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Homework Help: Prove that d(x,y)/(1+ d(x,y)) is another metric

  1. Aug 17, 2007 #1
    I have the next problem:

    Let [itex]d(x,y)[/itex] be a metric in [itex]\mathbb{R}^n[/itex]. If we define [itex]\tilde{d}(x,y)=\frac{d(x,y)}{1+d(x,y)}[/itex], proof that [itex]\tilde{d}(x,y)[/itex] is another metric.

    I have proven that is not degenerated (i.e. [itex]\tilde{d}(x,y)=0 \longleftrightarrow x=y[/itex]) and symmetric. But I cant proof the triangle's inequality, I only get that is equivalent to proof that

    [itex]\frac{d(x,y)}{1+d(x,y)}\leq \frac{d(x,z)}{1+d(x,z)}+\frac{d(y,z)}{1+d(y,z)}[/itex]

    I need some help, please, because if the denominators wouldn't exist it will be easy. That's I need to know if the denominators in the right member of inequality are lesser than the one on the other member. I hope can you help me.
    Last edited: Aug 17, 2007
  2. jcsd
  3. Aug 17, 2007 #2


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    Did you try to multiply through to clear all of the denominators? [multiply the inequality by the product of all of the denominators]
    It seems that there will be numerous terms that can be grouped.
    Make use of the fact that d already satisfies the inequality.
  4. Aug 17, 2007 #3


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    This is called the bounded metric (associated with d) because it is bounded by 1.

    I did this problem earlier this summer, and I found it helpful to prove this lemma first:

    "If [itex]0\leq a \leq b[/itex], then [tex]\frac{a}{1+a}\leq \frac{b}{1+b}[/tex]"

    Then knowing that d already satisfies the triangle inequality, the answer is very near.
  5. Aug 18, 2007 #4
    Quasar is essentially is saying that [itex]f(x)=\frac{x}{1+x}[/itex] is increasing on [itex]\mathbf{R}^+[/itex]. You can prove this algebraically, which I believe quasar is hinting at, but it "easier" to do it with calculus.
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