Deeper than primes

[doronshadmi]

Prove it, then. Should be easy if everything is so self-evident to you.

By the way, are there more, less, or the same number (I use that term loosely) of elements in your list as there are 0's and 1's in each list item?

(The correct answer, under your latest set of assumptions, is more. And just like the case with 3 bits of 8 list items, the inverse of the diagonal does in fact exist in the list...just not in the first three. Not everything that's self-evident is the truth.)

An inverse of P(N) diagonal <0,1> form exists in P(P(N)), an inverse of P(P(N)) diagonal <0,1> form exists in P(P(P(N))), etc. ... ad infinitum, where the set of all powersets does not exist ( please look again at http://www.internationalskeptics.com/forums/showpost.php?p=6800559&postcount=14053 ).

In other words, given any set or powerset (whether it is finite or not) there is an inverse of the diagonal of that set or powerset, that exists not in the range of the given set or powerset, and since the set of all powersets does not exists we can conclude (by explicitly provide a <0,1> form that is not in the range of any given set or powerset) that any given set or power set is incomplete.

Prove it, then.

A self evident truth does not need any proof.

 

[doronshadmi]

Here read this:

It does not work that way epix.

You have to provide the reference of any quoted material that is not your material but it is used by you.

Please provide the link to your last quote.

 

[zooterkin]

In other words, given any set or powerset (whether it is finite or not) there is an inverse of the diagonal of that set or powerset, that exists not in the range of the given set or powerset, and since the set of all powersets does not exists we can conclude (by explicitly provide a <0,1> form that is not in the range of any given set or powerset) that any given set or power set is incomplete.

Only for bizarre definitions of 'incomplete'.

 

[epix]

In other words, given any set or powerset (whether it is finite or not) there is an inverse of the diagonal of that set or powerset, that exists not in the range of the given set or powerset, and since the set of all powersets does not exists we can conclude (by explicitly provide a <0,1> form that is not in the range of any given set or powerset) that any given set or power set is incomplete.

Are you up to destroying the world of atheism? If any finite set, such as

LIFE = {birth, 1, 2, 3, 4, ..., death}

is incomplete, then Catholicism and its idea of Heaven is correct, but there was never a rigorous proof of that until now.

You should visit next door
http://www.internationalskeptics.com/forums/showthread.php?t=198526
or start a new thread announcing the proof that Heaven exists.

Why do you think that Cantor harbored a particular belief?

Cantor believed his theory of transfinite numbers had been communicated to him by God.

http://en.wikipedia.org/wiki/Georg_Cantor

(I love it. I love when the atheists have to memorize at schools of higher education all the power set formulas that "delusional" Cantor came up with. LOL.)

Well, Cantor was too dumb to figure out God's intervention on behalf of the self-evident truth denied to the rest of the world by closed-minded folks. But God didn't give up and made sure that mathematical giant Doron would prove some 150 years later that finite sets are actually incomplete and therefore the counting of natural numbers called "years of life" will resume in Heaven and continue forever and ever and more and more in the divinely divergent manner, unbound and unstoppable the way <0,1> prescribed it.

I think you will be the first person to win the Jesus Christ Medal for Relentless and Unstoppable Advancement of Roman Catholic Doctrine. The elevation to the sainthood is automatic and effective the moment you receive the medal in St. Peter's Basilica from the hands of the pope.

 

[jsfisher]

An inverse of P(N) diagonal <0,1> form exists in P(P(N))

Great, but that wasn't the thing you need to prove. You need to show that the element derived from the diagonal of the list isn't in the list. How simple is that?

A self evident truth does not need any proof.

Your made up crap can't be proven.

 

[doronshadmi]

You need to show that the element derived from the diagonal of the list isn't in the list. How simple is that?

It is indeed simple.

The axiom of the inverse of <0,1> diagonal form:

The inverse of the diagonal of any given collection of <0,1> distinct forms is different by at least one bit from any given <0,1> form of that collection.

By this axiom, the inverse of <0,1> diagonal form is not included in any collection of <0,1> distinct forms.

-------------------------

Some examples of this axiom:

Distinction is the main concept here, and in the following examples we are using <0,1> as the general form for both P(N) and P(P(N)) as follows:

The power set of N ( notated as P(N) ) that includes {},{1,2,3,...} and any finite or infinite subset between {} and {1,2,3,...}, is translatable to <0,1> form, for example:

{
00000000000... ↔ { },
11000000000... ↔ {1,2},
10000000000... ↔ {1},
10101010100... ↔ odd numbers {1,3,5,...},
10100000000... ↔ {1,3},
01010101010... ↔ even numbers {2,4,6,...},
01000000000... ↔ {2},
01100000000... ↔ {2,3},
00100000000... ↔ {3},
11111111111... ↔ N numbers {1,2,3,...},
...
}

where the <0,1> distinct form that is not in the range of P(N), starts with 1011101110..., in this case.

