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I want to calculate $\frac{\overline{2}}{\overline{3}}$, $\frac{\overline{1}}{\overline{9}}$, $\frac{\overline{32}}{\overline{9}}$ in $\mathbb{Z}_7$.

About first example my book says: $\overline{3}*\overline{5}=\overline{15}=\overline{1}$, then $\frac{\overline{1}}{\overline{3}}=\overline{5}$; and finally $\frac{\overline{2}}{\overline{3}}=\overline{10}=\overline{3}$. It's pretty clear, except first step? From where 5 and 3 came out?

Regards

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    The first step is showing that the inverse of $\bar 3$ is $\bar 5$. You know that such a thing exists and is unique, but it's nice to have it written as the class of some concrete integer.2012-02-02
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    $3$ came from the fact that you are trying to calculate $\frac{\overline{2}}{\overline{3}} = \overline{2}\times\frac{1}{\overline{3}}$, so you are looking for a solution to $\overline{3}x = \overline{1}$ (in order to know what $\frac{1}{\overline{3}}$ is). As to where $5$ came from, possibly inspection, or you can express $1=\gcd(3,7)$ as a linear combination of $3$ and $7$; this quickly gives you $1 = 3\times 5 + 7\times(-2)$, so the $x$ you want is $\overline{5}$.2012-02-02
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    $\mathbb{Z}_7$ is tiny. To find the inverse of something, try stuff until you find something that works, though there are shortcuts, barely needed. What is the inverse of 9 aka 2? Can you think of something which when multiplied by $2$ gives $1$ more than a multiple of $7$? Surely $4$ jumps to mind. (Now you know the inverse of $2$ for any $\mathbb{Z}_n$ with $n$ odd. Others are less simple for $n$ big.)2012-02-02

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