Prove that the function $e(a,b)=\frac{d(a,b)}{1+d(a,b)} \ where \ a,b\in X$, is a metric

Question

I am looking to prove that the following function from the non empty group $X$ is a metric with $d$ already being a metric:

$$e(a,b)=\frac{d(a,b)}{1+d(a,b)} \ \text{where} \ a,b\in X$$

But I am not sure how to proceed with such equations.

UPDATE:

From the comments posted so far, I interpreted the following:

$$d(a,b) \geq 0$$ $$e(a,b) \geq 0 \implies \frac{d(a,b)}{1+d(a,b)} \implies d(a,b) \geq 0 \Rightarrow \text{true}$$
$$d(a,b) = 0 \implies \frac{d(a,b)}{1+d(a,b)} = 0\implies d(a,b) = 0 \implies a=b$$
$$d(a,b)=d(b,a) \implies e(a,b) = e(b,a) \implies \frac{d(a,b)}{1+d(a,b)} = \frac{d(b,a)}{1+d(b,a)}$$
$$d(a,c) \leq d(a,b) + d(c,d) \implies e(a,c) \leq e(a,b) + e(b,c) \implies \frac{d(a,c)}{1+d(a,c)} \leq \frac{d(a,b)}{1+d(a,b)} + \frac{d(b,c)}{1+d(b,c)}$$

Am I following the right path?

Metrics have three properties to check. For example, you need to check that $e(a,b)\geq 0$ with equality if and only if $a=b$. Have you tried checking each of the individual properties for $e$, using the fact that you know they hold for $d$? — Aaron, Nov 03 '17 at 11:29
So then I just do basic transposition in the equation? For example: $d(a,b)=1+d(a,b)$ and then simplify? — Omari Celestine, Nov 03 '17 at 11:34
First begin with : if $e(a,b) =0$ then... ? Secondly, If $a=b$, then what can you say about $e(a,b)$ ? — anonymus, Nov 03 '17 at 11:44
@OmariCelestine Have you tried to verify the 3 conditions of a metric — blancmange, Nov 03 '17 at 11:55
I updated my question, could you let me know if I am following the conditions correctly? — Omari Celestine, Nov 03 '17 at 12:17

score 1 · Accepted Answer · answered Nov 03 '17 at 15:59

You are almost there, but there are some issues about the order you've written things in.

Since $d$ is a metric, $d(a,b)\geq 0$ for all $a,b\in X$. It follows that $1+d(a,b)\geq 1$ and $e(a,b)=\frac{d(a,b)}{1+d(a,b)}\geq 0$ for all $a,b\in X$.

2. $$e(a,b)=0 \iff d(a,b)=0 \iff a=b$$ where the last equivalence comes from the fact that $d$ is a metric.

Since $d$ is a metric $d(a,b)=d(b,a)$ for all $a,b\in X$. It follows that $e(a,b)=e(b,a)$ for all $a,b\in X$.
Lastly, we need to prove that the triangle inequality holds. Take any $a,b,c\in X$. We must show $e(a,b)+e(a,c)\geq e(a,c)$. This is the hard part of this question. There is a nice solution here: If $d$ is a metric, then $d/(1+d)$ is also a metric

1 Answers1