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Re: Possible base-20 numeric system

From:Simon Richard Clarkstone <s.r.clarkstone@...>
Date:Saturday, October 23, 2004, 23:20
Trebor Jung wrote:

> Compounds: > > ketïs: twenty > koltïs: thirty > neltïs: forty > ötïs: fifty > sabtïs: sixty > haudïs: seventy > mbitïs: eighty > naetïs: ninety > tistïs: one hundred > üstistïs: one hundred and one > rastïs: one hundred and ten > üsrastïs: one hundred and eleven > töltïs: one hundred and twenty > ðaertïs: one hundred and thirty > ... > ketïstïs: two hundred > ... > tistïstïs: two hundred > ...
No, wrong, you have a bizarre dual-base system. Each of the quoted should be 20 more than its predecessor. In the next "place", each would be 20*20=400 more than its predecessor. Also, you must (well, should) only have one name for each integer, but you have: ketïs-tis = koltïs (assuming hyphens add the two numbers) A better scheme: nol = ^0 ... köt = ^19 I will epress the base-20 numbers in ways such as this: 7_2_8_1_2 then we have: nol = _0 = 0 üs = _1 = 1 ket = _2 = 2 kol = _3 = 3 ... eiv = _17 = 17 aum = _18 = 18 köt = _19 = 19 Then things slightly more complicated: üstïs = _1_0 = 1*20 + 0 = 20 ketïs = _2_0 = 2*20 + 0 = 40 koltïs = _3_0 = 3*20 + 0 = 60 . . . eivtïs = _17_0 = 17*20 + 0 = 340 aumtïs = _18_0 = 18*20 + 0 = 360 köttïs = _19_0 = 19*20 + 0 = 380 üstïstïs = _1_0_0 = 1*20*20 + 0*20 + 0 = 400 üstïstïstïs = _1_0_0_0 = 1*20*20*20 + 0*20*20 + 0*20 + 0 = 8000 üstïstïstïstïs = _1_0_0_0_0 = 1*20*20*20*20 + 0*20*20*20 + 0*20*20 + 0*20 + 0 = 160000 for complicated numbers: köttïstïstïs-naetïstïs-ðahtïs-nel = _19_9_16_4 = _19_0_0_0 + _9_0_0 + _16_0 + _4 = 19*20*20*20 + 9*20*20 + 16*20 + 4 = 155924 -- Simon Richard Clarkstone s.r.cl*rkst*n*@durham.ac.uk / s*m*n_cl*rkst*n*@hotmail.com

Replies

Danny Wier <dawiertx@...>
Rene Uittenbogaard <ruittenb@...>