Talk:Conway's base 13 function

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13

Is the exact number 13 important? Or could this have been done with 14, 17, 23? Crasshopper (talk) 16:10, 30 September 2011 (UTC)[reply]

The only importance of base 13 is that this particular encoding of real numbers uses 13 symbols. If you used an encoding with two symbols then it would be a base 2 function. David Radcliffe (talk) 04:39, 4 November 2011 (UTC)[reply]

Reference

Needs a better reference (94.212.42.237 (talk) 19:21, 5 October 2011 (UTC)).[reply]

Digits

It is more conventional, and would be less confusing, to use the digits 0,1,2,3,4,5,6,7,8,9,A,B,C. — Preceding unsigned comment added by 131.162.130.215 (talk) 15:07, 25 October 2011 (UTC)[reply]

I agree, and I've changed the article to match this. 128.86.179.86 (talk) 17:37, 25 January 2012 (UTC)[reply]

Error correction

"Indeed, $f$ takes on the value of every real number on any closed interval $[a,b]$ where $b>a$ ." This is false. For example, if $a=.++$ (base 13) and $b=.+++$ , then $f$ is constantly 0 on $[a,b]$ .

I edited the page accordingly, but am still not sure if the article %100 correct.

Quinn (talk) 03:34, 26 December 2011 (UTC)[reply]

(In the new notation, this example reads a=0.ACC, b=0.ACCC.)

The article was correct before, so I changed it back. I think you missed the fact that the expansion only has to end in the required form; I've now highlighted that in the definition. For example, the interval you've given includes 0.ACC0 up to 0.ACC1, so you can encode any decimal in this region e.g. f(0.AAC0B123A456...)=123.456.... Let me know if you want more details (I haven't completely figured out the algorithm for picking the analogue to 0.ACC0 for arbitrary a and b, but I don't think it's too hard). 128.86.179.86 (talk) 18:02, 25 January 2012 (UTC)[reply]

Algorithm: starting from the left, find the first digit in the base 13 expansion where a and b differ, say, digit n. Then b_n > a_n, if the differ by at least two than any number with expansion a_1, ..., a_n + 1, ... lies in the interval. Otherwise, take some m > 0 so that a_(n+m) < C, then any number with expansion a_1, ... , a_(n+m) + 1, ... lies in the interval. (We can always find such an m since we're avoiding expansions with c repeating.) 69.195.54.191 (talk) 18:34, 8 April 2012 (UTC)[reply]