# Boolean Logic cannot deal with infinitely many objects

1. Dec 10, 2003

### Organic

I realized that my last thread (named by "permutations") can’t be understood by professional mathematicians.

Because I don’t know how to write my idea in the common formal way, I am going to do it in a non-formal way, but I will do my best to write it in the clearest way.

So here it is:

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 full 01 list, combinations list.

Now, let us use Cantor's Diagonalization method on some finitely long combinations list, for example, the combinations list of number 3:

000
001
010
011
100
101
110
111

We can change the order of the rows, and then use Cantor's Diagonalization method, for example:

001
011
010
000
101
100
111
110

The input for Cantor's Diagonalization method in the first example is 000 and the output is 111.

The input for Cantor's Diagonalization method in the second example is 010 and the output is 101.

In both examples we find that the result is already in the combinations list, and this combination, which is already in the list, is one of the combinations that Cantor's Diagonal does not cover.

The number of the combinations, which are out of the range of Cantor's diagonal is:

2^n - n

Every column, which belongs to some combinations list is a sequence of 01 notations, based on some periodic frequency changes, for example:

the right column of number 3 combinations list, is based on 2^0(=1).

Therefore the periodic frequency changes are 1, and the result in this case is:
01010101.

The result of the middle column is based on 2^1(=2), therefore the sequence is:
00110011.

The result of the left column is based on 2^2(=4), therefore the sequence is:
00001111.

and we get the full combinations list of number 3:

000
001
010
011
100
101
110
111

We can get a combinations list of infinitely many places, by using the ZF Axiom of infinity induction, on the left side of our combinations list, by using the induction on the power_value of each column, for example:

2^0, 2^1, 2^2, 2^3, ...

In this stage we have proven, by induction, that Cantor's diagonal cannot cover any full 01 combinations list, finite or infinite.

Therefore its result is not a new combination (that has to be added to the list).

Because Cantor's diagonal cannot cover the full 01 combinations list (of aleph0 places for each combination) we can conclude that 2^aleph0 > aleph0.

But, because no diagonal's result is a new combination (and therefore not added to the list) each infinitely long sequence of 01 notations can be mapped with some natural number, for example:

...000 <--> 1
...001 <--> 2
...010 <--> 3
...011 <--> 4
...100 <--> 5
...101 <--> 6
...110 <--> 7
...111 <--> 8
...

Therefore we can conclude that 2^aleph0 = aleph0, and we come to contradiction.

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

One can say that at least the sequence ...111 is not in the list, for example:

...000 <--> 1
...001 <--> 2
...010 <--> 3
...011 <--> 4
...100 <--> 5
...101 <--> 6
...110 <--> 7
...111 <--> 8
...

Let us examine the infinite from another point of view.

When we have ...111 AND ...000 in an ordered combinations list, it means that the list is complete.

But this is the whole point, infinitely many objects cannot be completed, otherwise they are finitely many objects.

Therefore ...111 AND ...000 are not in the list of infinity many objects.

In other words [...000, ...111) XOR (...000, ...111] .

There are 2 possible structural types of infinitely many 01 notations:

(?...0]
(?...1]

We know how some infinitely long combination starts, but its left side is
unknown (can be 0 XOR 1) and this missing information is essential to the existence of the induction.

Therefore we can find a meaningful missing result by Cantor's Diagonalization method, only in a finite combinations list.

For more details please look at:

http://www.geocities.com/complementarytheory/RiemannsBall.pdf

Organic

Last edited: Dec 13, 2003
2. Dec 10, 2003

### HallsofIvy

As I said before, Cantor's Diagonalization method requires that you have as many digits in each number as you have numbers (the way it is normaly applied, this is a countable list of numbers each having a countable number of digits).

In addition, the "input" for Cantor's Diagonalization method is the list of numbers not any one number.

It is not a matter of not knowing how to "write" mathematics. You've never taken the time to understand the mathematics itself.

3. Dec 10, 2003

### Organic

Dear HallsofIvy,

Now i know that you don't really want to understand what you read.

And to understand what i wrote you have to do 2 basic things:

1) You have to read it from the first word until the last word.

2) After you read all of it, you have to check if you understand it.

3) After you understand it, than and only than, please write a detailed reply.

Thank you.

