About 2-adic representation of integers

Question

How would I express -3 in 2-adic representation? Is it just revercimal calculation of binary expression of -3?

like: -3 = -11 in binary, so using revercimal, -11. in binary = 01. ?

Answer 1

Just use the definition of those numbers.

For example, $-3$ is the number such that $(-3)+3=0$. Since $0=\sum_{n=0}^\infty 0\cdot2^n$ and $3=1\cdot2^0+1\cdot2^1+\sum_{n=2}^\infty 0\cdot2^n$, you have $-3=1\cdot2^0+0\cdot2^1+\sum_{n=2}^\infty 1\cdot2^n$.

I leave you the details.

Answer 2

If $n \in \Bbb{Z}_p$, then we have

$$-n = \lim_{k\to\infty} (p^k - n) = \lim_{k\to\infty} \big( (p^k-1) - (n-1) \big). $$

If we write $n-1 = \sum_{k=0}^{\infty} a_k p^k$ with $a_k \in \{0, \cdots, p-1\}$, then this observation allows us to compute the $p$-adic representation of $-n$ as the (diminished) radix complement of $n-1$:

$$-n = \sum_{k=0}^{\infty}(p-1-a_k)p^k. $$

Example 1. If $n = 3$, then $n-1 = 2 = \cdots00010_{(2)}$ and hence $-3 = \cdots11101_{(2)}$.

Example 2. In $\Bbb{Z}_7$, one of the square root $n$ of $2$ has the following expansion

$$n = \cdots6421216213_{(7)}.$$

Then $n-1 = \cdots6421216212_{(7)}$ and hence the other square root $-n$ can be expanded as

$$ -n = \cdots0245450454_{(7)}. $$