$(X,d)$ 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.

HINT: If $d(x,y)\ge\epsilon$, then
$$d'(x,y)=\frac{d(x,y)}{1+d(x,y)}=1-\frac1{1+d(x,y)}\ge 1-\frac1{1+\epsilon}=\frac{\epsilon}{1+\epsilon}\;.$$
If $d'(x,y)\ge\epsilon$, then $\epsilon<1$, and
$$\frac{d(x,y)}{1+d(x,y)}\ge\epsilon\;,$$
so $d(x,y)\ge\epsilon+\epsilon d(x,y)$, and 
$$d(x,y)\ge\frac{\epsilon}{1-\epsilon}\;.$$
Use these to show that if a sequence is not Cauchy in one metric, then it’s not Cauchy in the other metric, and use the fact (that I suspect you’ve already proved) that $d$ and $d'$ are topologically equivalent.
