Show that $\sup\{x\in\mathbb{Q}: x>0,x^2<2\}=\sqrt{2}$. Show that $\sup\{x\in\mathbb{Q}: x>0,x^2<2\}=\sqrt{2}$. It is problem 1.1 in Kaczor's "Problems in mathematical analysis 1".
I had big problems with this problem. And I even posted a question about the solution to this problem. But did not get answer. So here I write down my solution.

*

*This set is nonempty and bounded from above. Although it is not stated precisly the supremum must be found in real numbers. So this set has the supremum.


*Let $s=\sup\{x\in\mathbb{Q}: x>0,x^2<2\}$.


*If so then for ANY $\epsilon>0$ we must have $$(s-\epsilon)^2<2$$ and $$(s+\epsilon)^2>2.$$


*For the first inequality $$s^2-2s\epsilon<s^2-2s\epsilon +\epsilon^2<2$$ and we get $$s^2\leq 2$$ because if $s^2>2$ then we can take $\epsilon=\frac{1}{2s}$ (contradiction).


*For the second inequality $$s^2+2s\epsilon +\epsilon^2>2$$ and we get $$s^2\geq 2$$ because in other case we can take $\epsilon=s$ (contradiction).
Is it correct? If not then where is a problem? I worry about two things. First, are my strict inequalities in step 3 correct? Second, if this set consists of rational numbers can I take ANY $\epsilon$  or I must consider only rational numbers (in step 3)?
EDIT: I did not write down it but my assumption for $\epsilon$ is $s\geq \epsilon >0.$
 A: In general, if $A \subseteq \Bbb{R}$ and $s \in \Bbb{R}$, the basic approach to proving $s = \sup A$ (the least upper bound of $A$), is to prove the following:

*

*$A \neq \emptyset$ (you need this because $\sup\,\emptyset$ is not defined: condition 3 below is not satisfiable if $A = \emptyset$).

*$\forall a \in A (s \ge a)$ (i.e., $s$ is an upper bound for $A$).

*$\forall x \in \Bbb{R}((\forall a \in A(x \ge a)) \Rightarrow s \le x)$ (i.e., $s$ is less than or equal to any other upper bound $x$ for $A$).

If $A$ is a downwards closed subset of $\Bbb{Q}_{>0}$, (i.e., if $A \subseteq \Bbb{Q_{>0}}$, and $\forall a \in A, x \in \Bbb{Q_{>0}}(x \le a \implies x \in A)$), then conditions 2 and 3 are equivalent to saying that, for any positive $\varepsilon$, $s - \varepsilon \le a$ for some $a \in A$ and $s + \varepsilon \not\in A$. This is the fact that you are using in your approach. I think you would do better to follow the basic approach I suggested above, because otherwise you have some extra justification to do.
In your example, steps 1 and 2 are straightforward using the fact that $a < b$ iff $a^2 < b^2$ for $a, b > 0$. 3 isn't difficult, but how to go about it depends on what you are expected to know already (but amounts to showing that if $x^2 < 2$, then there is $a \in \Bbb{Q}$ such that $x^2 < a^2 < 2$).
