# Thread: Verify that it defines a distance

1. ## Verify that it defines a distance

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...

2. 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.
$\begin{gathered}
d(X,Y) = 0\text{ if and only if }X=Y \hfill \\
d(X,Y) = d(Y,X) \hfill \\
d(X,Y) \leqslant d(X,Z) + d(Z,Y) \hfill \\
\end{gathered}$

3. 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:
Make common denominators and combine the fractions: $\frac{a(1+b)(1+c)+b(1+a)(1+c)-c(1+a)(1+b)}{(1+a)(1+b)(1+c)}$.

You only have to show that the numerator is positive, as the denominator is clearly positive (because $a,b,c>0$).

Multiply out the numerator: $a(1+b)(1+c)+b(1+a)(1+c)-c(1+a)(1+b)$ $=a+ab+ac+abc+b+ab+bc+abc-c-ac-bc-abc$ $=a+b-c+2ab+abc$

Now we use the fact that $a+b-c\geq0$ to say that $a+b-c+2ab+abc\geq 2ab+abc$.

Clearly $2ab+abc\geq0$, because $a$, $b$, and $c$ are all positive. So we're done.

4. that was a lot easier than I thought it was, thanks! I think it was all the R's and the wording that kinda threw me off

5. actually, I'm still kinda lost on the third part... I don't really understand how to prove the triangle inequality