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- Feb 14, 2012

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- Thread starter anemone
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Given that $R_4$ and $R_5$ are rational, it follows that $R_8$ and $R_{10}$ are rational.

Next, $R_8R_2 = R_{10} + R_6$ and $R_4R_2 = R_6 + R_2$. So $R_8R_2 = R_{10} + (R_4-1)R_2$ and therefore $R_2 = \dfrac{R_{10}}{R_8 - R_4 + 1}$. Hence $R_2$ is rational, and so is $R_6 = (R_4-1)R_2$.

Finally, $R_5R_1 = R_6 + R_4$, so that $R_1 = \dfrac{R_6+R_4}{R_5}$, which is rational.