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Math Help - Absolute Value + Inequality Proofs

  1. #1
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    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....
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  2. #2
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    Quote Originally Posted by mikijo View Post
    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.
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  3. #3
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    Quote Originally Posted by mikijo View Post
    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|<a \Leftrightarrow  -a<x<a \Leftrightarrow  x > a or x < -a
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  4. #4
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    That's wrong.

    When you say -a<x<a\iff x>a or x<-a, there's an obvious contradiction.
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    That's wrong.

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

    Prove |x| > a iff x > a or x < -a
    <br />
|x| > a \Leftrightarrow x > a or -x > a \Leftrightarrow x > a or x < -a. Thus proving both directions.
    Last edited by GaloisTheory1; March 13th 2009 at 08:45 AM.
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