Strong approximation theorem and Chinese remainder theorem The main question in this post is: How to proof the Chinese remainder theorem (in elementary number theory, i.e. in $\mathbb{Z}$) using the strong approximation theorem in $\mathbb{Q}$ in valuation theory.
Any proof and references are welcomed! :)
We shall state the strong approximation theorem here. It is clearer to introduce the weak approximation theorem at first:

Weak approximation theorem: Given $n$ inequivalent nontrivial valuation $\vert \cdot \vert_i$, $i=1,\ldots,n$ of a field $k$, an arbitrary positive real number $\epsilon$, and $n$ arbitrary elements $a_i$, there exists an element $a \in k$ such that $$ \vert a - a_i \vert_i < \epsilon.$$

The strong approximation theorem of $\mathbb{Q}$ goes like this (keep using the notations above):

Strong approximation theorem: Let $k$ above be the rational number field $\mathbb{Q}$, and the valuations $\vert \cdot \vert_i$ be $p_i$-adic valuations, then not only does there exists an $a \in \mathbb{Q}$ such that $ \vert a - a_i \vert_i < \epsilon$ for $i=1,\ldots,n$, but $ \vert a\vert_p \leq 1$ for all other $p$-adic valuations as well.

The source of this question and my attempts: I have heard people said that the approximation theorem in valuation theory is somehow a generalization of Chinese remainder theorem. So I am wondering whether we can use the approximation theorem to prove the Chinese remainder theorem. By looking up many books, especially on G. Bachman’s Introduction to $p$-adic numbers and valuation theory, there is an exercise asking for proof of the Chinese remainder theorem using the strong approximation theorem.
 A: The Chinese Remainder Theorem has various equivalent fomulations, but let's take this one:

Let $n_1, ..., n_k$ be pairwise coprime positive integers, and $a_1, ..., a_k$ any integers. Then there exists an integer $a$, unique modulo $n:= \prod n_i$, such that for all $i$ we have $a \equiv a_i$ (mod $n_i$).

That such an $a$, if it exists, is unique modulo $lcm(n_1, ..., n_k) =n$ is easy to show by elementary means. The main thing to prove, for which we can use the strong approximation theorem, is existence of $a$.
Now notice, following user Berci's comment, that in the special case that each $n_i$ is a prime power $p_i^{k_i}$, this is almost literally the theorem as quoted by you, applied with $\epsilon := \min_i \{p_i^{-k_i}\}$. Namely, the $a \in \mathbb Q$ which now exists due to that theorem actually is an integer $a \in \mathbb Z$ (that is what $\lvert a \rvert_p \le 1$ for all primes $p$ means), and for each $i$, $\lvert a-a_i\rvert < \epsilon \le p^{-k_i}$ literally means that $a \equiv a_i$ (mod $p^\ell$) for some $\ell > k_i$ which is even stronger than $a \equiv a_i$ (mod $p^k_i$).
In the general case, we decompose each $n_i$ into its prime (power) factors; formally, let's say we have numerated all primes $p_1, p_2, p_3, ...$, then for each $i$ let $J(i)$ be the set $\{j \in \mathbb N: p_j \mid n_i \}$ of those primes which divide $n_i$, so that $n_i = \displaystyle\prod_{j \in J(i)} p_{j}^{k_{j}}$. Note that since the $n_i$ are mutually coprime, the sets $J(i)$ are mutually disjoint; let $J := \bigcup_i J(i)$ be their union and $a_j := a_i$ for all $j \in J(i)$. Now apply the theorem to the $a_j, j \in J$ and $\epsilon:= \min_{j \in J} \{p_j^{-k_j}\}$. Check that again the $a$ whose existence comes from the approximation theorem is an integer such that for each $i$, the difference $a-a_i$ is divisible by all $p_j^{k_j}$ for $j \in J(i)$, hence by their product $n_i$, in other words $a \equiv a_i$ (mod $n_i$).
