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Is this a metric (proof)

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data
    Let a function [tex]\rho:\Re^{2}x\Re^{2}\rightarrow \Re_{+}[/tex] be defined by:
    [tex]\rho((x_{1},y_{2}),(x_{2},y_{2})) = |x_{1} - x_{2}| + |y_{1} - y_{2}|[/tex]

    Prove that [tex]\rho[/tex] is a metric on [tex]\Re^{2}[/tex]


    2. Relevant equations
    To be a metric it must satisfy:

    1. d(x, y) ≥ 0 (non-negativity)
    2. d(x, y) = 0 if and only if x = y (identity of indiscernibles)
    3. d(x, y) = d(y, x) (symmetry)
    4. d(x, z) ≤ d(x, y) + d(y, z)


    3. The attempt at a solution

    I'm not going to give a full proof of each - I just want to see if my basic ideas are correct. I will exclaim that I am not very great at writing proofs just yet (so be critical but polite please :D )

    1. It's obvious (considering the absolute values) that this will be greater than zero (but that isnt a valid statement in a proof...).

    Should I do it case by case, ie:
    x1 > x2 > 0 implies |x1 - x2| > 0
    x2 > x1 > 0 implies x1 - x2 < 0 implies |x1 - x2| = -(x1 - x2) > 0
    so on and so on until I have all cases and then can assume that is correct?


    2. I'm kind of at a loss for this one... maybe(by contradiction)
    (x1,y1), (x2,y2) are real numbers and distinct and |x1 - x2| + |y1 - y2| = 0
    |x1 - x2| = d > 0
    |y1 - y2| = e > 0
    if this were to equal zero then
    d = -e which is a contradiction (one of these would be less than zero)


    3. This one would just be algebra

    4. This one is also just some algebra


    Obviously I need to formally state everything but otherwise does it seem correct?
     
  2. jcsd
  3. Aug 31, 2010 #2

    Office_Shredder

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    For 1 you don't need to do a case by case proof. [tex]|x_1-x_2| \geq 0[/tex] and [tex]|y_1-y_2| \geq 0[/tex] so [tex]|x_1-x_2|+|y_1-y_2|\geq 0[/tex].

    You can use this to prove part 2 by noting that the inequalities [tex]|x_1-x_2| \geq 0[/tex] and [tex]|y_1-y_2| \geq 0[/tex] only give you zeros if [tex]x_1 = x_2[/tex] and [tex] y_1=y_2[/tex]
     
  4. Aug 31, 2010 #3
    Alright... that's what I thought at first but for some reason that seemed too simple...

    could my methods be considered correct (albeit longer and not as simple)?
     
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