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]

I put in a reference to a conversation with Conway. I haven't found a written source, unfortunately, but in the absence of this, surely this is better than nothing? I'm also unclear how it is less reliable (especially the explanation of the reason for creating it) than a lecture about it, which is currently the only given reference. So I'll put it back in the general references for the time being, but won't cite it as an in-article reference. Julian Gilbey (talk) 22:56, 31 July 2013 (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]