Nim Addition and Split Extensions basis for

Nim Addition and Split Extensions basis for

ByLydia Njuguna and Benard Kivunge

Kenyatta University, Kenya

Third Mile High Conference on Nonassociative Mathematics 13th August 2013

onsn 2

A sequence of algebras over the field of real numbers can be constructed, each with twice the dimension of the previous one

The oldest method of constructing these algebras is the Cayley-Dickson formula and the algebras thus produced are known as Cayley-Dickson algebras

Introduction

• The algebra constructed by doubling complex numbers is the quaternions

• Next we have the octonions,constructed by

forming ordered pairs of quaternions

• The algebra immediately following the octonions is the sedenions

Real Real

Complex Complex

Quaternion Quaternion

Octonion Octonion

Sedenion Sedenion

onn 12 onn 12

onn 2

ons

22

02

12 ons

ons

32 ons

42 ons

Let L be a multiplicative subloop of the non-zero octonions. Then its sedenion extension

is the disjoint union within the sedenions.

Elements of this union are encoded as pairs with , by

The multiplication of elements of in general is given by the following equations:

Multiplication of Spit Extensions

0SL

),( a }1,1{0 S )1,(aa )1,( aaf

i). ,1 ,1 ,1x y xy

ii). ,1 , 1 , 1x y yx

iii). )1,()1,)(1,( yxyx

iv) , 1 , 1 ,1 x y xy

Consider the non-negative integers {0, 1, 2, 3,…}.

Nim addition and multiplication gives a way of defining addition and multiplication in to make it a field of characteristic 2.

The rules of Nim addition simplify to the form: The Nim-sum of a number of distinct powers of 2

is the ordinary sum. The Nim-sum of two equal numbers is 0 Example: 32 +16 + 4 + 1 = 53

Nim Addition

Z

Complex split extensions

Let )1,(),1,1(),1,(),1,1( 3210 iaaiaa

be basis elements in 0SC .The

multiplication of the elements gives rise to the

following table.

0a 1a 2a 3a

0a 0a 1a 2a 3a

1a 1a - 0a 3a - 2a

2a 2a - 3a - 0a 1a

3a 3a 2a - 1a - 0a

Table 1 Multiplication of Complex Split Extensions

0 1 2 3

0 0 1 2 3

1 1 0 3 2

2 2 3 0 1

3 3 2 1 0

The following table gives the Nim addition for the elements from 0 to 3

Table 2 Nim addition for the elements 0 to 3

This table is similar to the multiplication table of the complex split extensions (Table 1).The subscripts of the elements in Table 1 make up Table 2.

Observation:

mkmk aaa k, m = 0,1,2,3

Quaternion Split ExtensionsLet )1,(),1,(),1,(),1,( 33221100 abababab ,

)1,( 04 ab,

)1,( 15 ab,

)1,( 26 ab,

)1,( 37 ab

be basis elements in 0SH .

The multiplication of the elements gives rise to the following table

0b 1b 2b 3b 4b 5b 6b 7b

0b 0b 1b 2b 3b 4b 5b 6b 7b

1b 1b 0b 3b 2b 5b 4b 7b 6b

2b 2b 3b - 0b 1b 6b 7b 4b 5b

3b 3b 2b 1b 0b 7b 6b 5b 4b

4b 4b 5b 6b 7b 0b 1b 2b 3b

5b 5b 4b 7b 6b 1b 0b 3b 2b

6b 6b 7b 4b 5b 2b 3b 0b 1b

7b 7b 6b 5b 4b 3b 2b - 1b 0b

Table 3 Mult of Quaternion Split Extention

0 1 2 3 4 5 6 7

0 0 1 2 3 4 5 6 7

1 1 0 3 2 5 4 7 6

2 2 3 0 1 6 7 4 5

3 3 2 1 0 7 6 5 4

4 4 5 6 7 0 1 2 3

5 5 4 7 6 1 0 3 2

6 6 7 4 5 2 3 0 1

7 7 6 5 4 3 2 1 0

Table 4 Nim addition for 0 to 7

This table is similar to the multiplication table of the quaternion split extensions (Table 3). The subscripts of the elements in Table 3 make up Table 4.

Observation

mkmk bbb

k, m = 0, 1,...,7

Octonion Split Extensions

Let )1,( 00 bc , )1,( 11 bc , )1,( 22 bc , )1,( 33 bc , )1,( 44 bc ,

)1,( 55 bc , )1,( 66 be , )1,( 77 bc , )1,( 08 bc , )1,( 19 bc , )1,( 210 bc ,

)1,( 311 bc , )1,( 412 bc , )1,( 513 bc , )1,( 616 bc , )1,( 715 bc

The multiplication of the elements give rise to the following table (only subscripts shown)

be basis elements in 0SK

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 1 0 3 2 5 4 7 6 9 8 11 10 13 12 15 14

2 2 3 0 1 6 7 4 5 10 11 8 9 14 15 12 13

3 3 2 1 0 7 6 5 4 11 10 9 8 15 14 13 12

4 4 5 6 7 0 1 2 3 12 13 14 15 8 9 10 11

5 5 4 7 6 1 0 3 2 13 12 15 14 9 8 11 10

6 6 7 4 5 2 3 0 1 14 15 12 13 10 11 8 9

7 7 6 5 4 3 2 1 0 15 14 13 12 11 10 9 8

8 8 9 10 11 12 13 14 15 0 1 2 3 4 5 6 7

9 9 8 11 10 13 12 15 14 1 0 3 2 5 4 7 6

10 10 11 8 9 14 15 12 13 2 3 0 1 6 7 4 5

11 11 10 9 8 15 14 13 12 3 2 1 0 7 6 5 4

12 12 13 14 15 8 9 10 11 4 5 6 7 0 1 2 3

13 13 12 15 14 9 8 11 10 5 4 7 6 1 0 3 2

14 14 15 12 13 10 11 8 9 6 7 4 5 2 3 0 1

15 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0


The table is similar to the multiplication table of the octonion split extensions 5. The subscripts of the elements in Table 5 make up Table 6

