Talk:Existence of Square Roots of Positive Real Number

Possible mistake
There is a comment on Math Stack Exchange stating that a part of this proof is incorrect. Basically, $u + \frac 1 n$ may not be rational. Is anyone willing to check it out?--Julius (talk) 11:13, 7 October 2022 (UTC)


 * Yes I see now, that last line $u + \dfrac 1 n \in S$ is a mistake. But it's simple to fix, we just need to invoke the fact that because $u + \dfrac 1 n < r$ then $\exists m \in S$ such that $u + \dfrac 1 n < m < r$ which can be done from Between two Real Numbers exists Rational Number. Oh, and there's also a couple of instances of $4 u$ which I think may need to be $2 u$, but I haven't studied it carefully enough to be sure about that.


 * I don't have the time to do this immediately (shouldn't really be doing this, I'm supposed to be debugging an app right now) so if anyone else is up for it, let them. --prime mover (talk) 11:30, 7 October 2022 (UTC)


 * I do not get why you are considering $\Q$ at all. I could review it later. --Usagiop (talk) 12:46, 7 October 2022 (UTC)


 * I have no idea why it was $S = \set {x \in \Q: x^2 < r}$ before. With $S = \set {x \in \R: x^2 < r}$ we do not have such problem. Now the proof should be correct. --Usagiop (talk) 19:16, 7 October 2022 (UTC)


 * I think I might have been getting confused with something I'd read about Dedekind cuts and the proof of the existence of the square root of $2$ from the existence of the rational numbers. Or something. --prime mover (talk) 21:18, 7 October 2022 (UTC)


 * OK, not wrong but just needlessly complex, assuming I haven't overlooked anything. --Usagiop (talk) 22:31, 7 October 2022 (UTC)