Are These Rules new Conjectures in Set Theory?

In the reference with article of “A Case of Accounting Control System Solved by a New Idea” posted on link: http://emfps.blogspot.com/2015/10/a-case-of-accounting-control-system.html, we can easily find out below rules in set theory:

1.  Let consider set “A” and power set of A which is set “C” as follows:

A = {a1, a2, a3, a4… an}

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

We can find set “B” in which each member of set B is generated by adding members of each subset of power set C below cited:

B = {{}, {a1}, {a2}, {a3}, {a4}, {a1+ a2}, {a1 + a3}, {a2 + a3 + a4)….{an}}

Rule 1: If SB = total Sum of members of set B and SA = total sum of members of set A, we will have:

SB / SA = 2^ (n-1)               ,         n = number of members set A

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

B = {{}, {1}, {2}, {3}, {4}, {1+2}, {1+3}, {1+4}, {2+3}, {2+4}, {3+4}, {1+2+3}, {1+2+4}, {1+3+4}, {2+3+4}, {1+2+3+4}}

SB = 0+1+2+3+4+3+4+5+5+6+7+6+7+8+9+10 = 80

SA = 1+2+3+4 = 10

SB / SA = 2^ (n-1) = 2 ^ (4-1) = 80 /10

2. Let consider A as set of Arithmetic Progression where:

d = 1,      a1 = 1   and    an = a1 + (n - 1) d, n = 1, 2, 3,…..

In this case, we have:

A = {a1, (a1+1), (a2 +1),……. (a1 + (n - 1) d)}

And power set of A which is set “C” as follows:

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

We can find set “B” in which each member of set B is generated by multiplying members of each subset of power set C below cited:

B = {{}, {a1}, {a2}, {a3}, {a4}, {a1* a2}, {a1*a3}, {a2*a3*a4)….{an}}

Rule 2: If SB = total Sum of members of set B and SA = total sum of members of set A, we will have:

(SB+ 1) / SA = 2n!/n               ,         n = number of members set A

Or     SB = [(2n!/n)*SA] - 1 

Since for Arithmetic Progression of A, we have: SA= n (n+1)/2   then, this rule can be simplified to:

 SB=2n! n ∗n(n+1)2 −1=n!(n+1)−1=(n+1)!−1    

 SB = (n+1)!-1       

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

B = {{}, {1}, {2}, {3}, {4}, {1*2}, {1*3}, {1*4}, {2*3}, {2*4}, {3*4}, {1*2*3}, {1*2*4}, {1*3*4}, {2*3*4}, {1*2*3*4}}

SA = 1+2+3+4 = 10   , SB = 1+2+3+4+2+3+4+6+8+12+6+8+12+24+24 = 119

(SB +1) / SA = (2*(1*2*3*4)/4) = 12 = (119 +1)/10

3. Let consider set “A” as subset of R and power set of A which is set “C” as follows:

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

We can find set “B” in which each member of set B is generated by average of members of each subset of power set C below cited:

B = {{}, {a1}, {a2}, {a3}, {a4}, {(a1+ a2)/2}, {(a1 + a3)/2}, {(a2 + a3 + a4)/3}….{an}}

Rule 3: If SB = total Sum of members of set B:

Then: SB = ((2^n)-1)*Average (A) , n = number of members set A

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

B = {{}, {1}, {2}, {3}, {4}, {(1+2)/2}, {(1+3)/2}, {(1+4)/2}, {(2+3)/2}, {(2+4)/2}, {(3+4)/2}, {(1+2+3)/3}, {(1+2+4)3}, {(1+3+4)/3}, {(2+3+4)/3}, {(1+2+3+4)/4}

SB = 0+1+2+3+4+1.5+2+2.5+2.5+3+3.5+2+2.333333+2.6666666+3+2.5 = 37.5

Average (A) = (1+2+3+4)/4 = 2.5

SB = ((2^ 4) -1)*2.5 = 37.5

4. Let consider set “A” and power set of A which is set “C” as follows:

A = {a1, a2, a3, a4… an}

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

Rule 4: If N = number of sets included in C which are only contained sequence of members A except sets with one member

