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How can we write $1/3$ in $\mathbb{Z}_5$ as series? We can write it as $\frac{1}{5-2}=-\frac{1}{2-5}=-(2+2*5+2*5^2+\dots)$ but $-2$ are negative coefficients...Please explain if there is a flaw in my understanding.

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    $\begingroup$ What's wrong with having negative coefficients? You can always substitute $-2 = 3-5$ if it really bothers you that much. Incidentally, you got the geometric series formula wrong. $\endgroup$ – Hurkyl Apr 23 '14 at 8:59
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Here is another version:

$$\frac 13=2-\frac 53=2+\frac {40}{1-25}=2+5\cdot\frac 3{1-25}+5^2\cdot\frac 1{1-25}$$

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My handy $p$-adic expander gives the $5$-adic expansion of a third as $\cdots313132;$, where $\cdots dcba;$ means $a+bp+cp^2+dp^3+\cdots$. This notation is nice because it’s simply the $p$-ary expansion you know from elementary school (except that $10$ is not prime). This means that, $5$-adically, $1/3=2+3\cdot5+5^2+3\cdot5^3+5^4+\cdots$, in other words $1/3=1+16/25+16/25^2+\cdots$, and the geometric series part of this evaluates to $16/(1-25)=-2/3$. There you are.

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