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# Tj'a-ts'a~n counting system

From: Christophe Grandsire Monday, February 22, 1999, 14:01
```        I think that Tj'a-ts'a~n counting system is one of the strangest feature
of Tj'a-ts'a~n (at least for Westerners). I don't know if any other conlang
or natlang has a system resembling it, and would be very happy to know that.

The counting system is based upon two numbers: 8 (the Sky People have 4
fingers per hand) and 7 (as they have 3 toes per foot, one foot plus one
hand equals 7 fingers). The basic numbers from 1 to 8 are the names of the
elements:

1: sjem /Sem/ (daily sky)
2: bor /bor/ (wood)
3: new /new/ (air)
4 wir /wir/ (water)
5: pi /pi/ (plant spirit)
6: nja /njQ/ (animal spirit) (/Q/ is a back rounded a)
7: sum /Sum/ or /Zum/ (earth)
8: jer /jer/ (fire)

You also have 56: pse /pse/ (big, tall), 64: roj /roj/ (human) and 448:
pse.roj (56*64!). The Sky People (at the time when I begin their history)
have no concept of number zero.

The counted object is an attributive at the partitive case, completing a
group of juxtaposed nouns corresponding to the numbers. Juxtaposition
corresponds to addition, compounding to multiplication (with the multiplier
-the smallest number- after the multipliee -the biggest number-). The
gender of a number is j- ("quality") except if the counted number is not
present. In that case, the number takes the gender of the omitted object.

To count beyond 8, you must compound and juxtapose numbers. Here is how it
works (compounding is multiplication *, juxtaposition is addition +):
9 = 8 + 1: j-jer tj-sjem
10 = 8 + 2: j-jer j-bor
11 = 8 + 3: j-jer nj-new
12 = 8 + 4: j-jer j-wir
13 = 8 + 5: j-jer tj-pi
14 = 2*7: j-sum.bor /j@-Zum-bor/
15 = 2*7 + 1: j-sum.bor tj-sjem
16 = 2*8: j-jer.bor
17 = 2*8 + 1: j-jer.bor tj-sjem
18 = 2*8 + 2: j-jer.bor j-bor
19 = 2*8 + 3: j-jer.bor nj-new
20 = 2*8 + 4: j-jer.bor j-wir
21 = 3*7: j-sum.new
22 = 3*7 + 1: j-sum.new tj-sjem
23 = 3*7 + 2: j-sum.new j-bor
24 = 3*8: j-jer.new
25 = 3*8 + 1: j-jer.new tj-sjem

As you can see, to name a number, you take the biggest multiple of 7 or 8
smaller than the number and you add the remainder if any. So when you
continue you have also:

48 = 6*8: j-jer.nja
49 = 7*7: j-sum.sum /j-ZumZum/
...
55 = 7*7 + 6: j-sum.sum nj-nja
56 = 56: tj-pse
57 = 56 + 1: tj-pse tj-sjem
...
63 = 56 + 7: tj-pse j-sum
64 = 64: j-roj
65 = 64 + 1: j-roj tj-sjem

With bigger numbers, you begin to seek the biggest multiple of 56 or 64
smaller than the number, and then you add the remainder whose name is
created with the first rule I explained.

111 = 64 + 6*7 + 5: j-roj j-sum.nja tj-pi
112 = 2*56: tj-pse.bor
113 = 2*56 + 1: tj-pse.bor tj-sjem
...
127 = 2*56 + 2*7 + 1: tj-pse.bor j-sum.bor tj-sjem
128 = 2*64: j-roj.bor
129 = 2*64 + 1: j-roj.bor tj-sjem
...
168 = 3*56: tj-pse.new
...
192 = 3*64: j-roj.new
...
447 = 7*56 + 7*7 + 6: tj-pse.sum j-sum.sum nj-nja
448 (= 8*56 = 7*64): tj-pse.roj (definitely a mistake, as 56*64=3584, but
as the order between pse and roj is not the good one -pse: 56 is smaller
than roj: 64, we must see that number as immotivated, not really as a
compound).

For bigger numbers, you take the biggest multiple of 448 and then add the
remainder. For instance:

1500 = 3*448 + 2*64 + 4*7: tj-pse.roj.new j-roj.bor j-sum.wir

NOTE: the order between juxtaposed terms can be as you want (addition is
commutative), but the one I propose is the most used.

Well, as usual: what do you think of it?

Christophe Grandsire
|Sela Jemufan Atlinan C.G.

"Reality is just another point of view."

homepage : http://www.bde.espci.fr/homepage/Christophe.Grandsire/index.html
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