Proving a property of Rationals using definition of Rationals (as an ordered field) I'm trying to find more behind the theorem:
Given $c,q \in \mathbb{Q}$ and $c,q > 0$ and $n \in \mathbb{N}$. If $q^n > c$ then $\exists \epsilon \in \mathbb{Q}$ with $0 < \epsilon < 1 $ such that $(q-\epsilon)^n > c$.
Is this a common theorem? Does it have a name? Is there a proof lying around. I cannot for the life of me find how to prove this on my own, so even if you have any tips on how to approach this, I'd greatly appreciate it. Thanks.
 A: It's not important enough to have a name. It can be proved easily enough
using analysis: it's true for small positive real $\epsilon$, and so
by the Archimdean property it's true for some $\epsilon = 1/N$, but you want to avoid analysis.
The inequality is equivalent to $(1-\epsilon/q)^n>cq^{-n}=1-d$
where $0<d<1$. By Bernoulli's inequality it will hold provided that
$n\epsilon/q <d$, that is if $\epsilon <dq/n$.
(One form of) Bernoulli's inequality states that
$$(1-x)^n\ge 1-nx$$
for $0<x<1$ and $n\in\Bbb N$. It's readily proved by induction
on $n$.
A: Just note that $$q^n-(q-h) ^n=h\{q^{n-1}+q^{n-2}(q-h)+\dots\} $$ and if $0<h<q$ then the RHS of the above equation is less than $nhq^{n-1}$ and we just need to ensure this to be less than $q^n-c$ and our job is done. In concrete terms any $h$ with $$0<h<\min\left(1,q,\frac{q^n-c}{nq^{n-1}}\right)$$ works. An example is $c=1000,q=2,n=10$ and $(q^n-c) /nq^{n-1}=3/640$ and hence $h=1/1000=0.001$ works fine and you can check that $1.999^{10}>1000$. 
The above uses $h$ instead of $\epsilon$ to save typing effort. 
