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Prove Absolute Value Property

  1. Feb 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that for every two real numbers x and y
    ##|x+y| \leq |x| + |y| ##


    2. Relevant equations



    3. The attempt at a solution

    There are three cases. The easiest ones is when they are both positive and negative.
    The third one I have problems with.
    The numbers have different sign. Say x>0 and y<0
    Divide this into two subcases:
    case 3.1
    ## x+y \geq 0##


    ## |x| +|y| = x+(-y) = x-y##
    Now, so far so good, but my book states the following.
    ## |x| +|y| = x+(-y) = x-y > x+y = |x+y|##
    How is it possible that x-y be ever greater than x+y?

    case 3.2
    ## x+y < 0 ##
    This one is easy too.
     
  2. jcsd
  3. Feb 26, 2014 #2

    pasmith

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    If [itex]-y > y[/itex], so that [itex]0 > 2y[/itex], ie. [itex]y < 0[/itex].
     
  4. Feb 26, 2014 #3
    ## -y > y ## would be false even if ## y<0##. They would be equal not greater than each other.

    I don't see the connection to the proof. Thanks for trying.
     
  5. Feb 26, 2014 #4

    pasmith

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    Let [itex]y = -1[/itex]. Do you agree that [itex]-y = -(-1) = 1 > -1 = y[/itex]?
     
  6. Feb 26, 2014 #5

    Ray Vickson

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    "How is it possible that x-y be ever greater than x+y?" Try x = 1 and y = -1. What is x-y? What is x+y?
     
  7. Feb 27, 2014 #6
    Thank you, I was confused.
     
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