Show that there is no least upper bound for $A=\{x: x^2<2\}$ in $\mathbb{Q}$.
Suppose $\alpha \in \mathbb{Q}$ is the least upper bound of $A$. Then either $\alpha^2 < 2$ or $\alpha^2 > 2$. I know the proof of the former case and why we can't have equality. So here is my attempt at the latter case ($\alpha^2 > 2$):
Since $\alpha \le x$ for all $x \in A$, for any $\epsilon > 0$ there is an $x' \in A$ such that $x'> \alpha - \epsilon$. Clearly $1\le\alpha\le 2$, so choose $\epsilon=\dfrac{\epsilon^2}{2\alpha}<1$. Now we have $x'^2>\alpha^2 - \epsilon^2 + \epsilon^2 = \alpha^2>2$. Contradiction.
Is that acceptable?
