Showing that $\mathbb{Q}$ is not complete 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?
 A: This might be slightly long winded, but it is a nice exercise in my opinion. Verify the following
$(1)$ Let $r=\dfrac pq $ be a positive rational such that $$r^2<2 $$
Then $$r'=\frac{3r+4}{2r+3}$$ is larger than $r$; and $$r'^2<2$$
$(2)$ Let $r=\dfrac pq $ be a positive rational such that $$r^2>2 $$
Then $$r'=\frac{3r+4}{2r+3}$$ is smaller than $r$; and $$r'^2>2$$
$(3)$ Prove that if we take any rational $r>1$ such that $r^2<2$ (or $>2$) then the sequence $$\langle r_n\rangle $$ defined by $$\begin{cases} r_0=r\\r_{n+1}=r_n^\prime \end{cases}$$
is such that $r_n^2\to 2$ monotonically from below (or above).  
Hint Prove that if we write $$r=p/q$$ with $p<q$ so $$r_n=\frac{p_n}{q_n}$$ with $$p_{n+1}={3p_{n}+4q_{n}}\;\;\; ; q_{n+1}={2p_n+3q_n}$$ the sequence of integers $p_n>q_n$ is such that for each $n$, $$p_n^2-2q_n^2=-1$$ and $q_{n+1}>5^nq_0$
This in turn tells you each approximation gives you one extra decimal place of $\sqrt 2$, and gives you an infinite set of pairs that solve the Diophantine equation $$x^2-2y^2=-1$$
A: Your first clause ("Since $\alpha\le x$ for all $x\in A$") is incorrect. The inequality needs to go the other way. Now, you can say "Since $\alpha$ is the least upper bound of $A$, then for any...."
Next, you say that $1\le\alpha\le2$, which is true, but you should be cautious about saying "clearly" in a proof unless you can actually prove the putatively clear statement. (I assume that you can.)
Now, when you say "choose $\epsilon=\dfrac{\epsilon^2}{2\alpha}$" you're really just saying "choose $\epsilon=0$ or $\epsilon=2\alpha$". The former can't be what you mean (since you're looking for $\epsilon>0$), so just say "choose $\epsilon=2\alpha$." Then it's true that $\epsilon=\dfrac{\epsilon^2}{2\alpha}$ and $\epsilon>0.$ Then we can conclude that there is some $x'\in A$ such that $$x'>\alpha-\epsilon=-\alpha.$$ Since $\alpha>0$, we can't draw any further conclusion from this alone regarding which of $x'^2$ and $\alpha^2$ is larger. Your approach breaks down completely at this point. In order to derive a contradiction from $x'>\alpha-\epsilon$ and $\alpha^2>2$, you'd need to know that $x'+\epsilon\in A$, so that $2>(x'+\epsilon)^2>\alpha^2>2$. You'll need to choose your $\epsilon>0$ and your resulting $x'\in A$ more carefully to guarantee that.
A: Hint: Show that if $\alpha$ is an upper bound, then so is $\frac{\alpha^2+2}{2\alpha}$ and show that if $\alpha^2>2$ this is a lower upper bound.
Details.
First, if $u^2<2$ and $u\leq\alpha$ then $$\frac{\alpha^2+2}{2\alpha}-u = \frac{\alpha^2-2\alpha u + 2}{2\alpha}=\frac{(\alpha-u)^2 + 2-u^2}{2\alpha}>0$$
Second, if $\alpha>0$ an $2<\alpha^2$, then $2+\alpha^2<2\alpha^2$ and hence $$\frac{\alpha^2+2}{2\alpha}<\alpha$$
