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I thought about the expansion of $\sqrt{3}$ as a series. But I didn't get anything useful. Also, I thought about a set with irrational numbers instead of rational numbers.

What is the general idea to attack this kind of problems? Since is the first one I'm doing of this type.

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  • $\begingroup$ Hint: Consider $$(\sqrt 3,2) \cap \mathbb{Q}$$ $\endgroup$ – Story123 Mar 30 at 7:47
  • $\begingroup$ As per this question, it is possible to build a sequence (and a set) for any $\sqrt{n}$. $\endgroup$ – rtybase Mar 30 at 9:26
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Hint: Apply Newton's Method to the function $f(x)=x^2-3$, starting with your favorite rational number that you are sure is $> \sqrt{3}$.

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  • $\begingroup$ Thanks, I will try it. $\endgroup$ – ClaraGarcía Mar 30 at 5:07
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A simple example of a subset of $\Bbb Q$ with infimum $\sqrt3$, and which does not presuppose existence of $\sqrt3$ or of any irrational numbers is $$\{a\in\Bbb Q:a>0\quad\text{and}\quad a^2>3\}.$$

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  • $\begingroup$ yep this is probably the simplest construction $\endgroup$ – qwr Mar 30 at 5:32
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This isn't very educational, but $a_n = \text{ceil}(\sqrt{3} * 10^n)/10^n$ would work. So

$a_1 = \text{ceil}(1.73205... * 10)/10 = \text{ceil}(17.3205...)/10 = 18/10 = 1.8$ $a_2 = \text{ceil}(1.73205... * 100)/100 = \text{ceil}(173.205...)/100 = 174/100 = 1.74$ $a_3 = \text{ceil}(1.73205... * 1000)/1000 = \text{ceil}(1732.05...)/1000 = 1733/1000 = 1.733$

This would obviously work for any irrational number in the same way.

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  • $\begingroup$ Thanks!! Now, to prove it, it seems a little weird to use the definition of infimum. Let $S$ be the set, so for every $s \in S$ we have: $s \geq \sqrt{3} + \varepsilon$ and now $a_n$ is our $s$, right? $\endgroup$ – ClaraGarcía Mar 30 at 4:55
  • $\begingroup$ Yep, that makes sense. $\endgroup$ – user2825632 Mar 30 at 5:01

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