Showing the rationals are not complete We consider the set 
$$A=\{x:x \in \mathbb{Q} \text{ and } x^2&lt;2\} $$
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 
$$r^2&lt;2, r&gt;0 \qquad 0&lt;\delta&lt;1, \delta&lt;\frac{2-r^2}{2r+1} $$
We wish to show that $(r+\delta)^2&lt;2$.

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:
$$\begin{align}
(r+\delta)^2&amp;=r^2+\delta(2r+\delta)\\\\
&amp;&lt;r^2+\delta(2r+1)\tag{$\delta&lt;1$}\\\\
&amp;&lt;r^2+2-r^2\tag{$\delta&lt;\tfrac{2-r^2}{2r+1}$}\\\\
&amp;=2
\end{align}$$
Assume there is $y\in \mathbb{Q}$ such that $y=sup\{x\in \mathbb{Q} : x^2&lt;2\}$. then $y^2\leq 2$. Since the square root of $2$ is not rational $y&lt;2$, i.e. $y\in \{x\in \mathbb{Q} : x^2&lt;2\}$. To get a contradiction we now use the tip. Let $0&lt; \delta \in \mathbb{Q}$ be such that $\delta &lt;\frac{2-y^2}{2y+1}$. Then certainly $y+\delta \in \mathbb{Q}$ and furthermore $(y+\delta)^2&lt;(y+\frac{2-y^2}{2y+1})^2=(\frac{y^2+y+2}{2y+1})^2 \leq 2$. Contradiction.

Edit: The last inequality is equivalent to $(y^2-2) (y^2+2 y-1)\leq 0$ which is true because $y^2&lt;2$ 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 $(y^2-2) (y^2+2 y-1)\leq 0$.