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Showing something is a metric

  1. Sep 13, 2009 #1
    d is a metric show d'= d/d+1 is a metric

    I know d(x,z)[tex]\leq[/tex]d(x,y)+d(y,z). And have been trying to make it fall out of this.

    I have been fooling around with the terms but it has not provided to be useful. Any direction would be helpful.
     
  2. jcsd
  3. Sep 13, 2009 #2

    Office_Shredder

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    Which parts of the definition of metric have you been able to demonstrate?
     
  4. Sep 13, 2009 #3
    d'(x,x)=0
    d'(x,y)>0
    d'(x,y)=d'(y,x)

    They follow immediately from d being a metric. But the triangle inequality is providing to be more difficult.
     
  5. Sep 13, 2009 #4
    I actually saw this problem a bit ago as well. If we let a = d(x,z), b = d(x,y) and c = d(y,z), then we know [itex]a\leq b + c[/itex] and it comes down to proving [itex]\frac{a}{1+a} \leq \frac{b}{1+b} + \frac{c}{1+c},[/itex] which is easy if you just multiply out, I think. Anyways there is probably a nicer way to do this, but I think the above works (I didn't really double check) and suits me.

    *EDIT* By the way, I think awhile ago you were trying to prove all of axioms of a metric spaces using only 2 of them. At first I said you had to have a rearrangement of the triangle inequality, but later I realized you could do it using the triangle inequality by switching around a few variables. If you're still interested, I could show you.
     
  6. Sep 13, 2009 #5
    I figured that one out... eventually. But thanks I'm impressed that you remember.
    Ya, that does work, I guess I missed the most obvious approach.
     
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