Finding the bounds (if they exist) and the maximum and minimum of the group

Question

I'm trying to find the supremum and infimum of a group $C={\{\frac pq| p \in Z , 1 \le q \in N, {p^2}\le 2 {q^2}\}}$

I started by using the ${p^2} \le 2{q^2}$ and got to $-\sqrt2 \le \frac pq \le \sqrt2$

but now I'm trying to prove that $\sqrt2$ is the supremum, I know I need to assume there's an m $\lt$ $\sqrt2$ where m $\gt$ $\frac pq$ but I'm not sure how to do that

Answer 1

I think, in your second inequality, you mean $-\sqrt{2}\leq \frac{p}{q}\leq \sqrt{2}$, but then it is correct.

A nice way to continue the proof is using Archimedean property: assume you have some $m$ such that $\frac{p}{q}<m<\sqrt{2}$ for all $\frac{p}{q}\in C$. By the Archimedean property, there exists some fraction $\frac{k}{n}$ satisfying $m<\frac{k}{n}<\sqrt{2}$. Now, what can you say about that fraction, and how is it a contradiction?

Same tactic will work for the infimum.

Answer 2

To show that $\sqrt 2$ is the supremum, one has to show that for each $\varepsilon>0$, there is a rational number $\frac{p}{q}$ such that $$ \sqrt2-\varepsilon<\frac pq<\sqrt 2 $$ which is true since $\mathbb Q$ is dense in $\mathbb R$. You can use the same argument for $-\sqrt2$.