Organic

Last edited: Dec 11, 2003
4. Dec 10, 2003

### Staff: Mentor

Frankly, Organic, the only thing that would keep me from laughing at the mess you made with that last thread would be my respect for you as a person (until proven otherwise, I assume everyone is worthy of my respect). Posts like this one diminish that. You won't get very far here unless you drop your attitude. And you will get even further if you first learn some real math before trying to invent your own new math.

5. Dec 11, 2003

### Organic

Dear russ_watters,

First you have to show that you understand what i wrote in this thread, and you can do in this way:

1) You have to read it from the first word until its last word.

2) After you read all of it, you have to check if you understand it.

If you can't follow these 3 steps, then you did not show to the persons that read your last raply, that you have any meaningful thing to say about what i wrote.

Yours,

Organic

Last edited: Dec 11, 2003
6. Dec 11, 2003

### NateTG

Perhaps you should try to understand Cantor's diagonal argument before you start claiming that it is invalid.

Usually the thing you call 'combinations list' in your post is referred to as $$2^A$$.

Cantor's argument is that $$|2^A| \neq |A|$$. This should painfully clear if $$A$$ is a finite set.

A particularly choice abuse is:
Your example pairings are either not well defined, or for example, if the ...'s are all zeros will never map any sequence that contains an infinite number of 1's.

Last edited: Dec 11, 2003
7. Dec 11, 2003

### suyver

I've understood what you are trying to say / prove. However, I do not agree with it.

Your reasoning is flawed. Cantor's argument ONLY works for a list of n numbers, each containing exactly n digits. Since your list does not follow this requirement, you can't use Cantor.

8. Dec 11, 2003

### Guybrush Threepwood

Re: Combinations

and I also didn't understand what do you mean by this......

9. Dec 11, 2003

### Organic

To NateTG,

Please show in what i wrote how do you come to the conclusion that what you call A, is finite.

10. Dec 11, 2003

### Organic

To suyver,

You are right, Cantor's diagonal is limited to aleph0^2(=aleph0), and this is exactly the reason why he can find infinitely many new numbers, which are not in this aleph0^2(=aleph0) list.

The complete list of all R numbers is 2^aleph0.

I proved, by using the ZF axiom of infinity induction, that 2^aleph0 > aleph0.

Then i proved that 2^aleph0 = aleph0 by this:

Each infinitely long sequence of 01 notations can be mapped with some natural number, for example:

...000 <--> 1
...001 <--> 2
...010 <--> 3
...011 <--> 4
...100 <--> 5
...101 <--> 6
...110 <--> 7
...111 <--> 8
...

Therefore we can conclude that 2^aleph0 = aleph0, and we come to contradiction.

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

Last edited: Dec 11, 2003
11. Dec 11, 2003

### Guybrush Threepwood

12. Dec 11, 2003

### Organic

To Guybrush Threepwood,

You wrote:
Here it is again:

Now, let us use Cantor's Diagonalization method on some finitely long combinations list, for example, the combinations list of number 3:

000
001
010
011
100
101
110
111

We can change the order of the rows, and then use Cantor's Diagonalization method, for example:

001
011
010
000
101
100
111
110

The input for Cantor's Diagonalization method in the first example is 000 and the output is 111.

The input for Cantor's Diagonalization method in the second example is 010 and the output is 101.

In both examples we find that the result is already in the combinations list, and this combination, which is already in the list, is one of the combinations that Cantor's Diagonal does not cover.

The number of the combinations, which are out of the range of Cantor's diagonal is:

2^n - n

13. Dec 11, 2003

### Guybrush Threepwood

Organic, you don't have to repaste the original message.

If the combination is already on the list, what do you mean is not covered?

And also the argument HallsOfIvy made about the incorrect use of the diagonalization method in this case stands.

14. Dec 11, 2003

### Organic

I proved it for n, i proved it for n+1, therefore i proved it for infinitely many cases.

15. Dec 11, 2003

### Organic

16. Dec 11, 2003

### HallsofIvy

Now, that's just bad elementary mathematics!

17. Dec 11, 2003

### Guybrush Threepwood

I ask again how?? I provided a link to the ZF set of axioms earlier. I just don't see it....

no, you forgot at least the all '1' sequence.....

18. Dec 11, 2003

### Organic

Dear HallsofIvy,

19. Dec 11, 2003

### Organic

Let us examine the infinite from another point of view.

When we have ...11111 AND ...00000 in an ordered combinations list, it means that the list is complete.

But this is the whole point, infinitely many objects cannot be completed, otherwise they are finite.

Therefore ...11111 AND ...00000 are not in the list of infinity many objects.

In other words [...000, ...111) XOR (...000, ...111] .

There are 2 possible structural types of infinitely many 01 notations:

(?...0]
(?...1]

We know how some infinitely long combination starts, but its opposite side is
unknown (can be 0 XOR 1) and this missing information is essential to the existence of the induction.

Therefore we can find meaningful missing result by Cantor's diagonal method, only in a finite combinations list.

For more details please look at:

http://www.geocities.com/complementarytheory/RiemannsBall.pdf

Last edited: Dec 11, 2003
20. Dec 11, 2003

### HallsofIvy

The statement "I proved it for n, i proved it for n+1, therefore i proved it for infinitely many cases." doesn't make any sense.
If you mean "n" as a single specific integer, then "I proved it for n, i proved it for n+1" only proves it for those two particular numbers.
If you meant n to be any natural number, then after "I proved it for n" it isn't necessary to prove it for n+1!
If you meant this as "proof by induction", then it is "Assuming the statement is true for n, prove it is true for n+1" and is only part of what is necessary for a proof by induction.

21. Dec 11, 2003

### Organic

Hi HallsofIvy,

Please go to the frist post of this thead, find the part of my proof by induction, and then please tell me if this part is proof by induction or not, and if not, why not.

Thank you,

Organic

22. Dec 12, 2003

### NateTG

Might be a poorf, definitley not a proof.

That one misses sequences like:
...01010101010101
which contain infinitely many ones and zeros.

23. Dec 12, 2003

### Organic

Thank you NateTG, I deleted it.

Please show in what i wrote how do you come to the conclusion that what you call A (look in page 1), is finite.

Organic

Last edited: Dec 12, 2003
24. Dec 12, 2003

### HallsofIvy

Okay, I did go back and read it. You look at 2 bit sequences and show that the number you get by applying Cantor's method to the sequence is in the sequence. Then you look at 3 bit sequences and show that the number you get by applying Cantor's method to the sequence is in the sequence. You then assert that you have proved "by induction" that this will be true for any finite or infinite.
The simplest thing that is wrong with that is that you did not show that "if it is true for N then it is true for N+1".

Actually my objection was to your statement "I proved it for n, i proved it for n+1, therefore i proved it for infinitely many cases."
Even assuming you meant "I proved that IF it is true for any positive integer n, then it is true for n+1, therefore I proved it true for infinitely many cases." that would not be correct- you still have to prove it is true for onecase.(And by the way, induction does not just prove "it is true for infinitely many cases", it proves the statement (which depends on the positive integer n) is true for all n. Saying something is true for all positive integers n is far different than saying it is true for infinitely many of them.)

Somewhat less simple is this: induction proves something is true for all positive integers- not for infinity. You referred to "ZF axiom of infinity induction" but you did not use that. A proof by infinite induction would have to appeal to Zorn's lemma or something equivalent.

Finally, you basic concept of Cantor's method is flawed. There are 4 binary numbers with 2 bits. In applying (your version of) Cantor's method, you used only 2 of them. Of course, the number you got was one of the other 2. There are 8 binary numbers with 3 bits. In applying Cantor's method, you used only 3 of them. Of course, the number you got was one of the other 5. In any list of numbers with n bits, there are 2n numbers, of which you would use the first n to produce a "new" number. It should be no surprise that that number is in the other 2n- n.

Cantor applied his method to a list of real numbers, each of which was represented by an infinite (countable) number of digits (or bits if you want to use binary notation). In constructing his new number, with an infinite number of bits, he used all of the numbers in the list. Since the nature of the construction guarenteed that the new number could not be any of those used in its construction, and all numbers in the list were so used, it follows that the new number cannot be on the list. That is the argument that does not apply to any finite list of numbers.

25. Dec 12, 2003

### Organic

Dear HallsofIvy,

The ZF Axiom of infinity simply says: if n exists than n+1 exists.

I use this built-in induction on the power_value of 2^power_value,
and the result of using the built-in induction of the ZF Axiom of infinity on the power_value, cannot be but 2^aleph0.

Please show me some mistake in what i wrote in this post.

Yours,

Organic

Last edited: Dec 12, 2003