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Which of the following is a metric on S? d^2 or d^(1/2)

  1. Mar 16, 2004 #1
    (For every set S and every metric d on S)

    The answer is d^(1/2)

    How do you prove this mathematically?
  2. jcsd
  3. Mar 16, 2004 #2


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    For [tex]d[/tex] to be a metric you need to show that:
    [tex]d(a,b) = 0 \iff a=b[/tex]
    (which is easy in this case)
    [tex]d(a,b) \geq 0[/tex]
    (also easy)
    [tex]d(a,c) \leq d(a,b)+d(b,c)[/tex]
    Which is the only one that really needs any looking into in this case.

    The does not necessarily hold for [tex]d'=d^2[/tex] since if you have [tex]d(a,b)=1[/tex] and [tex]d(b,c)=1[/tex] and [tex]d(a,b)=2[/tex], then the triangle inequality does not hold for [tex]d'[/tex].

    To prove that the triangle inequality holds for [tex]d'=d^{\frac{1}{2}}[/tex], start with the triangle inequality for [tex]d[/tex], complete the square on the RHS, and take the square root of both sides.
  4. Mar 16, 2004 #3
    that makes perfect sense.
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