Multiplicative version of the principle of Archimedes 
Any clear proof of the above theorem is greatly appreciated.
 A: Suppose first that $y\ge 1$. We prove that the set $S=\{x^n\}$, $n=0,1,2,\dots$ is unbounded. Suppose to the contrary that $S$ bounded. Let $b$ be the supremum. Then for some $n$, we have $x^n \gt \frac{b}{(x+1)/2}$. Then $x^{n+1}\gt b\frac{x}{(x+1)/2}\gt b$. This contradicts the fact that $b$ is an upper bound of $S$. 
Thus the set of positive integers $k$ such that $x^k\gt y$ is non-empty. Let $n$ be the smallest element of this set. Then $x^{n-1}\le y$, and we are finished.
For $0\lt y\lt 1$, the easiest way to proceed is to show that there is a non-negative  integer $m$ such that $x^m \lt \frac{1}{y}\le x^{m+1}$. The argument for this is almost the same as the previous one. 
A: The set $S= \{x^i\ \vert\ i \in \mathbb{Z}\}$ has a greatest lower bound $c$. It can be easily shown that $0\leq c$.
If $c$ is a lower bound of $S$, so is $xc$, and hence $xc \leq c$ and $(x-1)c\leq 0$. Since $x-1>0$ and $c$ is nonnegative, it follows that $c=0$.
Now $y>0=c$ so it is not a lower bound of $S$, and $x^{n} < y$ for some $n$ (*). 
Similarly $x^{n} < y^{-1}$ for some integer $n$, hence $y < x^{-n}$ for some $n$. (**)
Finally let $T$ be the set of integers $m$ for which $x^m > y$. $T$ is nonempty (**), bounded below (*), and hence has a least element $k$. This $k$ satisfies
$$x^{k-1} \leq y < x^{k},$$ 
as desired.
