$p$(ain)-adic number sequence

Question

I am trying to figure out how $p$-adic numbers work and currently am having trouble wrapping my head around how they work, so I made a pun! HAH!

Jokes aside, I am working on this question

Show that the sequence $(3,34,334,3334,.....)$ is equal to $2/3$ in $\hat{\mathbb{Z}}_5$

I assume they mean it converges to $2/3$ in the $5$-adic integers by the question. My big issue is that I am struggling to properly attack it. One question immediately comes to mind is: does $334$ mean

$334=3\cdot 5^0+3\cdot 5^1+4\cdot 5^2$

as all $p$-adic numbers can be written as such, or what does it exactly mean? If so, how would I go on exactly demonstrating this? Their initial peculiarity, especially with increased number decreases distance is throwing me off quite a bit.

Answer 1

The sequence $(a_n)=(4,34,334,3334,\ldots)$ satisfies $3a_n-2\to 0$ for $n\to \infty$ in the $5$-adic topology. You can "see" this by considering \begin{align*} 3\cdot 4 & = 12 \\ 3\cdot 34 & = 102 \\ 3\cdot 334 & = 1002 \\ 3\cdot 3334 & = 10002 \end{align*} etc. Hence for the metric $d$ we have $d(a_m,a_n)=\frac{1}{5^n}$ for all $m>n$, so that $a_n$ is a Cauchy sequence, which converges to $2/3$.

Answer 2

Here, the notation is in the usual base $10$, so the number $a_n = 33\ldots34$ simply denotes

$$a_n = 3 \times 10^n + 3 \times 10^{n-1} + \ldots + 3 \times 10 + 4 = 1 + 3\sum_{k = 0}^n 10^k = 1 + \frac{10^{n+1}-1}{3}$$

Now, in the $5$-adic topology, the sequence $(10^n)_{n \ge 0}$ goes to $0$ at infinity because $\left|10^n\right|_5 = \frac{1}{5^n} \longrightarrow 0$, so you get

$$\lim_{n \to \infty} a_n = 1 + \frac{0-1}{3} = \frac{2}{3}$$