Here is the question I want to answer:
Show that $\mathbb{Z}[\sqrt{3}]$ is dense in $\mathbb{R}.$
I got this hint:
Hint: Find $u \in \mathbb{Z}[\sqrt{3}]^*$ with $u > 1$ and consider $lim_{n\rightarrow \infty} u^{-n}.$
My question is:
How can we find this $\zeta$? how can I prove that it is a unit? and why we are taking it from the set of units?should this unit be greater than 1 or less than 1?
You don't really need a unit $\zeta$, just an element $\zeta$ such that $0 < \zeta < 1$. You can start with any element $r$ in your ring that is not integral. Then $\zeta =\{r\} = r - [r]$ is in the interval $(0,1)$, and can be used in your nice argument.
$\bf{Added:}$ For example, consider $r=\sqrt{3}$. We have $[r]=1$. Take $\zeta= \sqrt{3}-1$, in $(0,1)$, but not a unit ( since its norm $(\sqrt{3}-1)(\sqrt{3}+1)=2>1$).