|
Spoonwood
|
read my profile
sign my guestbook
Name: Spoonwood
Birthday: 8/21/1979
Gender: Male
Interests: Fuzzy sets, mathematics, fuzzy numbers, fuzzy logic, neutrosophic sets, possibility theory, soft computing, type-2 fuzzy logic, hyperreal analysis, Doctor Who, Star Trek.
Expertise: Reversing the polarity of the neutron flow.
Occupation: Sonic screwdriver technician a
Industry: U.N.I.T.
Message: message me
Website: visit my website
Member Since:
12/14/2001
|
|
| I've started posting here http://dougspoonwood.blogspot.com/
| | |
| Watching My Own Worst Enemy tonigth reminded me of The World of Null-A. They have some similarities in terms of theme.
| | |
| Ordering Sets of Real Numbers Let's consider the power set of the real numbers P(R). We can partially order this set by inclusion. More interestingly though, on this set we have what have gotten called interval numbers. We define an interval number A as [a_1, a_2]={x:a_1<=x<=a_2, x in R. If a1=a2, then we have I={x:a_1<=x<=a_1}= the set of x, where x indicates a real number. Many texts which use interval numbers, such as Bojadiev and Bojadiev's Fuzzy Sets, Fuzzy Logic, Applications, take {r} where r indicates a real number, as the real number r. On p. 2 they specifically say "If in particular a_1=a_2=a, the interval number A given by [this definition {x:a_1<=x<=a_2, x in R}] reduces to the real number a=[a, a] which is called a point interval or singleton." Since we do have an ordering among real numbers, we can then say that if a<b, then {a}<{b}, since we can treat the real number 'a' as a point interval.
Somewhat similarly, then, we can say that for a set of real numbers X={x_1, ..., x_n}<Y={y_1, ..., y_n} if all x_n of X are less than all y_n of Y. For instance, {1, 2, 3}<{4, 5, 6} since 1<4<5<6, 2<4<5<6, and 3<4<5<6. We could also write that for sets of real numbers, X and Y, X<Y if max(X)<min(Y), since if max(X)<min(Y) all x_n in X<all y_n in Y. This definition makes it much easier to check if X<Y, since we only need to check the relation max(X)<min(Y) than the relation of '<' for all x_n and all y_n, as we did for {1, 2, 3}<{4, 5, 6}. This definition also applies to interval numbers, since [a_1, a_2]<[b_1, b_2] if max[a_1, a_2]<min[b_1, b_2]. But, how do we order sets with partial overlap, such as C={1, 2, 3} and D={2, 3, 4} by '<'?
Well, we know 1<2, 1<3, 1<4, 2<3, 2<4, 3<4. So, we have six pairs where c<d. There exist nine pairs of c and d. So, we can then say that C<D to 6/9=2/3 degree. In other words, C has the less than relation to a degree of 2/3, or C is 2/3 less than D. We also know 2=2, and 3=3, and so we can say that C is 2/9 equal to D. Lastly, 3>2, so C is 1/9 greater than D. Here the pair (C, D) has degree of 2/3 in <. This comes as an instance of a fuzzy relation, even though the numbers C={1, 2, 3} and D={2, 3, 4} qualify as crisp numbers in the sense that for every element c of C and d of D we can tell whether or not c belongs to C or d belongs to D. In general, for any discrete real number, in other words for any set of discrete real numbers, we can compute E<F, E=F, E>F much like the above. Let E={e_1, ..., e_n}, F={f_1, ..., f_n}. The degree that (E<, =, >F)=(number of pairs ((e_n), (f_n)) where e_n<, =, >f_n))/(number of pairs of (e_n, f_n)).
When E is a proper subset of F things seems of note. For instance, {2, 3, 4} is 6/15 less than {1, 2, 3, 4, 5}, 3/15 equal to {1, 2, 3, 4, 5}, and 6/15 greater than {1, 2, 3, 4, 5}.
Let d(E<F) indicate the degree to which E<F and d(e>f) indicate the degree to which e>f. For all real numbers r, s we have either r<s or r=s or r>s. We can view r<s as indicating that r is less than s to a degree of 1, or r is 1 less than s, since if r<s, (r, s) belongs to the membership relation of '<'. If (r, s) does not belong to = or > or <, then r does not equal s, greater than s, or less than s respectively. In this sense if r<s, then (r, s) has degree of equality of 0, and (r, s) has degree of membership of 0 in > relation. So, for real numbers r and s it becomes clear that d(r<s)+d(r=s)+d(r>s)=1, since one of those relations will hold, while the others don't impying that either d(r<s)=1 or d(r=s)=1 or d(r>s)=1 and that if d(r<s)=1, then d(r=s)=d(r>s)=0, if d(r=s)=1, then d(r<s)=d(r>s)=0, if d(r>s)=1, then d(r=s)=d(r<s)=0. In other words, if one of the degrees of membership of the relation for the pair (r, s) equals 1, the other two relations have degree of membership of 0 for (r, s), so the sum of all three degrees of membership of the relations equals 1. For discrete real numbers, we will similarly have +(d(R<S), (d(R=S), d(R>S))=1. This seems straightforward, but I haven't figured out how to show that yet, so I'll leave it as an exercise.
