# Difficulty with nonarchimedean valuation exercise

I'm reading through a book on algebraic number theory, and in the opening section on valuation theory I've been stumped by one of the problems. The problem is in two parts; the first says

"Let $\varphi_1,\ldots,\varphi_n$ be nontrivial inequivalent nonarchimedean valuations of $F$, and consider $a_1,\dots,a_n\in F^\times$; then [prove] there exists $a\in F^\times$ such that $\varphi_i(a)=\varphi_i(a_i), i=1,...,n$."

I solved this by using the approximation theorem to find an $a$ such that $\varphi_i(a-a_i)<\epsilon$; since $\varphi_i$ is nonarchimedean, if $\varphi_i(a)\neq\varphi_i(a_i)$ then $\varphi_i(a-a_i)=\max(\varphi_i(a),\varphi_i(a_i))$, so by picking $\epsilon<\varphi_i(a_i)$ for all of the $i$, this equality cannot hold true and it must be that $\varphi_i(a)=\varphi_i(a_i)$.

However, I don't know what to do for the second part, which asks me to prove that "If $\epsilon_i>0$ is a value taken on by $\varphi_i$ for $i=1,\ldots,n$, then there exists $a\in F$ such that $\varphi_i(a-a_i)=\epsilon_i, i=1,\ldots,n$." I presume I have to use methods similar to the first problem, but I don't see how to extend it. Where should I start?

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Let $b_i$ be such that $\varphi_i(b_i)=\epsilon_i$. You can find $a$ such that $\varphi_i(a-a_i-b_i)<\epsilon_i$ for all $i$, which implies $\varphi_i(a-a_i)=\epsilon_i$.