I have no idea how to do this question. Can anyone give any suggestions?
Given a rational sequence $a: \mathbb{N} \rightarrow \mathbb{Q}$, we say that $\underline{a}$ converges to $l$ (here $l$ is a rational number) p-adically", when for any $\epsilon > 0$, there exists an $N>0$, such that, if $n>N$, then $|a_n - l|_p<\epsilon$. Show that the rational sequence, $a_n = 2^n$, converges to 0, in the sense of $2$-adic absolute value. Thus, we would say that, $\lim_{n \rightarrow \infty} 2^n=0$.
We have two definitions involved this question.
First, the notion of p-adic absolute value defined by $|x|_p=p^{-v_p(x)}$ for $x\in \mathbb{Q}$ where $v_p$ is the p-adic valuation and also the notion of convergence with this absolute value. This is the convergence associated to the metric distance $d_p(x,y)=|x-y|_p$. In other words, $x_n\rightarrow x$ iff $|x_n-x|_p\rightarrow 0$ as a sequence of real numbers.
Now to prove that $2^n\rightarrow 0$ with this notion of convergence you just have to prove that $|2^n-0|_2\rightarrow 0$ as a sequence of real numbers. But this is easy because $|2^n|_2=2^{-v_2{2^n}}=2^{-n}$.