Are These Rules new Conjectures in Set Theory?

In summary: The solution to finding the number of subsets with precise members number is to use the Binomial Coefficient.
  • #36
Gh. Soleimani said:
Here is another rule:

6. Following to item 5 and rule 7 (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 8: number of sets with specific member number can be calculated by below formula:

N' = n - r +1

Where: n = number of members set A, r = specific member number
What does "specific number member" mean? Even with the example below, I can't figure out what you're trying to say.
Gh. Soleimani said:
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'1 = 4 and sets are: {1}, {2}, {3}, {4}
Why are you writing N'1 in this and the following examples?
Gh. Soleimani said:
r = 2, N'1 = 3 and sets are: {1, 2}, {2, 3}, {3, 4}
Why aren't {1, 3}, {1, 4} and {2, 4} included? "Specific number member" can't mean "subsets of the power set with a specific number of members" since you are omitting three of the subsets with two members.
Gh. Soleimani said:
r = 3, N'1 = 2 and sets are: {1, 2, 3}, {2, 3, 4}
Why isn't {1, 2, 4} included?
Gh. Soleimani said:
r = 4, N'1 = 1 and set is: {1, 2, 3, 4}
 
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  • #37
I referred you to my previous rule. Please carefully read it again ( Following to item 5 and rule 7 (previous rule)).
I considered only sequence of members A or as Samy _A said to us:

Samy_A said:
What you call a sequence is defined by its first and last element
 
  • #38
Hi Mark 44,
I wrote an article of “How Can a Simple Rule in Set Theory help us to derive strategies in Strategic Management?” In this article, I used above rule (N' = n - r +1) to solve a case in excel. Maybe the application of above rule in excel will be useful for all members in this group.

Am I allowed to post this application here in this group?
 
  • #39
Gh. Soleimani said:
I referred you to my previous rule. Please carefully read it again
Which "you" are you replying to? If this refers to the questions I asked in post #36, please respond to those questions. If you are responding to someone else, please use the Quote button to identify the person who is making the remarks as well as what you're replying to.

The notation in each rule should be self-explanatory. If your notation is not clear, then your rules are not helpful.
For your convenience, here are the questions I asked:
  • What does "specific number member" mean? Even with the example you gave, I couldn't figure out what you're trying to say.
  • Why are you writing N'1 in your examples (in post #36)?
  • When r = 2, why aren't {1, 3}, {1, 4} and {2, 4} included? "Specific number member" can't mean "subsets of the power set with a specific number of members" since you are omitting three of the subsets with two members.
  • When r = 3, why isn't {1, 2, 4} included?
 
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  • #40
Gh. Soleimani said:
Hi Mark 44,
I wrote an article of “How Can a Simple Rule in Set Theory help us to derive strategies in Strategic Management?” In this article, I used above rule (N' = n - r +1) to solve a case in excel. Maybe the application of above rule in excel will be useful for all members in this group.

Am I allowed to post this application here in this group?
If the thrust of your article is in deriving strategies for management, it probably doesn't belong here in the Mathematics section.
 
  • #41
Hi Mark44,

Please accept my apology, if I was not able to write a clear notation.

Your first question is about “Specific number member” while I wrote “Specific member number” in post #35 (rule 8). My meaning of “Specific member number” is “specific size of members” for instance we have a sequence set with one member or a sequence set with two members or three members…

For answering to your other questions, let me completely change the sentence of rule 8 in post #35 and also change formula N' = n - r +1 to N'r = n - r +1 as follows:

6. Following to item 5 and rule 7 (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 8: 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'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'1 = 4 and sets are: {1}, {2}, {3}, {4}

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

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

r = 4, N'4 = 1 and set is: {1, 2, 3, 4}
 
  • #42
Gh. Soleimani said:
For answering to your other questions, let me completely change the sentence of rule 8 in post #35 and also change formula N' = n - r +1 to N'r = n - r +1 as follows:
OK, now I understand what you're trying to say.
Gh. Soleimani said:
6. Following to item 5 and rule 7 (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 8: 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'r = n - r +1
Just for clarity, the context here is P({1, 2, 3, ..., n}), the power set of {1, 2, ..., n}
Might I suggest this notation:
N(n, r) = n - r - 1,
where N(n, r) represents the number of subsets of P{1, 2, 3, ..., n} of size r, in which the elements of each subset are in sequential order.
Gh. Soleimani said:
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'1 = 4 and sets are: {1}, {2}, {3}, {4}

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

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

r = 4, N'4 = 1 and set is: {1, 2, 3, 4}
I can see that this works in your example, and I have tried it out with a slightly larger set, {1, 2, 3, 4, 5}. Can you prove that your formula gives the right results?

