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Evidently, the weird number 13 turning up in the definition of this function is just so there's 3 extra digits, in addition to the 10 decimal ones. But 10 itself sure is pretty arbitrary here, and just used for human-habit reasons?

If we mapped, instead of base-10, simply to base-2, we would only need base-5 as input, and the analogue construction would sure work as well. That would look a bit less "magic-number"-like.

Or am I mistaken here? What might perhaps not work?

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Yes this works with bases $b$ and $b+3$ in general.

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  • $\begingroup$ Conway's base 13 always confused me, I never understood why +3 is used rather than +2. Surely he just needs one extra digit for the decimal point, and one extra digit to mark the end of the base-13 prefix? So why didn't we end up with Conway's base 12 function? $\endgroup$
    – Marcel Besixdouze
    Nov 25, 2015 at 5:56
  • $\begingroup$ @MarcelT. With sufficiently sophisticated encoding one might even come up with base 3 encoding, e.g. if $x$ in base $3$ after removal of the period matches the pattern $[012]^*2\underbrace{[01]}_a\underbrace{[01]^*}_b2\underbrace{[01]^*}_c$ (with the last Kleene star actually meaning infinitely many digits), we might map this number to $(-1)^a\cdot2^y\cdot z$ where $y$ is the natural number represented in binary by $b$ and $z$ is the real number represented in binary by $0.c$; you could surely even formulate a base 2 only method, but whatfor? The base-13 explanation is just less convoluted $\endgroup$
    – Hagen von Eitzen
    Nov 25, 2015 at 12:59
  • $\begingroup$ Oh thank you, I think I get it now, the extra symbol is to code all the negative real numbers. As a logician, I often forget what the "reals" are e.g. that they aren't just $[0,1]$ or $\omega^\omega$ or $2^\omega$ etc. $\endgroup$
    – Marcel Besixdouze
    Dec 1, 2015 at 5:26

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