Complete extensions of valuations from Q to R. This is somewhat related to the question and the answers here:
Is completeness of a field an algebraic property?
My question is (to which I believe the answer must have been known), does every extension of of a non-trivial non-archimedean valuation on $\mathbf{Q}$ to $\mathbf{R}$ fail to make $\mathbf{R}$ into a complete field? (Valuations considered here are general Krull valuations.)
I guess the notion of Cauchy sequence should be generalized a little bit, considering that $\mathrm{cf}(2^{\aleph_0}) > \aleph_0$, i.e. one should allow a Cauchy sequence of any well order type.
Thanks! 

Edit: I apologize that I did not make the question as clear as it should have been. Let me explain it furthermore here.


*

*All the valuations here are Krull's general valuations, which could be of any rank, not necessarily absolute values. 

*For a valued field with higher-rank value group $\Gamma$, the definition of Cauchy sequences (or completeness) depends on the value group, in the sense that  a Cauchy sequence must be of length $\mathrm{cf}(\Gamma)$. (hence my second paragraph above). The definition could be found in Engler and Prestel's book Valued Fields. More specifically, let $\kappa=\mathrm{cf}(\Gamma)$, and a sequence $(a_\nu)_{\nu<\kappa}$ is said to be a Cauchy sequence if for every $\gamma\in \Gamma$, there exists some $\delta <\kappa$ such that for all $\mu, \nu \in (\delta, \kappa)$, one has 
$v(a_\mu-a_\nu)>\gamma$. 

*And the valued field is said to be complete if every Cauchy sequence has a limit, meaning there exist some $a$ in the field, such that $v(a_\nu-a)>\gamma$ for any $\gamma$ and sufficiently big $\nu$, which says that $a$ is a limit point of the sequence in the topological sense.
 A: Any nontrivial non-archimedean valuation  on the field of rational numbers $Q$ is essentially $p$-adic for some prime $p$, i.e., there exists a prime $p$ such that $|p|<1$ and one may  check that such $p$ is unique: $|q|=1$ for all other primes $q$. (This is a classical theorem of Ostrowsky, see first pages of Koblitz's p-adic numbers, p-adic analysis and zeta-functions.)  So, if R is complete wrt some extension of the valuation then it must contain the $p$-adic completion of $Q$, i.e., the field $Q_p$ of $p$-adic numbers. More precisely, the field of real numbers $R$ must contain a subfield isomorphic to $Q_p$. In particular, all rational numbers that are squares in $Q_p$ are squares in $R$. But this is not the case: for example, $1-p^3$ is a square in $Q_p$ but not in $R$, since it is negative. This proves that $R$ does not contain a subfield isomorphic to $Q_p$. This implies that in the case of usual valuations the answer to your question is positive in the following sense: any extension of of a nontrivial non-archimedean valuation on $Q$ to $R$ fails to make $R$ a complete field. 
Of course, if the original non-archimedean valuation  on $Q$ is trivial. i.e., every nonzero rational number has norm 1 then one may extend it to the trivial valuation on $R$; every field is complete wrt the trivial valuation, including $R$.
