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

Question

Here is the question I want to answer:

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?

This was my answer! https://math.stackexchange.com/a/3889452/774222 You don't need $\zeta$ to be a unit, per se; what's important in the argument is that $|\zeta| < 1$, so that the union $\cup_n \zeta^n \Bbb{Z}$ is dense in $\Bbb{R}$. — Rivers McForge, Nov 06 '20 at 03:15
It seems like you got angry with using your answer @RiversMcForge do you want me to delete it? — , Nov 06 '20 at 03:40
@RiversMcForge I deleted your answer ..... could you please delete your comment so that the question becomes compatible? thank you — , Nov 06 '20 at 09:50
I don't thinks he's annoyed by it, but anyhow, I thinks his comment is worth staying, since it gives a reference to a good answer/hint. — Darsen, Nov 06 '20 at 10:12
@Confusion It didn't bother me at all. I'm glad it got you thinking about this! — Rivers McForge, Nov 06 '20 at 10:13

orangeskid · Accepted Answer · 2020-11-06T03:39:14.950

2

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

edited Nov 06 '20 at 03:39

answered Nov 06 '20 at 03:10

orangeskid

53,909

so we need its norm to be greater than 1? why? – Nov 06 '20 at 03:33
why $x$ is at most that distance? why dividing by 2 in the second hint in my question? – Nov 06 '20 at 03:36
@Confusion: units in a quadratic ring have norm $\pm 1$. Therefore $\sqrt{3}-1$ is not a unit. – orangeskid Nov 06 '20 at 03:36
Why is this a quadratic ring? – Nov 06 '20 at 03:37
can you please answer my above 2 comments? – Nov 06 '20 at 03:38
What do you mean by integral? des it means integer? – Nov 06 '20 at 03:40
Quadratic ring: something like $\mathbb{Z}[\sqrt{d}]$. If you take the integer multiples of some real number $t$, they form a sequence of dots on the real line. Every number $x$ will be between two such dots. The distance between consecutive dots is $t$. So this number $x$ will closer to one of these dots, so closer than $\frac{t}{2}$. Yeah, integral meant there integer. – orangeskid Nov 06 '20 at 03:43

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

1 Answers1