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?

`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 ??
