An application of the Weak Approximation theorem - Artin-Whaples Approximation Theorem Let us recall the weak approximation theorem from Valuation theory in Algebraic Number Theory.
Let $K$ be a field, and let $|\cdot|_{1},\dots, |\cdot|_n$ be pairwise non-equivalent nontrivial archimedian or non-archimedian absolute values. For each $a_1,\dots,a_n \in K$, and for each $\epsilon\gt 0$, there exists $x\in K$ such that $|x-a_i|_{i}<\epsilon$ for all $i = 1,\dots, n$.
    So my question is the following:

Let $K$ be a field, and let $v_1, \dots, v_n$ be nontrivial, pairwise non-equivalent non-archimedian valuations. Given $a_1,\dots,a_n \in K^\times$ prove that there exists some $x\in K$ such that 
  $$ v_i(x) = v_i(a_i)$$ for all $i=1,\dots,n$.

A hint is given for this exercise. You can use the following property: If $x_1, x_2 \in K^\times$ satisfy $$v(x_1) \neq v(x_{2}),$$ then $$v(x_{1} + x_2) = \min \lbrace(x_1), v(x_2) \rbrace.$$
So I don´t actually understand how to proceed, I´ve tried using the equivalence of valuations and absolute values to no avail. Am I missing something?
 A: I can answer my own question. 
Given $\epsilon \gt 0$.
We can pick an $x$: $|x - a_i|_i \lt \epsilon \iff v_i(x - a_i) \gt \eta$
So at this point assume that we are given $\eta \gt 0$ and remember that in the definition of a valuation we have $v(a_i) = v(-a_i)$. So we can say that there exists a x such that $$v_i(x - a_i) \gt \eta$$
Let us fix i. And we can remark that we have the following two possibilities
\begin{equation}v_i(x) \neq v_i(a_i) \Rightarrow v_i(x - a_i) = \min\lbrace v_i(x), v_i(a_i)\rbrace  \;\;\;\;\;\;\; (1)\end{equation} 
or$$v_i(x) = v_i(a_i) \;\;\;(2)$$
We note at this point that(2) is what we wish to prove. For all i. 
So we proceed by proving that (1) is impossible. 
If (1) is true. Then $\eta \lt v_i(x-a_i) = \min\lbrace v_i(x), v_i(a_i)\rbrace$
 If we have chosen $\eta \gt v_i(a_i)\;$ this implies that the minimum is $v_i(x)$ which is equivalent to saying that $v_i(x) \leq v_i(a_i) \lt \eta$ which is a contradiction. This implies that (2) holds. 
We then just pick a larger set than just i = 1, and the proof comes from finiteness. 
