Show that $A = \{p\in\mathbb{ Q}^+\mid p^2<2\}$ contains no largest number and $B= \{p\in\mathbb {Q}^+\mid p^2>2\}$ contains no smallest number [duplicate]

Solution 1:

This is a "good old-fashioned" textbook which tells you the answer and proves it is true but never gives you a clue as to where the answer came from in the first place. It might have been something like this.$\def\eps{\varepsilon}$

We have a positive rational $p$ such that $p^2<2$; we want a rational bigger than $p$, such that its square is still less than $2$. So we want, let's say, $$(p+\eps)^2<2$$ with $\eps$ a positive rational. Expanding and rearranging, $$\eps^2+2p\eps<2-p^2\ .$$ Now we could solve this to find $\eps$, but it would be rather messy. More importantly, the formula would contain a square root, so it would be very hard to be sure that $\eps$ was rational. So we have to be a bit more careful. Write the above as $$\eps(2p+\eps)<2-p^2\ ;\tag{$*$}$$ we can get rid of the second $\eps$ by noting that we will certainly need $p+\eps<2$. So if $$\eps(p+2)\le2-p^2\ ,$$ then $(*)$ will be true, because its LHS is less than this LHS. Thus we can take $$\eps=\frac{2-p^2}{p+2}$$ and hence $$p+\eps=p+\frac{2-p^2}{p+2}\ ,$$ which is Rudin's $q$.

Solution 2:

Equation 2 is the definition of $q$. Rudin (or somebody) had the inspiration that this $q$ would work. Adding some steps:$$\begin{align} q &= p - \frac{(p^2 - 2)}{p + 2} &(2)\\ &=\frac {p(p+2)}{p+2}-\frac{(p^2 - 2)}{p + 2}\\ &=\frac{p^2+2p-p^2+2}{p+2}\\ &=\frac{2p + 2}{p + 2} &(3)\\ \Rightarrow q^2 - 2 &=\frac{4p^2+8p+4}{p^2+4p+4}-\frac{2p^2+8p+8}{p^2+4p+4}\\ &=\frac{2(p^2 - 2)}{(p + 2)^2} &(4) \end{align}$$