Continuous p-adic function similar to q-adic norm Given two distinct primes $p,q$, I am looking for some notion of a $q$-adic valuation of $p$-adic numbers.  Obviously, I can define $f:\Bbb{Q}_p \to \Bbb{Q}_p$ by $f(x) = \begin{cases}|x|_q & x\in \Bbb{Q}_q \\ 1 & \text{otherwise}\end{cases}$.  I was hoping that I could redefine $f$ to be continuous.  In other words, does there exist a continuous function $f \in C(\Bbb{Q}_p,\Bbb{Q}_p)$ such that for all $x \in \Bbb{Q}\setminus \{0\}$, $f(x) = |x|_q$?  Even better if this is true for any $x \in (\Bbb{Q}_p\cap \Bbb{Q}_q)\setminus \{0\}$. I know the q-adic norm is not continuous at zero in $\Bbb{Q}_q$, but since $\left||x|_q\right|_p=1$, I was hoping I might even be able to get continuity at zero, so long as $|f(0)|_p = 1$.
 A: 
Does there exist a continuous function $f:\Bbb Q_p\to\Bbb Q_p$ such that $f(x)=|x|_q$ for all $x\in\Bbb Q^\times$?

Continuity implies sequential continuity: $f(\lim\limits_{n\to\infty} x_n)=\lim\limits_{n\to\infty}f(x_n)$.
Given any $p$-adic number $x\in{\Bbb Q}_p$ (in particular, any rational), there exists a sequence $(x_n)_{n\ge1}$ of rationals such that $x_n\to x$ in the $p$-adic topology but $|x|_q$ does not converge in $\Bbb Q_p$. 
Proof. Pick any sequence of rationals $(x_n)_{n\ge1}$ such that $x_n\to x$ and $v_q(x_n)=0$ bounded from below, then form the new sequence $y_n=x_n+(p/q)^n$. We have $y_n\to x$ but $|y_n|_q=q^n$ for all sufficiently large $n$, which does not converge.

Given two distinct primes $p,q$, I am looking for some notion of a $q$-adic valuation on $\Bbb Q_p$.

Their topologies are incompatible as we've seen, but if we forget about $\Bbb Q_p$'s topology we can put an absolute value on in that extends $|\cdot|_q$ on $\Bbb Q$, though not constructively (meaning we won't be able to write it out explicitly, we just know such an absolute value exists). Let $K$ be the underlying field of $\Bbb Q_p$ and let $\overline{K}$ be its algebraic closure. Since there is a unique algebraically closed field of every cardinality and characteristic up to isomorphism, there exists an isomorphism $\overline{K}\cong\overline{\Bbb Q_q}$. If we apply this isomorphism to the embedding $K\hookrightarrow\overline{K}$ we get an embedding $K\hookrightarrow\overline{\Bbb Q_q}$ and thus by transport of structure we can put a $q$-adic absolute value on $\Bbb Q_p$.