The above procedure for obtain the degree of membership of <, =, or > doesn't generalize to continuous real numbers so easily, as it would end up with infinities in the numerator and denominator. We can say that
[1, 2]<[3, 4] we would said {1, 2}<{3, 4} in that every member of [1, 2]<every member of [3, 4] as well as we can say that max[1, 2]<min[3, 4]. But, to what degree does [1, 3]<[2, 4] hold? Let us first consider how much [1, 3]=[2, 5]. We know that these intervals overlap on the interval [2, 3]. We also know that they don't overlap on [1, 2) and (3, 5]. Let me define min2(a, b, c, d) as the least number greater than the minimum of a finite set... I would say the second least number of {a, b, c, d}, and max2(a, b, c, d) as the greastest number less than the maximum member of a finite set {a, b, c, d} or the second greatest number of {a, b, c, d}. In this sense we can say that we partitioned [1, 3]U[2, 5] into [min(1, 3, 2, 5), [min2(1, 3, 2, 5)), max2(1, 3, 2, 5)], (max2(1, 3, 2, 5), max(1, 3, 2, 5)] or more simply [1, 2), [2, 3], (3, 5]. We do know that [1, 2)<[2, 5] since every number of [1, 2) is less than every number of [2, 5]. We also know that [1, 3]<(3, 5] since every member of [1, 3] is less than every member (3, 5]. The degree of membership for <, =, and > preferably will take this into account.
First, for two interval numbers F=[a, b] and G=[c, d] we'll first partition the union of them into
[min(a, b, c, d), min2(a, b, c, d)), [min2(a, b, c, d), max2(a, b, c, d)], (max2(a, b, c, d), max(a, b, c, d)]. We'll then suppose [min(a, b, c, d), min2(a, b, c, d))<[c, d] to degree 1, [min2(a, b, c, d), max2(a, b, c, d)]=[c, d] to degree 1, and (max2(a, b, c, d), max(a, b, c, d)]>[c, d] to degree 1. Then, I propose that
d(min(a, b, c, d), min2(a, b, c, d)))*d(c, d)/(d(a, b)*d(c, d)) indicates the degree to which
F<G,
d(min2(a, b, c, d), max2(a, b, c, d))*d(c, d)/(d(a, b)*d(c, d)) indicates the degree to which F=G,
d(max2(a, b, c, d), max(a, b, c, d))*d(c, d)/(d(a, b)*d(c, d)) indciates the degree to which F>G. I think that will work, but I'd need to check more examples to feel more sure that it will.
| | |
| Versatile Numbers and Partitions of Sets Suppose we have a function f which assigns a number of factors measure for a natural number. 2 has 2 factors, so f(2)=1. 4 has 3 factors so f(4)=3. Then, a versatile number v consists of a natural number such that for all natural numbers n less than v the number of factors measure for n is less than the number of factors measure for v. For all naturals n<v, f(n)<f(v). For example let v=6. We have {1, 2, 3, 6} as the set of factors, so f(v)=f(6)=4. We know v as versatile since f(1)=1, f(2)=2, f(3)=2, f(4)=3, f(5)=2.
n<v implies f(n)<f(v) might look trivial. Versatile numbers and prime number both have a "natural" (readily available) number of factors measure f, since they get defined in terms of the number of their factors. This counting measure f reflects (preserves) the order of the natural numbers since n<v implies f(n)<f(v), while it distorts (reverses) the order of the natural numbers for prime numbers since n<p implies f(n)>=f(p), for n>1. In fact, only when (n=p1)<p2 does f(n=p1)=f(p2). If n equals a composite number c, then c<p implies f(c)>f(p). Since there generally exists more composite numbers than prime numbers in a given list of natural numbers {2, ..., n}, more often than not if n<p, then f(n)>f(p) for a randomly given natural number n>1. In other words, the probability of n<p implying f(n)>f(p) is greater than 1/2 for a randomly given natural number n>1. And again, if n<p doesn't imply f(n)>f(p), then n<p implies f(n)=f(p)... both relatonships do not reflect or distort the order of the natural numbers.
Now, consider a set of objects {a1, ..., an}=A. A partition of such a set A indicates a set S of subset(s) s1 ... sn of A such that union of the subsets (s1 v ... v sn) equals A and no two subsets have a common member,
(si ^ sj='0' where '0' denotes the empty set, for all i, j in {1, ... n}). I'll call a partition equinumerous if all subsets of the partition have the same number of members. For instance for the set {1, 2, 3} the partition {{1}, {2}, {3}} qualifies as equinumerous. The partition {{1}, {2, 3}} does not, since {1} has 1 member, while {2, 3} has 2 members. Now, for a finite set {a1, ..., an}=A, if n equals a versatile number v, then A has more equinumerous partitions than any proper subset of A. Also, if n equals a prime number for A={a1, ..., an}, then there exists only one equinumerous partition. So, a versatile number v maximizes the number of equinumerous partitions for all natural numbers n<v, while a prime number p minimizes the number of equinumerous partitions for all natural numbers.