From tables 5 and 6, the following observations can be made: There are 4 cases

Observation

Case 1: 7,7 mk

mkmkmkmk cbbbcc )1,()1,()1,(

Case 2: 8,7 mk

km

km

km

km

km

mkmk

cbbb

bbbbcc

)1,()1,()1,(

)1,()1,()1,(

8

8

8

8

8

Case 3 : 78 mk

km

mk

mk

mk

mk

mk

mkmk

cbb

bbbb

bb

bbcc

)1,()1,(

)1,()1,()1,(

)1,()1,(

8

8

8

8

8

8

Case 4: 88 mk

km

mk

mk

mk

mk

mk

mk

mkmk

c

b

b

bbb

bbbb

bbcc

)1(

)1,(

)1,()1,(

]1),)([()1,(

)1,()1,(

,16

16

88

88

88

88

88

In general,. mkmk ccc k, m = 0, 1,...,15

3.4 Sedenion Split Extensions

)1,( 0cdo , )1,( 11 cd , )1,( 22 cd , )1,( 33 cd , )1,( 44 cd , )1,( 55 cd , )1,( 66 cd ,)1,( 77 cd , )1,( 88 cd , )1,( 99 e , )1,( 1010 cd , )1,( 1111 cd , )1,( 1212 cd , )1,( 1313 cd , )1,( 1414 cd

, )1,( 1515 cd , )1,( 016 cd , )1,( 117 cd , )1,( 218 cd , )1,( 319 cd , )1,( 420 cd , )1,( 521 cd ,)1,( 622 cd , )1,( 723 cd , )1,( 824 cd , )1,( 925 cd , )1,( 1026 cd , )1,( 1127 cd ,)1,( 1228 cd , )1,( 1329 cd , )1,( 1430 cd , )1,( 1531 cd

Using the three results of the previous Section the multiplication of these elements can be summarized in the following 4 cases

Case 1: 15,15 mk

km

km

km

mkmk

dcc

ccdd

)1,()1,(

)1,()1,(

16,15 mk

km

km

km

km

km

mkmk

dccc

ccccdd

)1,()1,()1,()1,(

)1,()1,(

16

16

16

16

16

Case 3: 15,16 mk

km

mk

mk

mk

mk

mk

mk

mkmk

d

ccc

cccc

cc

ccdd

)()1,()1,()1,(

)1,()1,(

)1,()1,(

16

16

16

16

16

16

16

Case 4: 1616 mk

mk

mk

mk

mk

mk

mk

mk

mkmk

dcc

ccc

cccc

ccdd

)1,()1,(

)1,()1,(

]1),)([()1,(

)1,()1,(

32

32

)16()16(

1616

1616

1616

1616

In general,

mkmk ddd k, m = 0, 1,...,31

Consider the split extension basis elements of dimension ,n = 1,2,…given by

4. Main Theorem

n2

where...,,,12210 nffff

12,...,12,2)1(

12,...,2,1,0),1,(

11,2

1

1nnn

i

ni

i

ig

igf

n

Then, mkmk fff for k ,m = 0, 1,…. 12 n

Proof:Let the split extension basis elements be given by

12,...,12,2)1(

12,...,2,1,0),1,(

11,2

1

1nnn

i

ni

i

ig

igf

n

(1)

By induction hypothesis, suppose it’s true for n-1 with basis elements

12210 1...,,,,ngggg . We prove that it’s true for n with basis elements

12210 ...,,,,nffff defined in equation 2 above.

If it is true for n-1 with basis elements 12210 1...,,,,

ngggg , then

mkk ggg m k, m=0,1,… 12 1 n

There are 4 cases

Case 1: 12,12 11 nn mk

mk

km

km

mk

mkmk

fgggg

ggff

)1,()1,(

)1,()1,()1,(

Case 2: 11 2,12 nn mk

km

km

km

km

km

mkmk

f

g

g

g

gg

ggff

n

n

n

n

n

)1,(

)1,(

)1,(

)1,(

)1,()1,(

1

1

1

1

1

2][

2][

2

2

2

Case 3: 12,2 11 nn mk

km

mk

mk

mk

mk

mk

mkmk

f

g

g

gg

gg

gg

ggff

n

n

n

n

n

n

)1,(

)1,(

)1,(

)1,(

)1,(

)1,()1,(

1

1

1

1

1

1

2][

2

2

2

2

2

Case 4: 11 2,2 nn mk

mk

mk

mk

mk

mk

mk

mk

mkmk

f

g

g

g

gg

gg

gg

ggff

n

n

nn

nn

nn

nn

nn

)1,(

)1,(

)1,(

)1,(

]1),)([(

)1,(

)1,()1,(

2

2

)2()2(

22

22

22

22

11

11

11

11

11

Therefore, mkmk fff for k ,m = 0, 1,… 12 n .

The end

Thank you

Nim Addition and Split Extensions basis for

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