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?

Use the second factorization:
$$\begin{align}
(r+\delta)^2&=r^2+\delta(2r+\delta)\\\\
&<r^2+\delta(2r+1)\tag{$\delta<1$}\\\\
&<r^2+2-r^2\tag{$\delta<\tfrac{2-r^2}{2r+1}$}\\\\
&=2
\end{align}$$
