Verify that it defines a distance

**thaopanda** · September 26th 2009, 01:54 PM

Verify that d : $R^3 \times R^3 \rightarrow R$ given by d(X,Y) := $\frac{|X-Y|}{1+|X-Y|}$ for $X,Y \in R^3$ defines a distance on $R^3$

So, I haven't actually learned this yet in class... so I don't really understand what this is saying...

**Plato** · September 26th 2009, 03:24 PM

Originally Posted by thaopanda

Verify that d : $R^3 \times R^3 \rightarrow R$ given by d(X,Y) := $\frac{|X-Y|}{1+|X-Y|}$ for $X,Y \in R^3$ defines a distance on $R^3$
So, I haven't actually learned this yet in class... so I don't really understand what this is saying...

You must show three properties.
$\begin{gathered} d(X,Y) = 0\text{ if and only if }X=Y \hfill \\ d(X,Y) = d(Y,X) \hfill \\ d(X,Y) \leqslant d(X,Z) + d(Z,Y) \hfill \\ \end{gathered}$

**redsoxfan325** · September 26th 2009, 06:54 PM

Originally Posted by thaopanda

Verify that d : $R^3 \times R^3 \rightarrow R$ given by d(X,Y) := $\frac{|X-Y|}{1+|X-Y|}$ for $X,Y \in R^3$ defines a distance on $R^3$

So, I haven't actually learned this yet in class... so I don't really understand what this is saying...

Originally Posted by Plato

You must show three properties.
$\begin{gathered} 1.~d(X,Y) = 0\text{ if and only if }X=Y \hfill \\ 2.~d(X,Y) = d(Y,X) \hfill \\ 3.~d(X,Y) \leqslant d(X,Z) + d(Z,Y) \hfill \\ \end{gathered}$

The first two are trivial. The third may be a bit tricky to prove if you get hung up on the absolute value signs. All you need to know is that $|X-Y|$ is a metric. It would be easiest to label them: $|X-Y|=a$ , $|Y-Z|=b$ , $|X-Z|=c$ . You know from the triangle inequality that $c\leq a+b$ .

So now here's what you have to prove:

$\frac{c}{1+c}\leq\frac{a}{1+a}+\frac{b}{1+b}\impli es\frac{a}{1+a}+\frac{b}{1+b}-\frac{c}{1+c}\geq0$ given that $c\leq a+b\implies a+b-c\geq0$ .

Spoiler: