Correctness of the proof that the set $\{x \in \mathbb{Q} : x>0 \text{ and } x^2>2\}$ does not have a smallest element 
Let $F=\{x \in \mathbb{Q} : x>0 \text{ and } x^2>2\}$.  I am asked to show that $F$ does not have a smallest element.  The hint is to simply prove the claim: 'If $p$ is a rational number in $F$ then $q=\frac{2p+2}{p+2}$ and satisfies $q < p$.  "

I need help editing my proof: see if it makes sense. It goes as follows:
Assume $p$ is a rational number in $F$.  This means $p$ is positive and $p>\sqrt2 $.  
Now consider $q=\frac{2p+2}{p+2}$.  It is positive and rational but to be part of set $F$, $q$ also needs to be greater than $\sqrt2$.  
So for $q>\sqrt2$, $q=\frac{2p+2}{p+2}>\sqrt2$.
After evaluating this expression, I get the inequality that $p>\sqrt2$ which it is, so $q$ belongs in $F$.
Now we need to show that $q$ is bounded above by $p$.  Assume by contradiction that $q=\frac{2p+2}{p+2} \geq p$.  After evaluating it, I get $p^{2} \leq 2$ which is a contradiction because in the beginning we assumed $p>\sqrt2$.  
Does this make sense?  If not, how should I correct it?
 A: I would avoid using the symbol $\sqrt{2}$ and stick to rational numbers only. I would also be more careful about proving the inequalities; how do you know that the manipulations that you did are reversible? Here's my version.
Suppose, towards a contradiction, that $F$ contains some smallest element, say $p$. Now consider the element:
$$
q = \frac{2p + 2}{p + 2}
$$
Notice that:


*

*$\boxed{q \in \mathbb Q}$: Since $p \in \mathbb Q$ and $p \neq -2$ and $\mathbb Q$ is closed under addition and multiplication and (nonzero) division, we know that $q \in \mathbb Q$.

*$\boxed{q > 0}$: This is immediate, since $2p + 2 > 2(0) + 2 = 2 > 0$ and $p + 2 > 0 + 2 = 2 > 0$.

*$\boxed{q^2 > 2}$: Observe that:
\begin{align*}
\left(\frac{2p + 2}{p + 2}\right)^2
&= \frac{4p^2 + 8p + 4}{p^2 + 4p + 4} \\
&= \frac{4p^2 + 8p + 4}{p^2 + 4p + 2 + 2} \\
&> \frac{4p^2 + 8p + 4}{p^2 + 4p + 2 + p^2} &\text{since } p^2 > 2\\
&= \frac{2(2p^2 + 4p + 2)}{2p^2 + 4p + 2} \\
&= 2
\end{align*}


Thus, we know that $q \in F$. But this is absurd! Notice that since:
\begin{align*}
\frac{2p + 2}{p + 2}
&< \frac{2p + p^2}{p + 2} &\text{since } p^2 > 2\\
&= \frac{p(p + 2)}{p + 2} \\
&= p
\end{align*}
it follows that $q < p$, which contradicts the minimality of $p$. Hence, we conclude that $F$ contains no smallest element, as desired. $~~\blacksquare$
A: I think you would have to be careful in your fifth paragraph  when you let $q > \sqrt{2}$. Asssuming you evaluated the inequality correctly, you showed that
$q > \sqrt{2}$ $\implies$ $p > \sqrt{2}$, but you really want to show the other way around. That is, you showed q may be in F when $p > \sqrt{2}$ but not that q must be in F.
I would like you to show the derivations of some of the inequalities. It would make our feedback more helpful.