Then, we will have:   N = C (n, 2)

Example: Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

{1,2}, {2,3},{3,4}, {1,2,3},{2,3,4}, {1,2,3,4}

N = C (4, 2) = 4! / [2! (4 -2)!] = 6

5. Following to item 4 and rule 4 (previous rule), if number of sets is equal to N = C (n, 2) +n, (added by sets with one member) then we will have below rule:

Rule 5: Number of sets in power set (C) which are included sequence of numbers and for each specific size of members (r) can be calculated by below formula:

N (n, r) = n - r +1

Where:   n = number of members set A, r = specific size of members in sets with sequence numbers

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

n = 4,

r = 1, N (4, 1) = 4 and sets are: {1}, {2}, {3}, {4}

r = 2, N (4, 2) = 3 and sets are: {1, 2}, {2, 3}, {3, 4}

r = 3, N (4, 3) = 2 and sets are: {1, 2, 3}, {2, 3, 4}

r = 4, N (4, 4) = 1 and set is: {1, 2, 3, 4}

6. Let consider “A1” as set of Arithmetic Progression where:

d = 1,      a1 = 1   and    an = a1 + (n - 1) d, n = 1, 2, 3,…..

In this case, we have:

A1 = {a1, (a1+1), (a2 +1),……. (a1 + (n - 1) d)}

One of permutations of set A1 is to invert members of set A1 as follows:

A2 = {(a1 + (n - 1) d),….,(a2 +1), (a1+1), a1}      

We can generate many sets which are the periodicity of set A1 just like below cited:

A3 = {a1, (a1+1), (a2 +1),……. (a1 + (n - 1) d)}

A4 = {(a1 + (n - 1) d),….,(a2 +1), (a1+1), a1}


Finally, we will have set B:


B = {A1, A2, A3, A4, …….An}     


Rule 6: If size members of set A1 or A2 orA3 or An, is equal to size members of set B, we will have a square matrix (Mn*n) to be generated by sets A1, A2, A3,…….An in which Eigenvalue of this matrix will be calculated by using of Binomial Coefficient as follows:

Eigenvalue (Mn*n) = λ = C (k, 2)

Where:   k = n+1, nϵ N, λ > 0


Example:

A1 = {1, 2, 3, 4}

A2 = {4, 3, 2, 1}      

A3 = {1, 2, 3, 4}

A4 = {4, 3, 2, 1}  


Matrix (M 4*4) = 
  
1 2 3 4

4 3 2 1

1 2 3 4

4 3 2 1

λ = C ((4+1), 2) = 10

7. Let consider set “A” as follows:

A = {x | x ϵ R}

Rule 7: Each type square matrix which has been generated from set “A”  just like below matrix:

a1 0 0 0 0 0 0 0 0….

a1 a2 0 0 0 0 0 0…..

a1 a2 a3 0 0 0 0 0 ….

a1 a2 a3 a4 0 0 0 0……

It will show us the eigenvalues which are just equal members set A which have been included in this square matrix (λ = a1, a2, a3, a4, …)


Example:

A = {0.67, 2, 43, 5, -23, 9, -2.3}

Assume we have set B which is a subset of A:

B = {43, -23, -2.3, 9)

Matrix N will be:

43   0  0  0  0

43  -23  0   0

43 -23 -2.3 0

43 -23 -2.3 9

Eigenvalues of Matrix N are: λ = 43, -23, -2.3, 9

8. Consider square matrices 2*2 as follows:

a    -a

b    -b

Or 

 a      b

-a    -b

a , b ϵ R

Rule 8: Eigenvalues of above matrices are equal to: λ = 0 and

λ = a – b

9. Consider square matrices 3*3 as follows:

 a     -a      a

 b     -b      b

 c     -c      c  

Or

  a     b       c

 -a    -b     -c

  a      b      c  

a , b , c ϵ R

Rule 9: Eigenvalues of above matrices are equal to: λ = 0 and

λ = (a + c)-b

10. Consider square matrices 4*4 as follows:

a      -a      a     -a

b     -b      b     -b

c     -c       c     -c

d     -d      d     -d

Or

  a     b      c      d

 -a    -b    -c     -d

 a      b      c      d

-a    -b     -c     -d

 a, b , c , d ϵ R

Rule 10: Eigenvalues of above matrices are equal to: λ = 0 and

λ = (a + c -d)-b

11. Consider square matrix “A” as follows:

            a       0      0

A =      a       b      0

            a       b      c  

If we turn this matrix around first column (as axis) and then we turn it around first row (as axis), we will have matrix “B”:

