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Thread: Defining a Distance

  1. #1
    Junior Member hoeltgman's Avatar
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    Defining a Distance

    We define a distance in $\displaystyle \mathbb{R}$ like the following:

    $\displaystyle d(x,y) = \dfrac{|x-y|}{1+ |x-y|} $

    A distance is defined by the following properties:
    1. d(x,y) >= 0
    2. d(x,y) = d(y,x)
    3. d(x,y) =< d(x,z) + d(z,y)

    I managed to prove the two first (it was rather easy). But I'm stuck on the third. Does anybody know how to do? I tried various things from expanding the first part to reducing the second part with the triangular inegality or even the inequation of Cauchy-Schwartz. I ended up always in something that I couldn't prove anymore.

    I would be thankful for any idea.
    Last edited by MathGuru; Nov 10th 2005 at 01:06 PM.
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  2. #2
    hpe
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    Quote Originally Posted by hoeltgman
    We define a distance in $\displaystyle \mathbb{R}$ like the following:

    $\displaystyle d(x,y) = \dfrac{|x-y|}{1+ |x-y|} $

    3. d(x,y) =< d(x,z) + d(z,y)
    Define $\displaystyle f(a) = \frac{a}{1+a}$ for non-negative a and prove two things:

    1. $\displaystyle f $ is increasing
    2. $\displaystyle f(a+b) \le f(a)+f(b)$ for $\displaystyle a, \, b \ge 0 $
    Using also the triangle inequality you should be able to complete the proof.
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  3. #3
    Junior Member hoeltgman's Avatar
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    Quote Originally Posted by hpe
    Define $\displaystyle f(a) = \frac{a}{1+a}$ for non-negative a and prove two things:

    1. $\displaystyle f $ is increasing
    2. $\displaystyle f(a+b) \le f(a)+f(b)$ for $\displaystyle a, \, b \ge 0 $
    Using also the triangle inequality you should be able to complete the proof.

    I've got a few things that don't seem really clear to me. The first point is clear. I derivate and find out that it's always positive....
    I do have problems to see how come from your second point to my original problem. If I say a = |x-y| then b could be |x-z| or |y-z|. Something is missing here I think. I tried your way and ended up with a + b + 1 -ab < 0. which doesn't really help since even if I transform back, I find can find the third distance, but I still have a term which keeps me from getting any good conclusion.
    Last edited by MathGuru; Nov 10th 2005 at 01:06 PM.
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  4. #4
    hpe
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    Quote Originally Posted by hoeltgman
    I've got a few things that don't seem really clear to me. The first point is clear. I derivate and find out that it's always positive....
    I do have problems to see how come from your second point to my original problem. If I say a = |x-y| then b could be |x-z| or |y-z|. Something is missing here I think. I tried your way and ended up with a + b + 1 -ab < 0. which doesn't really help since even if I transform back, I find can find the third distance, but I still have a term which keeps me from getting any good conclusion.
    a = |x-y|, b =|y-z|, c = |x-z|. You want to prove
    $\displaystyle f(c) \le f(a) + f(b)$
    and know from the ordinary triangle inequality that $\displaystyle c \le a + b$.
    So:
    $\displaystyle f(c) \le f(a+b)$ because f is increasing
    $\displaystyle ... \le f(a) + f(b)$ because of the second property. You are done. As to the proof of the second property...
    $\displaystyle f(a+b) = \frac{a+b}{1+a+b} = \frac{a}{1+a+b} + \frac{b}{1+a+b}$
    $\displaystyle ... \le \frac{a}{1+a} + \frac{b}{1+b} = f(a) + f(b)$
    Hope this helps
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  5. #5
    Junior Member hoeltgman's Avatar
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    Thanks a lot. I got it now.
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