# Show that $mathbb{Z}[sqrt{3}]$ is dense in $mathbb{R}.$

Mathematics Asked by Confusion on December 7, 2020

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 $$lim_{nrightarrow infty} u^{-n}.$$

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 $$(sqrt{3}-1)(sqrt{3}+1)=2>1$$).

Correct answer by orangeskid on December 7, 2020