Prove that for all $x\in\mathbb{Z}$ there exists $n\in\mathbb{N}$ such that $n|x|<2^n$ How can we prove using Archimedes that for all  $x\in\mathbb{Z}$ there exists $n\in\mathbb{N}$ such that $n|x|<2^n$?
 A: Hint: If the proposition is false, then there is necessarily some integer $x$ such that $2^n\le n|x|$ for all $n\in\Bbb N$. Equivalently, $|x|$ is an upper bound on the function $f:\Bbb N\to\Bbb R$ defined by $f(n)=\frac{2^n}{n}$.
Now, you should show that for every $m\in\Bbb N$ there is some $n\in\Bbb N$ such that $m\leq f(n)$, from which it will follow that $|x|$ would be an upper bound of $\Bbb N$. This is, of course, impossible by the Archimedean Property of the Reals. Hence the Proposition holds.

You should consider the following result, related to the (now deleted) answer from Shu Xiao Li. You can use it to get almost all the way to the result I suggested that you prove.
Lemma: For any positive integer $m\neq3$, we have $m^2\le2^m$.
Proof: This is simply a matter of verification for $m=1,2,4,5$, and when $m=5$, the inequality is strict.
If for some $m>4$ we have $m^2<2^m$, then $$\begin{align}(m+1)^2 &= m^2+2m+1\\ &< m^2+4m\\ &< 2m^2\\ &<2\cdot 2^m\\ &= 2^{m+1},\end{align}$$ so the Lemma holds by induction on $m$. $\Box$

It isn't really necessary to use the Archimedean Property for this result, as we can simply proceed by (eventual) induction on $|x|$ using the Lemma, since $x$ is an integer. If we allow $x$ to be an arbitrary real number, then the Archimedean Property becomes a great deal more convenient, and we can proceed as I outlined above.
