Prove that nested interval theorem doesn't hold for $\mathbb{Q}$

Question

I want to prove that the nested interval theorem doesn't hold for the rational numbers.

For this I want to find a closed and rational interval $I_n = [a_n,b_n] \cap \mathbb{Q}$, with $a_n,b_n \in \mathbb{Q}$ so that $\bigcap_{n=1}^\infty I_n = \{\sqrt{2}\}$. This would show the claim, because $\{\sqrt{2}\}$ is empty in $\mathbb{Q}$.

But I have difficulties finding such a closed rational interval with rational endpoints. I wanted to chose $I_n = [\sqrt{2}-\frac{1}{n},\sqrt{2}+\frac{1}{n}] \cap \mathbb{Q}$ but then the endpoints are not rational, right? And also the interval needs to be closed.

Answer 1

$[1,2] $, $[1.4,1.5] $, $[1.41,1.42] $, $[1.414,1.415] $, $[1.4142,1.4143] $ etc.

Answer 2

Use continued fractions (https://en.wikipedia.org/wiki/Square_root_of_2#Continued_fraction_representation):

$[1,1+\frac{1}{2}]$, $[1+\frac{1}{2+\frac{1}{2}},1+\frac{1}{2}]$, $[1+\frac{1}{2+\frac{1}{2}},1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}]$ etc.

$a_n$'s and $b_n$'s are the convergents of the continued fraction expansion of $\sqrt{2}$ and in each step we replace exactly one of them by the next convergent while keeping the other one unchanged.