For example for the list {a, b, c} there exists only two equinumerous partition since there exist 3 letters, namely {{a}, {b}, {c}} and {{a, b, c}}. Next, let us suppose we have the list {a, b, c, d}. The equinumerous partitions come as follows
{{a, b}, {c, d}}, {{a, c}, {b, d}}, {{a, d}, {b, c}}, {{a}, {b}, {c}, {d}}, and {{a, b, c, d}}. Notice that we have 5 possible partitions here, even though 4 has only 3 factors. Versatile numbers will still maximize the number of equinumerous partitions for all n<v, since there exists more possibilities for the number of subsets in the partition of a versatile set {a1, ... , av}=A than a proper subset of that versatile set. For example, with the versatile set {a1, a2, a3, a4} we can either have 2 subsets of 2 elements or 4 subsets of 1 element. In general for a set {a1, ..., an} for a partition we have c subsets of d elements or d subsets of c elements where c*d=n. For a versatile number, since it has more factors than any natural number smaller than it, the number of ordered pairs of c and d is greater than for a smaller natural number, n(c, d)<v(c, d) where n(c, d) indicates the number of ordered pairs c and d such that c*d=n, and v(c, d) indicates the number of ordered pairs such that c*d=v.
Note that the number of possible partitions for a prime number p equals the number of its factors, while for any composite number m the number of possible equinumerous parititions does not equal the number of its factors, since for any composite number m we can have c subsets of d elements such that c*d=g where neither c=1 nor d=1. This implies that we have an equinumerous partition
{{a1, ..., an}, {b1, ..., bn} ... {z1, ..., zn}} which we can use to produce at least one other partition by replacing a given a1 by some other b1 and acoordingly then replace the same b1 by the same a1 to yield another partition which has c subsets of d elements (this doesn't hold for a partition of a set S={a1, ... ap} where p indicates a prime number, since the order of elements for a set comes as immaterial), provided that such replacement doesn't yield a partition we've already listed. So, we only need suppose we only have one partition of the form {{a1, ..., an}, {b1, ..., bn} ... {z1, ..., zn}} where n>1. We can always produce a second (equinumerous) partition by replacing b1 for a1 and a1 for b1 accordinly as above. E.G. for the partition {{q, r}, {s, t}} we replace q by s, and accordingly then replace s by q to yield {{s, r}, {q, t}} another paritition which has c subsets of d elements. So, a composite number m has more possible equinumerous partitions than the number of factors of m f(m), or f(m)<e(m), with e(m) indicating the number of possible equinumerous partitions of m. Since for a versatile number v and any natural number n<v, f(n)<f(v) and since f(v)<e(v) (since any given versatile number belongs to the set of composite numbers), f(n)<e(v). Oh... I want e(n)<e(v), where n<v and v indicates a versatile number. Since I've written for a while and it's way past a reasonable time to go to sleep, I'll leave it as an exercise (in other words, the reasoning I foresee as needed to demonstrate such comes as more subtle than I expected.)
Online References:
http://en.wikipedia.org/wiki/Partition_of_a_set
http://www.earth360.com/math-versatile.html
| | |
| P(A|B) means the probability of the event (subset) A given the set B.
A^B indicates the intersection of A and B.
'*' stands for ordinary multiplication
'=' stands for equality.
Theorem: Every subset B has probabilistic independence with respect to its reference set X. Note the reference set X can qualify as an event within X, since X qualifies as a subset of X.
Proof: A set Z gets called independent of a set V if it does not depend on V. In other words, V has no effect on A. So, P(Z|V)=P(Z). The probability of a subset B P(B) equals P(B|X) since the definition gives us X with B as a subset of X. In other words P(B)=P(B|X). So, B qulaifies as independent of X. If we use the definitition that B is independent of X iff P(B^X)=P(B)*P(X), the theorem still holds, since P(B^X)=P(B) since B^X=B, and since P(X)=1, P(B)*P(X)=P(B)*1=P(B).
One can comment here and say that the whole notion of probabilistic independence suffers from a sort of semantic paradox if one doesn't stick to a formal notion of "independence." A subset B of a reference set X, does depend on its reference set in the sense that one can define the subset B in terms of a collection of indicator functions for its members with respect to the reference set X. E.G. for the reference set {1, 3, 5}=A we can define a subset B by using the indicator function I(xn) for each member of A. Iff I(xn)=1, then xn belongs to B, while iff I(xn)=0, then xn does not belong to B. For instance, for {1, 3} we can define {1, 3} in terms of an indicator function I(xn) which here yields I(1)=1, I(3)=1, I(5)=0. So, we can write {(1, 1), (3, 1), (5, 0)} for {1, 3} to make our reference set more explicit.
One can argue that the probabilisitic notion of "independence" doesn't refer to that sort of independence, but a mere formal definition, namely P(A^B)=P(A)*P(B) and this fits more with our notions of independence for proper subsets A, B, of the reference set X. For example, for the set {1, 2, 3, 4, 5, 6}, the subsets {1, 3, 5} and {2, 4, 6} have independent probabilities and this makes sense intituitively since neither have a connection like how a reference set and a subset of the reference set do.
| | |
|
|
