Dedekind cuts in Rudin's PMA I'm working on Appendix to chapter I of Rudin's Principles of mathematical analysis and I have the following problem: Given a positive cut $\alpha$ and a rational $x>1,$ how can I prove that there exists an integer $n$ such that $x^n\in\alpha$ and $x^{n+1}\notin\alpha$ (I used this intuitive fact to prove that $\alpha\cdot\alpha=1^*,$ which is left to the reader)?. I was thinking on consider the set $N$ of all integers $n$ such that $x^n\in\alpha$ (or the set of all $n$ such that $x^n\notin\alpha$) and then prove, somehow using the Archimedean property  and the Well ordering principle, sorry if it is too obvious but I have no idea how to proceed..
Any help is really appreciated!
 A: Even though I asked (and solved) this question a lot of time ago, here is a solution for future references.
Case I. $\alpha=1^*.$ This case is trivial since $x^{-1}\in\alpha$ and $x^0\notin\alpha.$
Case II. $\alpha\subsetneq1^*.$ Let $a\in\alpha\cap\mathbb Q_{>0}$ and let $m$ and $n$ be positive integers such that $n>a^{-1}$ and $x>1+(1/m).$ Then $$x^{mn}>\left(1+\dfrac{1}{m}\right)^{mn}\geqslant1+n>a^{-1}$$ and it follows that there exists some positive integer $t$ such that $x^{-t}\in\alpha$ and $x^{-t+1}\notin\alpha.$
Case III. $1^*\subsetneq\alpha.$ Let $a\in\mathbb Q\setminus\alpha,$ let $n$ be a positive integer such that $n>a$ and let $m$ be as in the first case. Then $$x^{mn}>\left(1+\dfrac{1}{m}\right)^{mn}\geqslant1+n>a$$ and it follows that there is some positive integer $t$ such that $x^t\notin\alpha$ and $x^{t-1}\in\alpha.$
Note that the solution that Tim suggests is incomplete (in fact, his "hint" was not helpful at all). As an example, one could argue that the floor function exists as follows.
Let $x$ be a real number. Since $\mathbb Z_{>0}$ is not bounded above, there is some $n\in\mathbb Z_{>0}$ such that $n>x.$ Therefore the set of all positive integers $t$ such that $t>x$ has a least element $s.$ It follows that $s-1\leqslant x<s.$ What happens if $x<0$?
A: Hint: Let $N=\{n\in\Bbb N: x^n\notin\alpha\}$.  This is non-empty, because $x>1$, so let $n+1=\min N$.
