# I Are These Rules new Conjectures in Set Theory?

Tags:
1. Nov 20, 2015

### Gh. Soleimani

We can easily find out below rules in set theory:

1. Let consider set “A” as follows:

A = {a1, a2, a3, a4… an} and also power set of A is set C:

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

Rule 1: To find the number of subsets with precise members number, we can use Binomial Coefficient

C (n, r) = n! / r! (n-r)!

Where:

n = number of members in set A

r = number of precise member for each subset

Example:

We have set A = {1, 2, 3, 4, 5, 6, 7}

The number of subsets with no member is: C (7, 0) = 1

The number of subsets with one member is: C (7, 1) = 7

The number of subsets with two member is: C (7, 2) = 21

The number of subsets with three member is: C (7, 3) = 35

The number of subsets with four member is: C (7, 4) = 35

The number of subsets with five member is: C (7, 5) = 21

The number of subsets with six member is: C (7, 3) = 7

The number of subsets with seven member is: C (7, 3) = 1

2. 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” below cited:

B = {x, y, z, t,…..}

Where:

x, y, z, t,… = functions of internal sum of each subset

For instance, x = a1, z = a1 + a2, t = a1 + a2 + a3 and ……..

Rule 2: 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}}

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

Now, the questions is:

- Are these rules new ones in Set Theory?

- Are These Rules new Conjectures in Mathematics?

2. Nov 21, 2015

### andrewkirk

What you are doing looks more like Combinatorics that Set Theory.
This needs to be explained better, because it's not at all clear what you intend B to be. You can't define a set with a 'for instance'. You need to define it using recognised, unambiguous set operations such as powerset, union, intersection, complementation or filtering on a logical criterion. Enumeration (listing elements) is fine, as long as it's completely clear what the elements are, but it's not OK if you use ellipses (the '....' symbol) as you have above. Enumeration has to be complete, with no ellipses and no 'etc'.

3. Nov 21, 2015

### Gh. Soleimani

Sorry, it is not the answer to my questions.

4. Nov 21, 2015

### Samy_A

5. Nov 21, 2015

### Gh. Soleimani

Please care again on rule1. Your answer is completely wrong. I say to you a problem: How can we find the number of subsets with precise members number? The solution is, to use Binomial Coefficient. For instance, if we have a set with 7 members, The number of subsets with four member will be 35. Have you any anouther solution for this problem? or Have you another reference which shows us this solution for this problem?

6. Nov 21, 2015

### Staff: Mentor

Samy_A was responding to what he thought you meant. What you stated was not clear, especially the part "with precise members number":
The binomial coefficients give the sizes of each of the subsets of a certain size.
Which is what Samy_A was saying.

7. Nov 21, 2015

### Gh. Soleimani

Thank you for your reply. But I did not see your quote (The binomial coefficients give the sizes of each of the subsets of a certain size) on this link: https://en.wikipedia.org/wiki/Binomial_coefficient#Combinatorics_and_statistics whereas, I need to have official reference (academic reference) because I am willing to know if your quote had been already proved or it is a conjecture. In the meanwhile, please be informed that wikipedia is not a reference because each moment its content has been edited.

8. Nov 21, 2015

### Staff: Mentor

It's in that section that Samy_A cited. For your example, with a set with 7 elements, the number of subsets with 4 elements is $\binom{7}{4}$ = 35.

If you have a set with n elements, the number of subsets with exactly k elements is $\binom{n}{k} = \frac{n!}{k!(n - k)!}$.
These formulas for the binomial coefficients have been around for a long time, and are very well known. Any textbook on combinatorics or probability will have this formula.

9. Nov 21, 2015

### Gh. Soleimani

Yes. Thank you. I found the proof of rule 1 on below papers:
http://www.maths.qmul.ac.uk/~pjc/notes/comb.pdf
http://math.mit.edu/~rstan/ec/ec1.pdf

10. Nov 21, 2015

### Gh. Soleimani

Let me tell you that the basic of above rules returns back to my case of accounting control such as below example:
Consider: Set A = {a1, a2, a3, a4,.....} and set B = {b1, b2, b3, b4,....]
Now,we want to know: which total sum of 1 or 2 or 3 or more of members of set A are just equal to total sum of 1 or 2 or 3 or more of members of set B? For instance,
a1 = b2+b3+b4+b5 or a2+a3+a7+a12 = b2+b3 and etc. How can we find all equilibriums?
I solved this case by using of a idea of power set and posted my solution on my website. Do you have any another idea to solve this case?

