TITLE: Define $S\equiv\{ x\in \mathbb{Q}\mid x^2<2\}$. Show that $\sup S=\sqrt{2} $.
QUESTION [6 upvotes]: Define $S\equiv\{ x\in \mathbb{Q}\mid x^2<2\}$. Show that $\sup S=\sqrt{2} $.


For this question, I think that I would use the completeness axiom. As $3$ is greater than $2$, so $S$ has a upper bound. To yield a contradiction, I think I need to find there exists an constant $c$ which is a least upper bound such that $b^2>2$, then there exists a number $r$ such that $b-r$ also in upper bound of $S$ which gives a contradiction since $b$ is the least upper bound of $S$. Then we can end the proof.

Does the idea right? If not, can anyone give a suggestion or a hit to write a better proof? Thanks.

REPLY [2 votes]: Hint: Here is an elementary argument. 
If you take any $x \in S$ and a rational number $r < 1$ such that $$0 < r < (2 - x^2)/(2x  + 1)$$
show that $x + r \in S$. This shows that $S$ has no maximum element. 
Consider $T = \{y \in \mathbb Q; y^2 > 2\}$ and similarly show that there is no minimum element in $T$. 
Conclusion: If $a = \sup S$ then $a^2 = 2$, $a > 0$. Use a simple contradiction.