# Cantor's Diagonalization function on the combinations list

THANK YOU.

Let us check these lists.

P(2) = {{},{0},{1},{0,1}} = 2^2 = 4

and also can be represented as:

00
01
10
11

P(3) = {{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8

and also can be represented as:

000
001
010
011
100
101
110
111

Let us call any 0,1 list, combinations list.

When we use Cantor's Diagonalization function on the combinations list of 2^power_value and get some output result, we find that this result is already somewhere in the list.

The formula that gives us the number of combinations , which are out of the range of Cantor's Diagonalization function, is:

2^n - n

Combinations are first of all structural changes, based on at least two parameters:

a)The number of different notations.
b)The number of places that have been given to permute these notations.

We get our list of infinitely many places, by using the ZF Axiom of infinity induction, on the left side of our combinations list (by changing power_value).

When we have infinitely many places to combine our two different notations, then the number of combinations, which are out of the range of Cantor's Diagonalization function is:

2^aleph0 - aleph0 = E where by E we mean that there are E possible combinations, which are out of the range of Cantor's Diagonalization function, where one of these combinations, is Cantor's Diagonalization function result.

Therefore Cantor's Diagonalization function result is not a new combination.

Because the aleph0 long Cantor's Diagonalization function result cannot cover the 2^aleph0 list, it means that 2^aleph0 > aleph0, but we can define a map between any unique combination and some natural number, therefore
2^aleph0 = aleph0.

Therefore (2^aleph0 >= aleph0) = {}, and we have a proof saying that Boolean Logic cannot deal with infinitely many objects.

Organic

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HallsofIvy
Homework Helper
1. These are not permutations.

2.
When we use Cantor's Diagonalization function on the permutations list of 2^power_value and get some output result, we find that this result is already somewhere in the list.
This makes no sense at all. Cantor's Diagonalization method (it's not really a function) "for the nth digit in the number choose a digit other than the nth digit of the nth number in the list" requires that each number have the same number of digits as there are number in the list. That cannot be applied to, for example, P(3), because there are 8 numbers each having only 3 digits.

Originally posted by Organic
Let us check these lists.
P(3) = {{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8
what you have here is the power set of {0, 1, 2}
the permutations of {0, 1, 2} are:
0 1 2
0 2 1
1 0 2
1 2 0
2 1 0
2 0 1
that means 3! = 6 permutations

The formula that gives the number of permutations is n! not 2n

Dear Guybrush Threepwood,

Thank you.

Organic

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HallsofIvy
Homework Helper
So you started calling them "permutations" because you didn't know what permutations were. Now you are calling them "combinations" and the only thing we learn from that is that you also do not know what "combinations" are.

Using mathematics terms that you do not understand only makes you look foolish.

The strange thing is that you seem to think not knowing any mathematics makes you capable of proving that whatever anyone else believes about mathematics is not true!

Dear HallsofIvy,

Thank you.

Organic

russ_watters
Mentor
This is the funniest thing I've seen all day.

Originally posted by Organic
Dear HallsofIvy,

Thank you.

Organic
You know, that isn't even necessary. If you write a 100 page proof of something in math, and someone finds a mistake on the first page, then your proof is wrong. And since you make a number of obvious errors, it would be a waste of time to take a detailed look at your proof.

HallsofIvy