Context : proof with an explicit expression of "closest integers" that $\mathbb{Z} \left[ \sqrt {2} \right]$ is euclidean.
When i draw a two dimensional network in the example of $\mathbb{Z} \left[ i\sqrt {2} \right]$ i find that , given $x+i\sqrt{2} y$ where $(x,y)\in \mathbb{R}^2 $ , we can find a closest element with compounds "x and y" integers given explicitely by : $$a=\lfloor x+\frac{1}{2}\rfloor,b =\lfloor \sqrt{2}y+\frac{\sqrt{2}}{2}\rfloor $$ (In the sense, $a+i\sqrt{2}b\in \mathbb{Z} \left[ i\sqrt {2} \right]$ and is the closest to $x+i\sqrt{2}y$ for conventional norm on $\mathbb{C}$).
If that's correct.
But in the case of $\mathbb{Z} \left[ \sqrt {2} \right]$ , i can't really find the explicit expression given $x+\sqrt{2} y$ where $(x,y)\in \mathbb{R}^2 $ of the closest element with integer compounds "x and y", i can't make a proper draw.
What's the proper explicit expression , and how to see it on a draw that seems to be one dimensional
There is no such closest element. For example, the diophantine equation
$$(2x-1)^2-2(2y)^2=1$$
has infinitely many solutions in positive integers $x,y$ and the corresponding elements $x-y\sqrt{2}\in \mathbb Z[\sqrt 2]$ tend to $\frac 12$ from the right. This means that there is no element of $\mathbb Z[\sqrt 2]$ closest to $\frac 12$ in the usual metric on the reals.