Find a completion of the incomplete metric space $(X,d)$, where
a) $X=\Bbb{Q}$, where $\forall{q,r}\in{X}$ $d(q,r) = \lvert {\arctan(q) - \arctan(r)}\rvert$
b) $X=(0,1)\setminus \mathbb{Q}$, where $\forall{x,y}\in{X}$ $d(x,y)=|x-1|+|y-1|$ for $x\neq y$, $d(x,x)=0$
For part a), I suspect that the metric $d$ is equivalent to the usual Euclidean metric $\lvert {q-r}\rvert$. If this is correct, then the completion of $X$ would be $\Bbb{R}$ I guess. To show the equivalence, I thought that we can use the sequential characterization of continuity along with the fact that the function $arctan$ is continuous, but I am not sure about this as the domain of $arctan$ is $\Bbb{Q}$, any idea is highly appreciated.
For part b), If $X$ was $(0,1)$ with the same metric, I would argue that any convergent sequence in $X$ converges to $1$, so the completion might be $(0,1]$. But $X=(0,1)\setminus \mathbb{Q}$, so should I still argue that the completion is $(0,1]$? My mind is very confused about this part.
a) for any bounded (in the usual sense) Cauchy sequence $\{q_n\}$, the uniform continuity of arctan on a closed interval guarantees that it is also Cauchy in the metric $d$, and with the same limit. For unbounded sequences, though, things are different. Consider $\{n\}$; this sequence is Cauchy for $d$, and its limit cannot be any real number. This suggests that the completion will require two more points, $\pm\infty$.
b) the only way that a sequence can be Cauchy is if it converges to $1$ in the usual metric. So only $1$ is needed as a new point to get the completion. For subsets that are removed from $1$ in the usual metric, the metric $d$ is discrete.