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Absolute value

  1. Jan 27, 2010 #1
    Hey folks, can someone quickly check my algebra?

    Given:

    [tex] d(x,A) \leq d(x,y) + d(y,A) [/tex]

    To show:

    [tex] \left|d(x,A) - d(y,A) \right| \leq d(x,y) [/tex]

    Proof:

    from given, [tex] d(x,A) - d(yA) \leq d(x,y); [/tex]

    and

    [tex] -d(x,A) + d(y,A) \geq -d(x,y); [/tex]

    [tex]\Rightarrow d(y,A) - d(x,A) \geq -d(x,y); [/tex]

    [tex]\Rightarrow -d(x,y) \leq d(x,A)-d(y,A) \leq d(x,y); [/tex]

    Therefore

    [tex] \left|d(x,A) - d(y,A) \right| \leq d(x,y) [/tex]
     
  2. jcsd
  3. Jan 27, 2010 #2

    arildno

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    Dearly Missed

    Your penultimate line doesn't follow.

    Instead, split up in the two cases a) d(y,A) less than d(x,A) and b) greater than

    For a), we have d(x,A)-d(y,A)=|d(x,A)-d(y,A)|, and hence, your first line of the proof may be written as:
    0<= |d(x,A)-d(y,A)|<d(x,y)

    you can manage b) on your own
     
  4. Jan 27, 2010 #3
    [tex]d(y,A) - d(x,A) \geq -d(x,y); [/tex]

    Does not imply

    [tex]-d(x,y) \leq -(-d(x,A)+d(y,A))[/tex]

    but the transition between the last two lines of your proof relies on it. You just switched the sides and direction of the inequality (which is fine) and multiplied one side by -1 (not good). If we could do that, it would be a two line proof. ;-)
     
  5. Jan 27, 2010 #4
    I knew there was something fishy. Thank you for your time y'all, I think I got it.
     
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