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

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.

• – dxiv Dec 14 '17 at 0:43
• @dxiv Thanks, but I wanted to prove it differently using the $\sqrt{2}$ approach. – philmcole Dec 14 '17 at 0:46
• That's fine as long as you are OK with pre-assuming knowledge of $\mathbb{R}$. However, the result is an intrinsic property of $\mathbb{Q}\,$, and can be proved without reference to $\mathbb{R}\,$, as one of the linked answers does. – dxiv Dec 14 '17 at 0:51

## 2 Answers

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

• How can I formalize this argument? Something like: $a_n$ is $\sqrt{2}$ up to the $n$-th decimal place and $b_n$ is $a_n$ but $+1$ on the last digit? Is there also a way to write the endpoints actually using "$\sqrt{2}$"? – philmcole Dec 14 '17 at 0:50
• Sort of. $a_n=10^{-n}\lfloor\sqrt{2}\cdot 10^n\rfloor$, $b_n=a_n+10^{-n}=10^{-n}\lceil\sqrt{2}\cdot 10^n\rceil$ – user491874 Dec 14 '17 at 0:54

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.