For example, if n = 5, and we're looking at subsets of size 2 in the power set, we have {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3. 5), (4, 5)}. The only member of this set in which the elements are in sequential order are {(1, 2), (2, 3), (3, 4), (4, 5)}. Other than by observation, how do you know not to count (1, 3), (1, 4) and the others that aren't in sequential order?
 
  • #43
Mark44 said:
For example, if n = 5, and we're looking at subsets of size 2 in the power set, we have {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3. 5), (4, 5)}. The only member of this set in which the elements are in sequential order are {(1, 2), (2, 3), (3, 4), (4, 5)}. Other than by observation, how do you know not to count (1, 3), (1, 4) and the others that aren't in sequential order?
I see it as follows: he only has to pick the smallest element of the subset, as the subset must contain the r-1 following elements. But there have to be r-1 elements available, so this only works for the n-(r-1) first elements.

No idea if that's how @Gh. Soleimani sees it.
 
  • #44
Here is another example for n = 5 due to your request:

A = {1, 2, 3, 4, 5}

C = {{}, {1}, {2}, {3}, {4}, {5}, {1, 2}, {1, 3}, {1, 4}, {1, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4}, {3, 5}, {4,5}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5},{1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}, , {1, 2, 3, 4}, {1, 2, 4, 5}, {1, 3, 4, 5}, {1, 2, 3, 5}, {2, 3, 4, 5}, {1, 2, 3, 4, 5}}

n = 5, N'r = n - r +1

r = 1, N'1 = 5 and sets are: {1}, {2}, {3}, {4}, {5}

r = 2, N'2 = 4 and sets are: {1, 2}, {2, 3}, {3, 4}, {4, 5}

r = 3, N'3 = 3 and sets are: {1, 2, 3}, {2, 3, 4}, {3, 4, 5}

r = 4, N'4 = 2 and set is: {1, 2, 3, 4}, {2, 3, 4, 5}

r = 5, N'5 = 1 and set is: {1, 2, 3, 4, 5}
 
  • #45
Samy_A said:
I see it as follows: he only has to pick the smallest element of the subset, as the subset must contain the r-1 following elements. But there have to be r-1 elements available, so this only works for the n-(r-1) first elements.
@Samy_A, can you elaborate on this -- I'm not following. For a given n (say 5) and given r (say 2), the subset consisting of 2-element members consists of {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)}.

What do you mean by "pick the smallest element of the subset" -- (1, 2)?
 
  • #46
Gh. Soleimani said:
Here is another example for n = 5 due to your request:

A = {1, 2, 3, 4, 5}

C = {{}, {1}, {2}, {3}, {4}, {5}, {1, 2}, {1, 3}, {1, 4}, {1, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4}, {3, 5}, {4,5}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5},{1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}, , {1, 2, 3, 4}, {1, 2, 4, 5}, {1, 3, 4, 5}, {1, 2, 3, 5}, {2, 3, 4, 5}, {1, 2, 3, 4, 5}}

n = 5, N'r = n - r +1

r = 1, N'1 = 5 and sets are: {1}, {2}, {3}, {4}, {5}

r = 2, N'2 = 4 and sets are: {1, 2}, {2, 3}, {3, 4}, {4, 5}

r = 3, N'3 = 3 and sets are: {1, 2, 3}, {2, 3, 4}, {3, 4, 5}

r = 4, N'4 = 2 and set is: {1, 2, 3, 4}, {2, 3, 4, 5}

r = 5, N'5 = 1 and set is: {1, 2, 3, 4, 5}
Yes, this is clear, and is an example that I've already worked out.

As already mentioned, your notation would be better (IMO) as N(n, r).
N'r is confusing, as it could mean N' * r.
Also, the prime on N (N') isn't needed, since N and n are different symbols.
 
