$(X,d)$ is a complete metric space iff $(X,d')$ is a complete metric space. I recently got asked this question on an exam and I wasn't able to give a solution. 
If $(X,d)$ is a metric space and we define
 $$d'(x,y)= \frac {d(x,y)}{1+d(x,y)}$$
(I've already proved that $(X,d')$ is a metric space.)
Prove that $(X,d)$ is a complete metric space iff $(X,d')$ is a complete metric space.

You can see that $d' = {d \over 1+ d} \le {d \over 1} = d$, hence a $d$-Cauchy sequence is a $d'$-Cauchy sequence.
Note that if $d'(x,y) \neq 1$, then since $d'(x,y) = { d(x,y) \over 1 + d(x,y) }$, then $d(x,y) = { d'(x,y) \over 1 - d'(x,y) }$.
Now suppose $x_n$ is a $d'$-Cauchy sequence, and choose $N$ such that if $m,n \ge N$, then $d'(x_n,x_m) \le {1 \over 2}$. Then for $m,n \ge N$, we have
$d(x,y) \le  2 d'(x,y)$ and so the sequence is a $d$-Cauchy sequence.
If $(X,d)$ is complete, and $x_n$ is $d$-Cauchy, then there is some $x$ such that $d(x,x_n) \to 0$. Since $d' \le d$, it follows that $d'(x,x_n) \to 0$ and hence $(X,d')$ is complete.
Similarly, if $(X,d')$ is complete, and $x_n$ is $d'$-Cauchy, then there is some $x$ such that $d'(x,x_n) \to 0$. Choose $N$ such that if $n \ge N$
then $d'(x,x_n) \le {1 \over 2}$, then we have $d(x,x_n) \le 2 d'(x,x_n)$ and
so $d(x,x_n) \to 0$ and hence $(X,d)$ is complete.
