Decimal expansion no calculator Find the decimal expansion of $\frac{9}{130} $ by finding the periodical and non periodical parts. am i supposed to be looking for the expansion of $\frac{1}{(9)(130)} $? cause my theorems don't seem to apply unless the top equals one?

$$ \begin{align*} \frac{9}{130} &= \frac{1}{10} \frac{90}{130} \\\ &= \frac{1}{10} \left( 0 + \frac{9}{13} \right) \end{align*} $$

Since $\frac{9}{13} < 1$, this tells us that $\frac{9}{130} = 0.0\ldots$.

$$ \begin{align*} \frac{9}{13} &= \frac{1}{10} \frac{90}{13} \\\ &= \frac{1}{10} \left( 6 + \frac{12}{13} \right) \end{align*} $$

So $\frac{9}{13} = 0.06\ldots$.

$$ \begin{align*} \frac{12}{13} &= \frac{1}{10} \frac{120}{13} \\\ &= \frac{1}{10} \left( 9 + \frac{3}{13} \right) \end{align*} $$

So $\frac{9}{130} = 0.069\ldots$. Now you have to figure out when it becomes periodic.