x| > a iff x > a or x < -a

I know it has a couple different cases...but from there...I'm lost! yikes!

2. Prove |x| - |y| < |x-y|

Again, I know it has to do with the triangle inequality...but everything I try doesn't seem to get me anywhere....
and for # 1,

$|x| $x > a$ or $x < -a$

4. That's wrong.

When you say $-aa$ or $x<-a,$ there's an obvious contradiction.

5. Originally Posted by Krizalid
That's wrong.

When you say $-aa$ or $x<-a,$ there's an obvious contradiction.
Here:

Prove |x| > a iff x > a or x < -a
$
|x| > a \Leftrightarrow$
$x > a$ or $-x > a \Leftrightarrow x > a$ or $x < -a$. Thus proving both directions.