Talk:Bijective numeration

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[edit] Talk:Decimal without a zero

I can't find any reference to this peculiar concept on the Web, nor does it make any sense; there really is a zero-equivalent as defined in this article -- it just is indicated with an "X" and functions a bit differently. I think perhaps this is gubbish? --Jpgordon 15:54, 28 Sep 2004 (UTC)

Well if you think ten is a zero-equivalent but functions a bit differently, then I suppose you are correct. Perhaps you haven't looked hard enough. Try "A Logical Alternative to the Existing Positional Number System" Volume 1 (Dec 1995) of SouthWest Journal of Pure and Applied Mathematics http://www.maths.soton.ac.uk/EMIS/journals/SWJPAM/vol1-95.html or try to decipher http://digilander.libero.it/ultimus2001/introd.htm --Henrygb 02:06, 6 Oct 2004 (UTC)
Thanks! Now I see the point (from the SWJPAM article). I wonder if anyone has followed his suggestion and gone back and looked at seemingly incorrect archeological arithmetic? --jpgordon 06:28, 6 Oct 2004 (UTC)

I'd like to merge this into bijective numeration, since decimal without a zero is just the case k=10 of the bijective base-k system described there. Does anyone object? 4pq1injbok 03:30, 6 August 2005 (UTC)

I have a question. Couldn't you solve the problem of there being an infinite way of expressing 2 when you expand the system to decimal fractions by making the base symbol associate to the left when on the left of the decimal place and associate to the right when on the right of the decimal place? This would make 1.A equal to 1.01. Then, you have .001, which, in the way described, would be, I think, -1.9A1. This simplifies it to .A9. The only problem I see with this is that .A9 x A becomes .A, which adds further confusion to a confusing system. --Some Random Guy 22:08, 16 January 2006 (UTC)

If I'm understanding you, you're wanting to overcome the fact that bijective base-k numeration does not "naturally" extend to numerals with digits to the right of the radix point. That is, if we define the notation in the natural way,
. d1 d2 d3 ... = d1 / k1 + d2 / k2 + d3 / k3 + ... (finite or infinite series)
then the system is not only not bijective, but there are infinitely-many representations of many real numbers; e.g. 1 = .A = .9A = .99A = ..., etc. Then, too, some reals have no such representation at all; e.g. .001, although .001 = 1.001 - 1 = .9A1 - 1 (not -1.9A1 as you wrote).
But the scheme you're proposing is still not bijective (e.g. still 1 = .A). Another problem is that your scheme — as I understand it — cannot represent most reals (e.g. irrationals), because it involves reversing the order of the digits to the right of the radix point and interpreting the result as an integer; e.g., you put .A9 == .001 because (9A)bijective decimal = (100)ordinary decimal. But if the number has infinitely-many digits in its representation, there is then no integer whose digits can be reversed as the scheme requires. --r.e.s. 19:14, 17 April 2007 (UTC)