Let $A=\{x\in \mathbb{Q}\mid x^2<2\}$. Then $A$ has no least upper bound in $\mathbb{Q}$.
The proof I read is: ''If it did, say $r$, then it can be shown that $r^2=2$. But we know that there is no such rational number.''
Question: How can it be shown that $r^2=2$ in the first sentence?
If $B$ is a set, and if $\sup B=k$, wouldn't it mean that $y\leq k$ for all $y\in B$ and there is no upper bound lesser than $k$? Or, is it using the fact that: ''There exists a real number $x$, such that $x^2=2$'', then replace $x$ by $r$, as supremum is unique?