TITLE: Showing that $\mathbb{Q}$ is not complete
QUESTION [8 upvotes]: 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?

REPLY [5 votes]: 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 $$x^2-2y^2=-1$$