Does $\langle3\rangle\cdot\{3x+2^{\nu_2(x)}:x\in\Bbb Z_2\}$ cover $\Bbb Z_2$? Does it fail to assign a $\nu_3(x)$ to any values?
Motivation: I've been shown before that there's no obvious 3-adic value (using the term "valuation" loosely) on the 2-adic integers $\Bbb Z_2$. However on $\Bbb Z$ we can define the function $f(x)=3x+2^{\nu_2(x)}$ and instantly see that the leaves of its graph are the multiples of $3$. Then knowing $2$ is coprime with $3$, we can define $f$ over $\Bbb Z[\frac12]/\langle2\rangle$ knowing the family of sets over which it surjects contains only numbers not divisible by $3$, and then with $\{3^n:n\in\Bbb Z\}\times \{3x+2^{\nu_2(x)}:x\in\Bbb Z[\frac12]/\langle2\rangle\}$ we have our valuation $\nu_3(x)=n$ for any given integer.
I don't see how this would fail to define the same valuation over the entirety of $\Bbb Z_2$:
$f:\Bbb Z_2\to\Bbb Z_2$ as $f(x)=3x+2^{\nu_2(x)}$
Then $\{3^n:n\in\Bbb Z\}$ times the set over which it surjects, gives us our 3-adic value over the whole of $\Bbb Z_2$.
Does this cover $\Bbb Z_2$, or does it fail to assign every number a value for any reason?