# Find a set of rational numbers where $\sqrt{3}$ is the infimum. Prove it

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.

• Hint: Consider $$(\sqrt 3,2) \cap \mathbb{Q}$$ – Story123 Mar 30 at 7:47
• As per this question, it is possible to build a sequence (and a set) for any $\sqrt{n}$. – rtybase Mar 30 at 9:26

## 3 Answers

Hint: Apply Newton's Method to the function $$f(x)=x^2-3$$, starting with your favorite rational number that you are sure is $$> \sqrt{3}$$.

• Thanks, I will try it. – ClaraGarcía Mar 30 at 5:07

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 $$\{a\in\Bbb Q:a>0\quad\text{and}\quad a^2>3\}.$$

• yep this is probably the simplest construction – qwr Mar 30 at 5:32

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.

• 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? – ClaraGarcía Mar 30 at 4:55
• Yep, that makes sense. – user2825632 Mar 30 at 5:01