I think I was wrong... Can you find it out and teach me how to prove (the title)?
If $c>0$ a real number. Define $E=\{x\in\mathbb R\mid x^2<c\}$. $E$ is nonempty because $0\in E$ (positivity axiom). Call $x_0=\sup E$. $E$ is also bound from above because $(c+1)^2>c>x^2 \Rightarrow c+1>|x|$. How to show $x_0^2=c$? The following is my reasoning.
Define a new set $U=\{y\in\mathbb R\mid y^2\ge c\}$, so all $y$ is an upper bound of $E$. $x_0\in U$ due to the completeness axiom. Therefore, $x_0^2\ge c$. We can exclude $x_0^2>c$ in the following way:
Suppose $x_0^2>c$. $\exists \epsilon>0$ s.t. $x_0^2-2x_0\epsilon+\epsilon^2>c$. So $x_0>x_0-\epsilon>x, \forall x\in E$, which is impossible because $x_0=\sup E$.
Is this correct?
There is always a trick to deal with this kind of problem. If we want to show $x_0^2=c$, we can first show $x_0^2\ge c$, as I did (if I was correct), then show $x_0^2\le c$. But in this problem, it seems to me that it is difficult to show the other half. Is there anybody who can use this trick?
You have to show that for small $\varepsilon$, $\sqrt n - \varepsilon$ is a member of the set and that no element greater than $\sqrt n$ is in the set. The second part should be easy.
For the first part, $(\sqrt n - \varepsilon)^2 = n +\varepsilon^2 -2\sqrt n \varepsilon = \sqrt n (\sqrt n + \frac{\varepsilon ^2}{\sqrt n} - 2 \varepsilon)$
For ultimately small $\varepsilon$, the second factor will be less than the first, implying the membership in the set. For any given $\varepsilon$, you can argue that there is an integer $k$ for which $1/k < \varepsilon$. Replace $\varepsilon$ with 1/k and I think you'll see it immediately.