A known approximation of $\pi$ is $\sqrt{2}+\sqrt{3}=\color{green}{3.14}\color{red}{6264...}$ But recently I came across a refinement of this approximation $$\pi\approx\sqrt{2}+\sqrt{3}+\frac{\sqrt{2}-\sqrt{3}}{68}=\color{green}{3.14159}\color{red}{0292...}$$ Knowing this, I tried to find interesting approximations of $\pi$ on $\mathbb{Q}(\sqrt{2},\sqrt{3})$ and I've been able to obtain $$\boxed{\pi\approx\frac{42(132\sqrt{2}+375\sqrt{3}-19^2)}{5\left(\left(5^2+2^2\cdot\sqrt{2}\right)\left(10^2+3^2\cdot\sqrt{3}\right)-2273\right)}=\color{green}{3.1415926535897}\color{red}{63336...}}$$ But this made me thinking:
Q: Similar to how we know that the best $\mathbb{Q}$-aproximations of a given real number $x>0$ are given via the partial fractions of its continued fraction representation, is there any notion of best $\mathbb{K}$-approximations where $\mathbb{K}$ is an algebraic number field (i.e. a finite field extension of $\mathbb{Q}$) like $\mathbb{K}=\mathbb{Q}(\sqrt{2})$ or $\mathbb{K}=\mathbb{Q}(\sqrt{2},\sqrt{3})$?
I know that, for $\mathbb{Q}$, we can say that $\frac{a}{b}\in\mathbb{Q}$ is a best approximation of $x>0$ if
$$\forall a'/b'\in\mathbb{Q}\text{ with }b'\leq b\text{ we have }\left|x-\frac{a}{b}\right|\leq\left|x-\frac{a'}{b'}\right|$$
However, we can't extend this to (say) $\mathbb{Q}(\sqrt{2})$ naively expressing numbers of $\mathbb{Q}(\sqrt{2})$ as $\frac{a+b\sqrt{2}}{c}$ since $\mathbb{Z}(\sqrt{2})=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}$ is dense on $\mathbb{R}$ so, for a fixed $c$, we can find $\frac{a+b\sqrt{2}}{c}$ arbitrarily close to $x>0$.
You can define best approximations in any $\mathbb Q$-vector space, and in particular in any finite field extension of the rational numbers as follows.
Let $K=\{q_0+a_1 q_1 +\dots + a_n q_n:q_i\in\mathbb Q\}$ be a $\mathbb Q$-vector space where $(1, a_1, \dots, a_n)$ is linearly independent so $K$ is $n+1$ dimensional. A vector $v\in \mathbb Z^{n+1}$ is a best approximation for the vector $(1, a_1, \dots, a_n)$ if $$|v\cdot (1, a_1, \dots, a_n)|< |w\cdot (1, a_1, \dots, a_n)|$$ for all $w\in\mathbb Z^{n+1}$ with $\lVert w\rVert<\lVert v\rVert$ for some norm $\lVert \cdot\rVert$. Since all the norms in $\mathbb R^d$ are equivalent, it doesn't really matter which one you choose.
This has a nice geometric interpretation, as you're looking for lattice points near a hyperplane going through the origin, and no other rational points. And you want the distance from your lattice points to the hyperplane to be small compared to the norm of the lattice point.
To approximate a number other than 0, say $\pi$ as in your example you can use $Q(\sqrt 2, \sqrt 3)=\{a+b\sqrt 2+c\sqrt 3+d\sqrt 6:a,b,c,d\in\mathbb Q\}=V$ so a best approximation is $x\in\mathbb Z^4$ such that $$|\pi-x\cdot (1, \sqrt 2, \sqrt 3, \sqrt 6)|<|\pi-y\cdot (1, \sqrt 2, \sqrt 3, \sqrt 6)|$$ for all $y$ with $\lVert y\rVert<\lVert x\rVert$.
You can find references for this type of thing by searching for approximation of linear forms. And there's an algorithm for producing good approximations called the Jacobi-Perron algorithm.