# Thread: 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

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

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,

$\displaystyle |x|<a \Leftrightarrow -a<x<a \Leftrightarrow$ $\displaystyle x > a$ or $\displaystyle x < -a$

4. That's wrong.

When you say $\displaystyle -a<x<a\iff x>a$ or $\displaystyle x<-a,$ there's an obvious contradiction.

5. Originally Posted by Krizalid
That's wrong.

When you say $\displaystyle -a<x<a\iff x>a$ or $\displaystyle x<-a,$ there's an obvious contradiction.
Here:

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