  • #47
Mark44 said:
@Samy_A, can you elaborate on this -- I'm not following. For a given n (say 5) and given r (say 2), the subset consisting of 2-element members consists of {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)}.

What do you mean by "pick the smallest element of the subset" -- (1, 2)?
In this rule he only counts subsets that consist of consecutive elements of the set (at least, that's how I interpret it). Let's call these subsets the "good" subsets (I'm tired, can't find a better term).
In your example, with n=5, r=2, {1,2} will be a good subset, because 1 and 2 are consecutive numbers. {2,4} will not be a good subset, as 2 and 4 are not consecutive numbers.
So once you know the first element of a good subset, say a, you know that the subset will be {a,a+1,...a+r-1}.
That's how I get n-r+1, as you can do this for all elements from 1 up to n-r+1. For n-r+2 there are not enough elements left in the set to form a good subset starting at n-r+2.

In the case n=5, r=2, there will be good sets starting at 1,2,3 and 4. That makes 4, and that is indeed n-r+1.

(I assume that the set is {1,2,3,...,n} for simplicity.)
 
  • #48
Samy_A said:
In this rule he only counts subsets that consist of consecutive elements of the set (at least, that's how I interpret it). Let's call these subsets the "good" subsets (I'm tired, can't find a better term).
In your example, with n=5, r=2, {1,2} will be a good subset, because 1 and 2 are consecutive numbers. {2,4} will not be a good subset, as 2 and 4 are not consecutive numbers.
So once you know the first element of a good subset, say a, you know that the subset will be {a,a+1,...a+r-1}.
That's how I get n-r+1, as you can do this for all elements from 1 up to n-r+1. For n-r+2 there are not enough elements in the set to form a good subset starting at n-r+2.

In the case n=5, r=2, there will be good sets starting at 1,2,3 and 4. That makes 4, and that is indeed n-r+1.

(I assume that the set is {1,2,3,...,n} for simplicity.)
"Good" subsets works for me.
Thank you for the explanation.
 
  • #49
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)}

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 9: 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
 
  • #50
As the sum of the elements in each row is constant (and equal to n(n+1)/2 =4*5/2=10 in your example), it is obvious that (1,1,...,1) will be an eigenvector of the matrix with eigenvalue n(n+1)/2 =##\binom {n+1}{2}##.

This has nothing to do with an arithmetic progression, by the way. It works for any matrix where the sum of the elements in a row is the same for each row. (1,1,1,...1) will be an eigenvector, and the eigenvalue will be the sum of all the elements in a row.

Example:
##A=\begin{pmatrix}
11 & 6 & 9 & 22\\
6 & 1 & 32 & 9\\
0 & 48 & 0 & 0\\
9 & 1 & 30 & 8\\
\end{pmatrix}##
For each row the sum of the elements is 48, and 48 is indeed a eigenvalue of A with eigenvector (1,1,1,1).
 
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  • #51
Gh. Soleimani said:
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}##.

Gh. Soleimani said:
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?
Gh. Soleimani said:
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##?
Gh. Soleimani said:
A4 = {(a1 + (n - 1) d),….,(a2 +1), (a1+1), a1}
How is this different from ##A_2##?
Gh. Soleimani said:
Finally, we will have set B:

B = {A1, A2, A3, A4, …….An}
 
  • #52
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
 
  • #53
Mark44 said:
n−1
Last member is equal:
a1 + (n - 1)d
a1 = 1 , d = 1 then it will be equal n
 
  • #54
Mark44 said:
Again, what is the purpose of including d?
It is the last member of A1
 
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  • #55
Mark44 said:
How is this different from A 1 A_1?
We can generate many sets which are the periodicity of set A1
 
  • #56
Gh. Soleimani said:
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##.
Mark44 said:
Again, what is the purpose of including d?
Gh. Soleimani said:
It is the last member of A1
That didn't answer my question. If d = 1, why do you write it as a factor?
Mark44 said:
How is this different from ##A_1##?
Gh. Soleimani said:
We can generate many sets which are the periodicity of set A1
But so what?
You aren't explaining your work very well.
 
  • #57
Gh. Soleimani said:
8. Let consider set “A” as follows:A = {x | x ϵ R}
Or more simply, ##A = \mathbb{R}##.
Gh. Soleimani said:
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.
Gh. Soleimani said:
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.
 
  • #58
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
 
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