Suppose $ A= \lbrace x\in \mathbb{Q}: x>0, x^2 < 2 \rbrace$. Show that $s=\sup{A}=\sqrt{2}$.
Here is the proof I came up with (after of course proving that $A$ has a least upper bound). Where am I going wrong here? Am I making unwarranted assumptions? If my proof makes sense, why is Kaczor’s much more involved? I am an amateur self studying analysis so any help appreciated…
Suppose $s < \sqrt{2}$. By density of rational numbers in the reals, there exists a rational number $a$ such that $s < a < \sqrt{2} \rightarrow s^2 < a^2 < 2 \rightarrow s \neq \sup{A}$. I would give a similar argument for the reverse inequality to prove equality.
For reference, here is the proof from Kaczor’s analysis problem book:
In response to fleablood’s comment about the line in Kaczor’s solution, I think the reason is:
$0 < [(n+1)s]-(n+1)s +1 \leq 1 \rightarrow s < w$
In general, I find these foundational type problems difficult as it’s unclear what’s assumed and what isn’t. I am guessing the issue here is I can’t assume the density of the rationals (but then again the author invokes a limiting argument before that concept is introduced…).
Thanks.