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Another Absolute Value Proof

  1. Feb 13, 2008 #1
    1. The problem statement, all variables and given/known data

    a and b are real numbers.

    Show l a-b l < l a l + l b l



    2. Relevant equations

    Well, I know la+bl < lal + lbl by the triangle inequality.

    3. The attempt at a solution

    If I can prove that la-bl < la+bl, then I'm done, but that most recent inequality almost seems too intuitive to write a formal proof. I can use the definition of absolute value to create cases perhaps, but I always get lost and seem to go nowhere

    Just to write the obvious stuff down.

    la-bl = a-b if a-b> 0 (or a>b)
    la-bl = b-a if a-b< 0 (or b>a)

    la+bl = a+b if a+b>0 (or a>-b) .
    la+bl = -a-b if a+b<0 ( or a<-b)

    I cant seem how to go somewhere with this.
     
  2. jcsd
  3. Feb 13, 2008 #2

    Dick

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    |a-b|=|a+(-b)|. Now you can use the triangle inequality.
     
  4. Feb 13, 2008 #3

    EnumaElish

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    By triangle ineq., distance between a and b < distance between a and 0 + distance between 0 and b.
     
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