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

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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