Dodecagonal Numbers

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Dodecagonal Numbers

Postby wendy » Thu Jan 10, 2019 12:07 pm

The Z-span of chords of a dodecagon x12o, forms an algberaic integer system that i have been playing with of late. These are the sorts of systems that I usually consult when i need to decide whether things are possible or not. This particular one looked interesting.

This is a system that algebraicly corresponds to \(z_1 + z_2 \sqrt{2}+z_3\sqrt{3} + z_4(\frac 12\sqrt{6}+\frac 12\sqrt{2})\). The name of this system is Z12, it is a class-four system, which means that numbers have four different +/- signs. The use of algebra is quite complex, the whole can be done with a coin-calculator on a board, one should have a supply of +1 and -1 chips for this exercise.

The design for the abacus is to have about 5 or so rows, and about 10 or so columns, with the '1' position in the centre column, and bottom row. The notation then follows that for a base, with 1,0 representing a unit to the left of the unit, 0;1 a unit to the right, (ie the semicolon is the radix), and 10 a unit upwards. So 100 is two units upwards, and 1,0,0 is two units to the left.

The positive numbers are represented by numbers, such as 0,1,2,3, and the negative numbers are represented by letters, A,B,C are -1,-2,-3 resp.

There are just two carries: 100 = 2, and 1,0,0 = 10,1. These equate to 10B = A02 = 1,A0,A = A,10,1 = 0. You can use these moves to eliminate large numbers of stones in any cell. It is possible to reduce a number to a positive or negative, with no more than one stone per column (of the same colour - both moves are on the same colour on a chess-board). While it is possible to reduce one number to positive on black and negative on white, these are not the 'four signs' mentioned above.

The basic system is somewhat larger than a union of Z4 \( 1, \sqrt 2\), and Z6 \((1, \sqrt 3\) and J10 \( 1, \sqrt 6\). since the dodecagon allows the inscription of polygons of {4}, {6} sides.

The principal units of the constituant systems are

Z4 = \(\alpha\) = \(1+\sqrt 2\) = 11
Z6 = \(w^2\) = \(2+\sqrt 3\) = 1,0,0 = (1,0)²
J10 = \(\beta^2\) =\(5 + 2\sqrt 6\) = 1,2,1;2,1 = (11;10)²

Not only are these units (that is, there is a reciprocal in the system), but there are others as well.

2+w= 1,2 = \(\sqrt{\alpha\beta/w}\)

Every element in Z is composite. The primes have at least two indirect factors, and some have four, according to their modulo 24 value.

1, 23 always have four distinct values
5, 19 have two factors of the form a sqrt(2)+b sqrt (3) or a+b sqrt(6).
7, 17 have two factors of the form a + b sqrt(2)
11,13 have two factors of the form a+b sqrt(3).

Indirect primes are ones whose powers are not divisible by the prime they divide, so (sqrt(6)+1)(sqrt(6)-1)=5, but 5 does not divide (sqrt(6)+1)². This remains true for any subsystem that J10 = z{1.r6} is part of.

2 and 3 are both direct primes here.

2 is represented by w+1 = \( 1 + \frac 12 \sqrt{6}+\frac 12 \sqrt{2}\), the fourth power is \(2 \alpha^2\beta^2\)
3 is represented by sqrt(3) = 1;10. It is an order-2 prime, the square gives 3.
6 is the square of w+1/w, or 1,0;1.
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