11. Nov 21, 2015

### Staff: Mentor

Counterexample :
Let A = {1, 2}
P(A) = { {}, 1, 2, {1, 2}}
Let B = { 2, 3, 5, 6}
P(B) = { {}, 2, 3, 5, 6, {2, 3}, {2, 5}, {2, 6}, {3, 5}, {3, 6}, {5, 6}, {2, 3, 5}, {2, 3, 6}, {2, 5, 6}, {3, 5, 6}, {2, 3, 5, 6}}
$\sum{P(B}$ =128

$\sum{P(A}$ = 1 + 2 + 3 = 6
$\frac{\sum{P(B}}{\sum{P(A}} = \frac{128}{6} \approx 21.33$. This is not a power of 2 for any integer n.

12. Nov 21, 2015

### Samy_A

It's not very clear what rule 2 exactly means.

My interpretation was as follows:
A is a finite set of integers with n elements. SA is the sum of all elements of A.
C is the power set of A.
Now, for each element of C one can compute the sum of its elements: these sums are the element of the set (actually the multiset) B.
SB is the sum of all elements of B.

Rule 2 states than SB=SA*$2^{(n-1)}$
I think that with this interpretation the rule is correct, but quite elementary.
As every element of A will be an element of exactly $2^{(n-1)}$ subsets of A, it is rather obvious that SB=SA*$2^{(n-1)}$

13. Nov 21, 2015

### Gh. Soleimani

Mark 44: Your way to get elements of B in your counterexmample is wrong. Because, we will have 0, 1, 2 and 3 for B then SB = 6 and SA = 3 , SB/SA = 2 = 2 ^ 2 -1
Samy A: you got true concept of this rule. Do you have any proof for this rule or it is a conjecture?

14. Nov 21, 2015

### Staff: Mentor

No, it is not wrong. You have not specified that the sets have to be consecutive integers starting with 1.
Rule 2 just mentions two sets A and B, and doesn't put any further restrictions on them.
It's up to you to provide more conditions.

15. Nov 22, 2015

### Samy_A

It is up to you to be clear. I was "lucky" in guessing what you meant with rule 2, mainly because I worked backward from the result.
This is just a consequence of the well known result that the power set of a set with m elements has $2^{m}$ elements, applied to m=n-1. (As already stated in my first post in this thread, by the way.)

16. Nov 23, 2015

### Gh. Soleimani

I deleted rule 1 and edited rule 2 and also added new rule (rule 3) as follows:

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 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 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

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

Now, the questions is:

- Are these rules new ones in Set Theory?

- Are These Rules new Conjectures in Mathematics?

17. Nov 23, 2015

### Samy_A

You have already been shown that Rule 1 (your former Rule 2) is a trivial consequence of an elementary fact about power sets.

Rule 2 is wrong without further conditions on A.
Example:
n=2
A={2,5}
SA=7
C={{},{2},{5},{2,5}}
B={0,2,5,10}
SB=17
(SB+ 1)/ SA =18/7
2n!/2=2

Last edited: Nov 23, 2015
18. Nov 23, 2015

### Gh. Soleimani

Yes. You are true.
A = a set of Natural number
Of Course, I do not consider zero as Natural number. I will say it Whole number if we consider zero.

Last edited: Nov 23, 2015
19. Nov 23, 2015

### Samy_A

My counterexample A={2,5} is a set of natural numbers, and doesn't contain 0. So your new rule 2 is still wrong.

20. Nov 23, 2015

### Gh. Soleimani

I did not tell you that A is a subset of natural number. It is ok. Let us consider N instead of A. It means that set of natural number. and then we have:
(SB+1) / SN = 2n!/n

21. Nov 23, 2015

### Samy_A

Can you please state clearly the conditions that the set A (or whatever you want to call it, the name of the set is irrelevant) has to satisfy in order for your rule 2 to be valid?

Clearly rule 2 is not valid for the set {2,5}.

If you don't state all the conditions for your rule 2 clearly, it is meaningless.

22. Nov 23, 2015

### Gh. Soleimani

please wait. I am busy now

23. Nov 23, 2015

### Staff: Mentor

I agree completely. If we decide to try your (Gh. Soleimani) rule on, say A = {1, 3, 5} and B = {2}, and find that it doesn't work, you can't come back and tell us we are wrong. As Samy_A said, if you don't specify clearly any and all conditions on the sets involved, the rules are meaningless.

24. Nov 24, 2015

### Gh. Soleimani

I edited my article and my assumtions as follows:

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

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

Now, the questions is:

- Are these rules new ones in Set Theory?

- Are These Rules new Conjectures in Mathematics?

25. Nov 24, 2015

### Samy_A

Thanks.

Let's simplify this for starters.
For a given n, A is simply the set of all integers from 1 to n: A={1,2,...,n}
So, SA=n(n+1)/2

Your rule then simplifies to $SB=\frac{2n!}{n}*\frac{n(n+1)}{2} - 1=n!(n+1)-1=(n+1)!-1$

So, for a given n, the rule states: SB=(n+1)!-1

I'm not immediately aware of this rule being known (someone else may be, of course), but it can be proven easily. I did it by mathematical induction, by the way.