If you take the base-$b$ expansion of a rational number where $b$ is irrational, do you get a non-terminating sequence of digits (assuming you pick the right(?) digits)? More informally, do rational numbers look irrational in irrational bases?
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What is a base-$b$ expansion when $b$ is not a positive integer? – Hagen von Eitzen Oct 21 '18 at 09:38
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I assume something like base-$\phi$ https://en.wikipedia.org/wiki/Golden_ratio_base – Eyob Tsegaye Oct 21 '18 at 09:39
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Rational numbers in decimal or binary or similar cases can have non-terminating recurring representations, for example $\frac13 = 0.3333333\ldots_{10}= 0.\overline{3}_{10}$. Presumably you want to include these
Your golden ratio base link shows that there are irrational bases in which all rationals can be represented with terminating or recurring expressions. For example $2 = 10.01_{\phi}$ and $\frac12 = 0.\overline{010}_{\phi}$
There are other bases for which this would not be the case - in particular with transcendental bases (since such a representation would imply a polynomial equation)
Henry
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Haha I didn't even notice. But could you possibly prove the transcendental base fact? That sounds really cool and I want to understand it better – Eyob Tsegaye Oct 21 '18 at 10:01
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1@EyobTsegaye Suppose I had something like $x=12.1\overline{02}_e$. Then I would know that $x=e^1+2e^0+e^{-1}+2e^{-3}+2e^{-5}+2e^{-7} +\ldots = e+2+e^{-1}+2e^{-3}/(e^{2}-1)$ so $e^6-e^4+(2-x)e^5+e^4-(2-x)e^3-e^2+2 =0$ and if $x$ were rational this would make $e$ algebraic which it is not – Henry Oct 21 '18 at 10:10