I Are These Rules new Conjectures in Set Theory?

[Mark44]

7. 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)}

Why does the last term have a factor of d? Above you set d to 1, so it is just extra baggage here that is likely to confuse someone who isn't reading carefully.

Also, the above doesn't make sense to me. Your set includes $a_1, a_1 + 1$, and then $a_2 + 1$, which of course is equal to $a_1 + 2$. Presumably the next term, according to your scheme, would be $a_3 + 1$, but for your last term you revert to $a_1 + (n - 1)d$, with d being unnecessary.

With the parameters you chose, your set is $A_1 = {1, 2, 3, \dots, n - 1}$.

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}

Again, what is the purpose of including d?

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)}

How is this different from $A_1$?

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

How is this different from $A_2$?

Finally, we will have set B:

B = {A1, A2, A3, A4, …….An}

 

[Gh. Soleimani]

8. Let consider set “A” as follows:A = {x | x ϵ R}

Rule 10: 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

 

[Gh. Soleimani]

n−1

Last member is equal:
a1 + (n - 1)d
a1 = 1 , d = 1 then it will be equal n

 

[Gh. Soleimani]

Again, what is the purpose of including d?

It is the last member of A1

 

[Gh. Soleimani]

How is this different from A 1 A_1?

We can generate many sets which are the periodicity of set A1

 

[Mark44]

Last member is equal:
a1 + (n - 1)d
a1 = 1 , d = 1 then it will be equal n

If you're referring to (n - 1)d, no, that is not equal to n. If d = 1, then $(n - 1)d = n - 1 \ne n$.

Again, what is the purpose of including d?

It is the last member of A1

That didn't answer my question. If d = 1, why do you write it as a factor?

How is this different from $A_1$?

We can generate many sets which are the periodicity of set A1

But so what?
You aren't explaining your work very well.

 

[Mark44]

8. Let consider set “A” as follows:A = {x | x ϵ R}

Or more simply, $A = \mathbb{R}$.

Rule 10: 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, …)

You need to define what A is, and redefine what $a_1, a_2,$ etc. are.
A reader shouldn't have to search back through this thread to figure out what these things mean.

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

Without proof, all you have are merely assertions.

 

[Mark44]

Since explanations of the questions I have asked don't seem to be forthcoming, I am closing this thread for moderation.

Edit: it will stay closed