Equivalent metric making rational complete I know of no necessary conditions for a metric d to be equivalent to the standard euclidean metric on Reals.Hence, I was facing difficulty in answering the following problem:
Does there exist d, equivalent to Standard metric on reals, s.t. it makes (Q,d) a complete metric space? 
 A: No.
Proof. Let $(x_n)_{n\geq0}$ be a sequence of rational numbers converging to $\sqrt{2}$ with repect to the standard metric in ${\mathbb R}$, and let $d:=d(\cdot,\cdot)$ be some other metric on ${\mathbb R}$, equivalent to the standard metric. Then $\lim_{n\to\infty} x_n=\sqrt{2}$ also with respect to $d$, whence $(x_n)_{n\geq0}$ is a Cauchy sequence with respect to $d$. If $({\mathbb Q},d)$ were complete then $(x_n)_{n\geq0}$, being a sequence of rational numbers,  would have to converge to some $\xi\in{\mathbb Q}$ with respect to $d$, but this would imply $\sqrt{2}\in{\mathbb Q}$ – a contradiction.
A: Here is some different aproach,
$\mathbb{Q}$ is countable so we can enumerate $\mathbb{Q}$ as $\{x_1,x_2,x_3,...\}$.
Suppose there exist a metric $d$ equivalent to the standard metric such that it makes $(\mathbb{Q},d)$ a complete metric space.
Now,  $\mathbb{Q}=\cup_{n=1}^{\infty}\{x_n\}$
$\{x_n\}$ is a finite set so it is closed in $\mathbb{Q}$ with respect to the standard metric as well as the $d$ metric.
Let, $\{x_n\}=A_n$  , $\forall n \in \mathbb{N}$
here , ${(A_n)}^{o}=\phi$ , i.e, the interior of the set $A_n$ is $\phi$ for all $n \in \mathbb{N}$.Since ${(A_n)}^{o}=\phi$ in $\mathbb{Q}$ with respect to the standard metric, which implies that it will be same in $(\mathbb{Q},d)$ since $d$ is equivalent to the standard metric.
Now according to $\textbf{Baire Category Theorem}$ ,

If $(X,d)$ be a complete metric space and $\{F_n\}_n$ be a sequence of non empty closed subsets of $X$ such that $X=\cup_{n=1}^{\infty}F_n$ . Then at least one of $F_n$'s has nonempty interior.In other words,a complete metric space cannot be a countable union of nowhere dense closed subsets.


nowhere dense set: A subset of a metric space $X$ is called nowhere dense in $X$ if the interior of the closure of the subset is empty. i.e., $(\bar{A})^{o}=\phi$.

Now $\{A_n\}_n$ is a countable sequence of nonempty closed subsets of $\mathbb{Q}$ and all $A_n$'s are nowhere dense subsets of $\mathbb{Q}$ and also $\mathbb{Q}$ can be written as countable union of nowhere dense subsets.
So it creates a contradiction since we assume that $(\mathbb{Q},d)$ is a complete metric space.
So there does not exist such metric $d$ which makes $(\mathbb{Q},d)$  a complete metric space where $d$ is equivalent to the standard metric.
here is a link https://en.wikipedia.org/wiki/Baire_category_theorem.
