1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Prove that d(x,y)/(1+ d(x,y)) is another metric
  1. Y''*y' = x(x+1) (Replies: 10)