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?

One Answer

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

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