Is there a method of turning a rational number in the recurring $ b $-ary expansion form into an irreducible fraction? I.e., given rational $0.\overline{d_1d_2...d_n}=0.d_1d_2...d_nd_1d_2...d_n...$, where each $ d_i $ is some digit of base-$ b $, I need to find an irreducible fraction $\frac{m}{k }$ (where $ m, k\in \mathbb Z $ represented in base-$b $) that equals to that rational. This question came to me out of curiosity, and I have no idea how to even start answering it :)
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For base-10, you just compute the sum, e.g., .abcabc...=(abc)/10^3+ (abc)/10^6+..... Should be similar in different bases. – user99680 Jun 26 '14 at 18:02
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See http://math.stackexchange.com/questions/295195/h0w-t0-prove-that-periodic-decimal-numbers-are-rational-a-1-a-kb-1b-2-b-l – mvw Jun 26 '14 at 18:06
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In your example, let this rational number be x.
$b^nx = {d_1d_2...d_n} + x$.
So $x = {d_1d_2...d_n}/(b^n-1)$. Now just reduce this.
Wonder
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I had taken it to mean we are looking at the string as a number instead of a product of the digits. But here I should probably remove it. – Wonder Jun 26 '14 at 18:12
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1That would probably be a good idea, but for now I am just going with the questioner's conventions. – Wonder Jun 26 '14 at 18:16
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The same process works as in base $10$. Multiply your rational by $b^n$ and subtract. If the repeat starts just before the point, you get $m=d_1d_2\dots d_n, k=b^n$ If there are digits before the repeat starts, you will need to work on the fraction a bit.
Ross Millikan
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