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A subset $S$ of $\mathbb{R}$ is said to be dense in $\mathbb{R}$ if $\forall$ $\epsilon > 0$ and $x \in \mathbb{R}$, $\exists$ $s \in S$ such that $| x - s| < \epsilon$.

Using this definition, I would like to try to prove that the set $S = \{ a + b \sqrt{3} : a ,b \in \mathbb{Z} \}$ is dense in $\mathbb{R}$.

Any help would be appreciated.

  • I am aware that the set $S$ is closed under addition and multiplication. I'm not sure if that helps.
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2 Answers 2

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Quick way to show that $S$ is dense in $\Bbb{R}$:

  • $S = \{ a + b \sqrt{3} : a ,b \in \mathbb{Z} \}$ contains the element $\zeta := 2 - \sqrt{3}$

  • Since $S$ is closed under addition and multiplication, $\zeta^n$ is in $S$ for every $n \geq 2$

  • For any $x \in \Bbb{R}$, $x$ is at most $\zeta^n/2$ away from some element of $\zeta^n \Bbb{Z} \subset S$

  • Since $0 < \zeta < 1$, $\zeta^n \to 0$ as $n \to \infty$.

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  • $\begingroup$ Thank you this helped. $\endgroup$
    – user486995
    Oct 31, 2020 at 23:14
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HINT: It follows from the fact that $\{nr\bmod 1:n\in\Bbb Z\}$ is dense in $[0,1)$ whenever $r$ is irrational. (Here $x\bmod 1=x-\lfloor x\rfloor$ is the fractional part of $x$.) This fact can be proved using the pigeonhole principle; I gave a proof here.

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  • $\begingroup$ Is there anyway to prove it straight from definition? The linked page went over my head. $\endgroup$
    – user486995
    Oct 31, 2020 at 22:21
  • $\begingroup$ @JordanReed: I can’t think of one offhand, I’m afraid. $\endgroup$
    – Brian M. Scott
    Oct 31, 2020 at 22:31

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