TITLE: Is it obvious that $q = \frac{2p+2}{p+2} > p$, how does one easily show $q > p$?
QUESTION [5 upvotes]: I was going through walter Rudin's first example when trying to show $A = \{p \in Q_+ : 0<p^2<2 \}$ has no smallest element (not $Q_+$ are positive rationals). In the first part of the proof he has the statement:
$$ q = p - \frac{p^2 - 2}{p+2} = \frac{2p+2}{p+2} $$
and the goal of it is to show $p < q$. For me its obvious just from the first statement that $q > p$ because $p \in A \implies p^2 < 2 \iff p^2 - 2 < 0 $. So it means $ - \frac{p^2 - 2}{p+2}$ has to be positive and therefore $q$ is a little bigger than $p$ since it starts off at $p$ and only increases. 
However, if I only had the second statement I would have no idea if $q > p$. Is there a way to see that the second statement is larger than $p$ also without appealing to the first statement? Or what is the purpose of the second statement on Rudin's example?

REPLY [3 votes]: Since $p^2<2$, we have $p^2+2p<2+2p$, so $p(p+2)<2p+2$, whence $p<\dfrac{2p+2}{p+2}$. (Note that $p+2>0$.)