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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
| | | Posted 9/11/2008 5:04 AM - 55 Views - 2 eProps - 4 comments
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