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This question follows on from this one about fractional bases. As is well known, numbers in integer bases have unique representations. The linked picture in the previous question shows clearly why integer bases yield unique representations and also implies that fractional bases can yield non-unique representations. The reddit commentator mentioned also claims this. However, it is very difficult to find an example of such duplicate representations. It is even difficult to find an exact representation of a number in a fractional base. The included program is meant to assist in such searches.

My question is - does the theory say anything about this?

3 Answers3

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I am slightly scared to raise this but even the most well known base, 10, has 1 = 0.999999... and many others.

badjohn
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I think Chris means the golden ratio $\frac{1+\sqrt{5}}{2}$:

$$\begin{align}\left(\frac{1+\sqrt{5}}{2}\right)^0 + \left(\frac{1+\sqrt{5}}{2}\right)^1 = 1 + \frac{1+\sqrt{5}}{2} &= \frac{3+\sqrt{5}}{2}\\ \left(\frac{1+\sqrt{5}}{2}\right)^2 = \frac{1 + 2\sqrt{5}+5}{4} = \frac{6 + 2\sqrt{5}}{4} &= \frac{3+\sqrt{5}}{2}\end{align}$$

  • So this is an irrational base that can represent an irrational number exactly in two different ways. But can a rational non-integer base do the same for a rational number? –  Mar 26 '17 at 01:34
  • What digits would be used in the rational base? – Olli Niemitalo Mar 26 '17 at 07:15
  • @selfawareuser1 That is an interesting question, which you should ask in a separate post for visibility! – Chris Culter Mar 26 '17 at 20:51
  • @selfawareuser1 In fact, I couldn't let it go, so I asked and answered the question here: http://math.stackexchange.com/questions/2204451/can-a-rational-base-represent-a-number-with-different-finite-expansions/2204452#2204452 – Chris Culter Mar 26 '17 at 22:04
  • @OlliNiemitalo If the base was expressed as a decimal it would be the integers less than its value. The script in the linked answer generates expansions accordingly. –  Mar 27 '17 at 01:02
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I'm not sure what you mean by exact representations, but I suspect this will do... There exists a non-integer base $b>1$ in which $100_b=11_b$. It's actually quite well-known!

Chris Culter
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