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Topic: How many?
Replies: 6   Last Post: Feb 13, 2007 6:10 PM

 Messages: [ Previous | Next ]
 Fede Ferreres Posts: 7 Registered: 2/13/07
Re: How many?
Posted: Feb 13, 2007 2:51 AM

On 12 feb, 12:35, "Kenshin" <rurouni_sohj...@hanmail.net> wrote:
> How many case of choosing n numbers is there ?
>
> 1,2,2,3,3,3,4,4,4,4,,,,,,,,n,n,n(n ns)
>
>
> Best regards,

I assume when you say "n" numbers, you mean "e" elements in the set.
If you choose the "number "2", you have 3 elements. If you choose 3,
you have five elements. You want the inverse, f a set has "e"
elements, how much numbers do I need, for e= any number of elements...

And the answer is ... if you use conventional math and want to test
for e = natural numbers, and want to find out how much "n"s you need,
that there is no answer, and that our math is wrong. For example, our
"n" could be the items in a equation using only multiplication and
prime numbers (after all, this should be enough). But if you add up
all the possible combinations of prime number, you arrive that there
are at lest 5.5% of numbers in n, that are NOT based on "knowable"
prime numbers...test for yourself:

1/2 + 1/2*1/3 + 1/2*2/3*1/5 + 1/2*2/3*4/5*1/7 + .... = 94.5XXXX%

The number of elements addressable with prime numbers converges to
0.945% (rounded) of the n! elements in the natural numbers domain (n
as large as you want) (this is what I believe the number of elements
is).

Why does it converge so something close to 94%? Because when you have
reached an element that is more than half the elements in the set,
then these prime numbers become "unknowable". For example, test it
with a finite set, that of 7! elements. The are nearly 367 primes
larger than half the set (something like 5040/2 is half the set). So
the primes lower than 2020, half the set, can only account to 93.9%
(close aproximation to the case of say the first 100 primes, or even
1000 primes). What is the explanation? I don't have one, but can state
that a prime larger than half the set cannot be divided by any number,
nor multiplied by any number. If you could divide it, it would not be
prime, and if you could multiply it, then it would mean there is a
number larger than the set, impling there are more numbers, than
elements in the set. We cannot allow this.

Another way to tackle the problem is to imagine the natural numbers 1,
2, 3, 4, ... 1/2e (half the elements), are ALL PRIMES, or that one
are. And therefore, the expansion:

1/2 + 1/2*1/3 + 1/2*2/3*1/4 + 1/2*2/3*3/4*1/5 + .... = 100%

And this is true, and would shield a result that you would need n!/2
"numbers. You would have to use a notation that would treat 2*4
different than 2*2*2 for operations ... and that would consider any
element in the set as a combination as two numbers.

Additionally, I have thought out a different notation that would
require much lesss symbols, or numbers, than elements. In this
notation, to identify each for the "e" elements in the set (we are
talking about n! here), you would have to solve:

(n-1)*n*(n+2)/2 = e

You would have to round up, and you could end up with some spare names
that you wont need.

1) It can't be known, or we don't know
2) It is 50% of "e"
3) It is the number resulting from solving (n-1)*n*(n+2)/2 = e

There may be other solutions, but we are talking about naming
conventions an axioms regarding how are we going to name, or identify,
or generate, the elements, and what operations and kind of numbers are
allowed. If you ask me, 1) is not a good answer, 2) and 3) are both
solutions, but 3) is not reliable as it has come out of my head
thinking of mubers as "parts" of pie, of which there can be infinite
pieces, up to 1 atom of pie on the limit.

Federico

Date Subject Author
2/12/07 Kenshin
2/12/07 kilian heckrodt
2/12/07 David Marcus
2/12/07 Henry
2/12/07 Dr. V I Plankenstein
2/13/07 Fede Ferreres
2/13/07 Michael Press