(H.W question) In Mathemaical Analysis of Rudin example 1.1 Pg 2 The author went on and proved that 
1) There exists no rational $p$ such that $p^2=2$ 
2) He defined two sets $A$ and $B$ such that if $p\in A$ then $p^2 <2$ and if $p\in B$ then $p^2>2$ and then constructed $$q=p - \frac{p^2-2}{p+2}$$ and $$q^2-2=\frac{2(p^2-2)}{(p+2)^2}$$ 
Then $A$ contains no largest element and $B$ contains no smallest element
Finally in the remark he said that the purpose of the above exercise was to show that the rational number system has certain gaps.
So my question is how is 2) used in arriving at this conclusion? I basically didn't understand the purpose of 2) in this discussion.
 A: Rudin doesn't give a #*@$ whether there is a rational square root of 2 or not. What he's trying to show is that you can divide all the rational numbers into two sets that exhaust the rationals; that one set can have every element larger than every element in the other yet it is possible to have no limits to either set; you can take infinitely many larger numbers in one set and never have a largest and you can take infinitely many for the other set and never have a lowest one yet the two sets can be infinitely close together approaching ... something... but that something not being anything that can exist in the rationals.
Thus we can say the rational are incomplete but that these incomplete "gaps" are infinitesimally small.
He shows this by quickly proving $(m/n)^2 = 2$ is impossible.  Ho-hum.  Too bad how sad. Who cares.  But once that has been shown comes the important part.
You can divide the rationals entirely into two sets.  All the $m/n$ where $(m/n)^2 > 2$ and all the $m/n$ where $(m/n)^2 < 2$.  And both sets are infinite and all can get infinitely close to the other set but neither set has a highest (or lowest) element despite clearly there being a "wall" they can't go past, but can never actually hit, either.
The only reason he cares that there is no rational square root of 2 is that of these two sets the option $(m/n)^2 = 2$ is simply not available.
That's the point.
A: Suppose $A$ contained a largest element $p$. Now we have $p^2-2<0$. Then we must have $q^2-2=\frac{2(p^2-2)}{(p+2)^2}<0$ as the numerator on the right hand side is negative and the corresponding denominator positive. 
So $q\in A$. However $q=p-(\mbox{negative quantity})$ i.e. $q=p+\mbox{positive quantity}$ by definition. So $q>p$ contradicting the $p$ was the largest element. So $A$ has no largest element. A similar argument works for $B$.
