Let $\epsilon \in \mathbb{Q}, \epsilon>0$ and $a\in\mathbb {Q}, a>1$, prove that exists $n_1,n_2 \in\mathbb {N} $ such that $$a^{-n_2}<\epsilon <a^{n_1}$$.
I saw someone using euclid's division somewhere to prove this but I'm completely lost...
note: You are only allowed to use rationals, therefore things valid for real numbers like limits can't be used.
If $\varepsilon=1$, then it's obvious that $a^{-1}<\varepsilon<a$.
Suppose $\varepsilon>1$. Set $a=1+t$; by Bernoulli's inequality (which just requires an easy induction) $$ a^n=(1+t)^n>1+nt=1+n(a-1) $$ for $n>0$, so you just need to find $n_1$ such that $$ \varepsilon<1+n_1(a-1) $$ and this means $$ n>\frac{\varepsilon-1}{a-1} $$ which is surely possible. Can you find $n_2$?
If $0<\varepsilon<1$< then $\varepsilon^{-1}>1$ and you can use the previous case.