Showing the rationals are not complete We consider the set 
$$A=\{x:x \in \mathbb{Q} \text{ and } x^2<2\} $$
We wish to show that this set has no largest member since this is all the rationals less than $\sqrt{2}$. I have been given the hint consider 
$$r^2<2, r>0 \qquad 0<\delta<1, \delta<\frac{2-r^2}{2r+1} $$
We wish to show that $(r+\delta)^2<2$.
I have tried expanding this in several ways and haven't been successful
$$(r+\delta)^2=r^2+2r\delta+\delta^2$$
We might factor in the two following ways
$$r(r+2\delta)+\delta^2 \qquad \text{ or } \qquad r^2+\delta(2r+\delta)$$
Any suggestions?
 A: Use the second factorization:
$$\begin{align}
(r+\delta)^2&=r^2+\delta(2r+\delta)\\\\
&<r^2+\delta(2r+1)\tag{$\delta<1$}\\\\
&<r^2+2-r^2\tag{$\delta<\tfrac{2-r^2}{2r+1}$}\\\\
&=2
\end{align}$$
A: `first of all the choise of the right set is important.:
So let A={$`x\in Q:x>0,x^2<2$}.
But this is not the only set that it does not have a Sup in Q,there is an infinite No of others..
Now since $A$ is a non empty,$1\in A$,bounded , $x\leq 2$ for all $x\in A$ subset of $Q$ it must have a Sup. call it $r$.
Assume now that this $r$  belongs to $Q$
.
Now our main efford is to show that this assumption will  lead us to a contradiction
By trichotomy we have :$r<\sqrt2$ or $r>\sqrt2$ or $r=\sqrt2$
Now by using the following two theorems:
a) if $r^2<2$,then there exists a $\delta>0$ such that $(r+\delta)^2<2$ and
b) if$2<r^2$,then there exists a $\delta>0$ such that $(r-\delta)^2>2$ and $r-\delta>0$
and the definition of the Sup$r$ we will conclude that the only possibility is:
...............................$r=\sqrt2$.....................................
But we know that $\sqrt2$ does not belong to $Q$ ,hence $r$ does not belong to $Q$ and $A$ has no SUP in $Q$.
Can you fill in the rest of the details ??
A: Assume there is $y\in \mathbb{Q}$ such that $y=sup\{x\in \mathbb{Q} : x^2<2\}$. then $y^2\leq 2$. Since the square root of $2$ is not rational $y<2$, i.e. $y\in \{x\in \mathbb{Q} : x^2<2\}$. To get a contradiction we now use the tip. Let $0< \delta \in \mathbb{Q}$ be such that $\delta <\frac{2-y^2}{2y+1}$. Then certainly $y+\delta \in \mathbb{Q}$ and furthermore $(y+\delta)^2<(y+\frac{2-y^2}{2y+1})^2=(\frac{y^2+y+2}{2y+1})^2 \leq 2$. Contradiction.
Edit: The last inequality is equivalent to $(y^2-2) (y^2+2 y-1)\leq 0$ which is true because $y^2<2$ and $y\geq 1$.
Edit 2: We want to prove that $(y^2+y+2)^2 \leq 2(2y+1)^2$ expanding both sides and cancelling yields $y^4+2y^3+2\leq 3y^2+4y$. Subtracting both sides by $3y^2+4y$ and factoring one sees that this is equivalent to $(y^2-2) (y^2+2 y-1)\leq 0$.
