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

  • Thread starter yxgao
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  • #1
yxgao
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(For every set S and every metric d on S)

The answer is d^(1/2)

How do you prove this mathematically?
 

Answers and Replies

  • #2
NateTG
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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.
 
  • #3
yxgao
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that makes perfect sense.
thanks
 

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