# Defining a Distance

• Oct 15th 2005, 03:16 AM
hoeltgman
Defining a Distance
We define a distance in $\mathbb{R}$ like the following:

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

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.
• Oct 15th 2005, 12:53 PM
hpe
Quote:

Originally Posted by hoeltgman
We define a distance in $\mathbb{R}$ like the following:

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

3. d(x,y) =< d(x,z) + d(z,y)

Define $f(a) = \frac{a}{1+a}$ for non-negative a and prove two things:

1. $f$ is increasing
2. $f(a+b) \le f(a)+f(b)$ for $a, \, b \ge 0$
Using also the triangle inequality you should be able to complete the proof.
• Oct 16th 2005, 12:55 AM
hoeltgman
Quote:

Originally Posted by hpe
Define $f(a) = \frac{a}{1+a}$ for non-negative a and prove two things:

1. $f$ is increasing
2. $f(a+b) \le f(a)+f(b)$ for $a, \, b \ge 0$
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.
• Oct 16th 2005, 04:01 PM
hpe
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
$f(c) \le f(a) + f(b)$
and know from the ordinary triangle inequality that $c \le a + b$.
So:
$f(c) \le f(a+b)$ because f is increasing
$... \le f(a) + f(b)$ because of the second property. You are done. As to the proof of the second property...
$f(a+b) = \frac{a+b}{1+a+b} = \frac{a}{1+a+b} + \frac{b}{1+a+b}$
$... \le \frac{a}{1+a} + \frac{b}{1+b} = f(a) + f(b)$
Hope this helps :)
• Oct 17th 2005, 08:17 AM
hoeltgman
Thanks a lot. I got it now.