          c      b     a

B =    0      b     a

          0      0     a  

a , b , c ϵ R

Rule 11: Eigenvalue of A + B is equal to λ = c

and eigenvalue of A – B or B - A is equal to zero (λ = 0)

Example:

If matrix A is:

            1.67     0        0

A =      1.67     4        0

            1.67     4       -3  

            -3     4      1.67

B =       0      4     1.67

             0      0     1.67  

                    -1.33     4        1.67

A + B =        1.67      8       1.67

                     1.67      4      -1.33  

λ =  -3

                4.67     -4       -1.67

A - B =    1.67      0       -1.67

                1.67      4       -4.67  

λ = 0

12. Here is another special matrix. Consider matrix A is a square matrix n*n just like below forms in which (a1, a3, a5, a7 a9…an) are diagonal of matrix as follows:

A =

a1   a2    0     0     0
0    a3    a4    0     0
0    0      a5   a6    0
0    0      0     a7   a8
0    0      0      0   a9
…………………....an

Or

A =

a1     0    0    0    0

a2    a3    0    0    0

0     a4   a5    0    0

0     0    a6   a7    0

0     0    0    a8   a9

…………………....an

a1, a2, a3,….an ϵ R  

Rule 12: Eigenvalues of matrix A are all members on diagonal of matrix A. In fact, property of matrix A to generate eigenvalues is just like Diagonal Matrix.

Example:

A =

-1.56    2     0
  0     -12     3
  0        0     4  

λ = -1.56, -12 and 4

Consequently, we can find out that inverse of previous special matrix (Rule 7) will have the same eigenvalues (members of diagonal). According to Rule 7, we had below matrix M:

M =

a1 0 0 0 0 0 0 0 0….

a1 a2 0 0 0 0 0 0…..

a1 a2 a3 0 0 0 0 0 ….

a1 a2 a3 a4 0 0 0 0……

Therefore, eigenvalues of M^-1 are all members on diagonal of M^-1.

13.

Consider "T" as set of vectors as follows:  

   

T = {V1, V2, V3, ….Vn}

Where:

V1 = a1i

V2 = a2 i + a3 j

V3 = a4 i + a5 j + a6 k

Vn = a7 i + a8 j + a9 k + a10 t …….. amtn   and     a1, a2, a3, a4, a5, …. am are members of R (real 

numbers)

Theorem 13:

If "A" is a square matrix n*n inferred from the set of above vectors just like below cited:

A = {(a1, 0, 0, 0,…), (a2, a3, 0, 0, 0, …), (a4, a5, a6, 0, 0, 0, …..), …....Vn}

In fact, Scalar amounts of V1, V2, V3, …Vn are respectively rows of matrix A.

Then:   Eigenvalues of (A^n) = all members on diameter of matrix A^n (n member of N).

A =

4	0	0	0	0
-6.5	1	0	0	0
5	2	7	0	0
3	9	-5	12	0
-3.45	9	43	15	72

A^4 =

256	0	0	0	0
-552.5	1	0	0	0
2210	800	2401	0	0
-13147.5	13995	-18335	20736	0
-249381	4740402	17264326	6713280	26873856

Eigenvalues are = 256, 1, 2401, 20736, 26873856

Questions:

1. Is this theorem a new one? Have you any counterexample or a proof?

2. If the answer to question (1) is positive, it means that we can produce many theorems every day. Therefore, I think these theorems are not very important but the most crucial thing is, what are the applications of these theorems in our real life?