2-adic expansion of (2/3)

Question

I have been asked in an assignment to compute the 2-adic expansion of (2/3). It just doesn't seem to work for me though. In our definition of a p-adic expansion we have $x= \sum_{n=0}^{\infty}a_np^n$ with $0 \leq a_0<p$.

So I use the same method as for similar questions which seems to work fine. This involves solving each congruence $\frac{2}{3} = \sum_{n=0}^{\infty}a_np^n \ (mod \ p^N)$ for $N=1,2,3,4,5$ and then we are supposed to guess the pattern and show it by summing the series.

Doing this, I get $\frac{2}{3} = 2 + 2^2 + 2^4 + 2^5 + ... \\ = (2+2^2)(1+2^3+2^6+...) \\ = 6(\frac{1}{1-2^3}) \\ = \frac{6}{-7}.$

Which is just completely wrong?

An example of me solving a congruence: $\frac{2}{3} \equiv a_0 \ (mod \ 2)$ so $0 \equiv 2 \equiv 3a_0 \ (mod \ 2)$ so $a_0 = 0$.

Answer 1

If I am not mistaken you can also do the following:

$\frac{2}{3}=1+\frac{1}{-3}$ where

$\frac{1}{-3}=\frac{1}{1-4}=\frac{1}{1-2^2}=\sum_{n=0}^{\infty}2^{2n}$ (geometric series expansion).

This yields the whole expansion of $\frac{2}{3}$.

Answer 2

I get $2^1+2^2+2^4+2^6+\ldots$, with no $2^5$. In the $2^4$ place, you have $3\times1+1=4=0$ carry $2$, so you need $0\times 2^5$ to give $0$ carry $1$