1/p in p-adic number system?

Question

to expand $1/p$, I tried first letting $1/p = a+b*p+c*p^2+d*p^3+...$ and it is $1=a*p+b*p^2+...$ but I guess there's no way to make the equality hold.

it's somewhat similar to dividing by 0. is it possible? I'm new to p-adic theory, everything looks just confusing.

Answer 1

The elements of $\mathbf Q_p$ are kinds of Laurent series in $p$: $$\sum_{k\ge k_0}a_k\mkern1.5mu p^k,\quad(k_0\in \mathbf Z)$$ the coefficients $a_k$ being subject to the condition $\;0\le a_k<p$. Hence $\dfrac 1p$ is but… $\dfrac 1p$.

Answer 2

My own convention is to write $p$-adic numbers as $p$-ary expansions extending (potentially) infinitely to the left, and only finitely to the right of the radix point, which I like to write as a semicolon. So, $5$-adically, five is $10;$, twenty-eight is $103;$, and one twenty-fifth is $0;01$. For infinite expansions, $-1=\dots4444;$ and one tenth is $\dots2222;3\>$. In particular, $1/p$ will always have the expansion $0;1$, just as one tenth in decimal is $0.1$

Answer 3

Suppose there exists an $\alpha$ such that $\alpha = \frac{1}{p}$, then $\alpha p = 1$. However, multiplying any integer in $\mathbb Z_p$ and $p$ gives an integer ending in $0$, which is a contradiction. So $\frac{1}{p}$ does not exist in $\mathbb Z_p$.