We are all familiar with the fact that every $q\in\mathbf{Q}$ gets represented in $\mathbf{R}$ as a repeating decimal, or whatever your favourite base is. (Here I count terminating decimals as decimals which eventually repeat $000...$.) The converse is of course true as well: repeating decimals are rationals.

Is this true in $\mathbf{Q}_p$ as well? That is, are the rationals precisely those $p$-adics which are eventually repeating (in the other direction, of course)? One direction, that repeating $p$-adics are rational, is pretty obvious: if $a\in p^{-t}\mathbf{Z}$ and $b\in\mathbf{Z}$ then

\[

a + b p^r + b p^{r+s} + b p^{r+2s} + \cdots = a + b \frac{p^r}{1 - p^s}

\]

is rational. What about the converse?

The converse seems trickier. How again did we do it in $\mathbf{R}$? I don't even remember: it's one of those things that we know so fundamentally (until very recently, $\mathbf{Q}$ was almost

*defined*in my brain as the reals which eventually repeat) that we forget how to prove it.

Who cares how to prove it? It is true, and it says, in base $p$, that if $q\in\mathbf{Q}$ then there exists $a\in p^{-t}\mathbf{Z}$ and $b\in\mathbf{Z}$ such that

\[

q = a + b p^r + b p^{r-s} + b p^{r-2s} + \cdots = a + b \frac{p^r}{1 - p^{-s}}.

\]

But hey, this implies that

\[

q = a - b \frac{p^{r+s}}{1 - p^s},

\]

which we already know is a repeating $p$-adic.

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