p-adic expansion of reciprocals

Question

I can't seem to find an explanation on how to find the p-adic expansion of a reciprocal, for example, the 5-adic expansion of $ \frac 1 {10}$. Would anyone be able to give me a general method for finding it/ work through an example? Cheers.

Answer 1

If you want to find the $5$-adic expansion of $\frac{1}{10}$, you really just need to find the $5$-adic expansion of $\frac{1}{2}$, and then multiply that by $\frac{1}{5}$ (which is already a $5$-adic expansion in $\Bbb Q_5$). Finding the $5$-adic expansion of $\frac{1}{2}$ isn't so bad, since $\frac{1}{2}\in\Bbb Z_5$.

Remember that we can define elements of $\Bbb Z_5$ as compatible systems modulo every power of $5$: say $\frac{1}{2} = \sum_{i = 0}^\infty a_i 5^i$. Then we have $1 = 2(\sum_{i = 0}^\infty a_i 5^i)$. So to find each of the $a_i$, we successively look mod powers of $5$. We have $1 \equiv 2a_0\pmod 5$, so that $a_0 = 3$. Looking mod $25$, we have $1\equiv 2(3 + 5a_1)\pmod{25}$, which simplifies to solving $0 \equiv 5 + 10 a_1 \pmod{25}$, or $0\equiv 1 + 2a_1\pmod{5}$. This gives $a_1 = 2$. If you keep repeating this process, you'll get the $5$-adic expansion for $\frac{1}{2}$.

In general to find a $p$-adic expansion of a rational number, you can factor out all negative powers of $p$ from your fraction to get a new $x$ which is in $\Bbb Z_p$, and then solve successive equations modulo powers of $p$ to determine the coefficients $a_i$ of $x$ (writing $x = \sum_{i = 0}^\infty a_i p^i$).

Answer 2

As Stahl points out, all you need is the expansion of $1/2$. I’m going to use standard $5$-ary expansion, so that $\dots dcba;$ stands for $a+b\cdot5+c\cdot5^2+d\cdot5^3+\cdots\,$. For negative powers of $5$, just put the digits to the right of the semicolon.

Now, when you subtract $1;$ from $0;$ you get $\dots4444;\,$, which you get by doing your standard subtraction (as you learned in elementary school), but in base $5$ instead of base $10$. The borrow-and-carry process takes you one place to the left each time you do it. And this is right, because $4+4\cdot5+4\cdot5^2+4\cdot5^3+\cdots$ is a convergent geometric series (ratio is the small number $5$), and the formula $a/(1-r)$ has $a=4$, $r-5$, and $4/(1-5)=-1$.

All the above was to persuade you that $-1$ has the $5$-adic expansion $\dots4444;\,$, and you see immediately that $-1/2$ has $5$-adic expansion $\dots2222;\,$. Add $1$, and get $\dots2223;\,$, the expansion of $1/2$. For $1/10$ you write $1/5$ as $0;1\,$, and multiply this by your $1/2$ to get the expansion $\dots2222;3\,$. If you must have an expansion as powers of $5$, that’s easily read off: $3\cdot5^{-1}+2+2\cdot5+2\cdot5^2+\cdots$.

You can actually make suitable modification to the long-division process you learned in elementary school to get the quotient of any two numbers written in $5$-ary but extending infinitely to the left, but it’s so much harder to explain the process in print than it is at a black (or white) board, that I won’t do it here. In most cases, you use other tricks, anyway.