Any given P(P(N)) subset is translatable to <0,1> form, for example:

{
0000000000... ↔ {},
1000000000... ↔ {{}},
0100000000... ↔ {{1}},
0010000000... ↔ {{2}},
0001000000... ↔ {{3}},
1100000000... ↔ {{},{1}},
1010000000... ↔ {{},{2}},
1001000000... ↔ {{},{3}},
1111111100... ↔ {{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}},
...
}

where the <0,1> distinct form that is not in the range of P(P(N)), starts with 111111111..., in this case.

Etc. ... ad infinitum.

 

[doronshadmi]

Only for bizarre definitions of 'incomplete'.

Unless it is a paradigm-shift of 'incomplete'.

 

[jsfisher]

It is indeed simple.

The axiom of the inverse of <0,1> diagonal form:

The inverse of the diagonal of any given collection of <0,1> distinct forms is different by at least one bit from any given <0,1> form of that collection.

Oh, if things were that simple. You could just assume the result you'd like, declare it to be an axiom, and life would be perfect, no?

Only one problem. That isn't an axiom. You really should look up what the term means. All this is simply your unsupported, incorrect assumption about what you'd like to be true. Unfortunately, it isn't.

 

[doronshadmi]

Unfortunately, it isn't.

You are invited to use your reasoning in order to support your "it isn't" argument about this axiom.

 

[jsfisher]

You are invited to use your reasoning in order to support your "it isn't" argument about this axiom.

First you need to describe with some specificity exactly how you'd identify the "diagonal". The obvious way would be to use the first bit from the first list item, then the second bit from the second list item, and so on, using the inverse of the i-th bit form the i-th list item for the i-th bit of the element constructed from the diagonal....but that would violate your conditions.

For that latter, what would you use for the diagonal of this:

000
001
010
011
100
101
110
111

I'd get 111, but that's just me.

 

[epix]

It is indeed simple.

We show that the finite set S of real AND counting numbers (R AND N) is incomplete by using the diagonal argument exactly the way Cantor did it with the sample of reals:

1.000000
2.000000
3.000000
4.000000
5.000000
6.000000
7.000000
--------------
Diagonal: 2.111111

The number 2.111111 is within the range; it is a real number, but not counting number. That contradicts the definition of the members of S; namely, Xi == R AND N. The proof failed. Too bad . . .

The set of natural numbers is infinite, but according to <0, 1> it is still not complete, as your binary diagonal method shows. That means there must be a natural number greater than infinity. How about proving the existence of God, Doron? He may be the Number that resides beyond infinity, and the number is very likely >10<. You must rigorously reverse <0, 1> and then check the Biblical "Seven Churches" to see if His Turbo-Wisdom is standing in the Church #8 staring at something.

 

[doronshadmi]

First you need to describe with some specificity exactly how you'd identify the "diagonal".

The obvious way would be ...

Exactly as you say, the construction of the diagonal is obvious, otherwise, it is not a diagonal.

For that latter, what would you use for the diagonal of this:

000
001
010
011
100
101
110
111

I'd get 111, but that's just me.

You are wrong, for 8 <0,1> distinct forms, each form has at least 8 <0,1> bits for each form, as follows:

00000000
00100000
01000000
01100000
10000000
10100000
11000000
11100000

So the <0,1> form that is not included in the collaction of distinct <0,1> forms above, is 11111111

You ignored all what is written in http://www.internationalskeptics.com/forums/showpost.php?p=6814651&postcount=14126 .

In other words jsfisher, please try again.

 

[doronshadmi]

The number 2.111111 is within the range;

So what?

1.00000000
2.00000000
2.11111111
3.00000000
4.00000000
5.00000000
6.00000000
7.00000000

In that case 2.11011111 is not in the range, so?

That means there must be a natural number greater than infinity.

Not at all.

That means that any collection (finite or not) of distinct <0,1> forms is incomlete.

 

[jsfisher]

Exactly as you say, the construction of the diagonal is obvious, otherwise, it is not a diagonal.

Excellent. So you have once again contradicted yourself. Your list is indexed. Doron, really, can't you get anything right?

You are wrong, for 8 <0,1> distinct forms, each form has at least 8 <0,1> bits for each form

And, again, you add index-based constraints. You are requiring the number of bits in each item equal the number of list items. My, oh, my, you are the master of the hidden assumption, aren't you?

 

[epix]

So what?

That means that any collection (finite or not) of distinct <0,1> forms is incomlete.

I think that it was zooterkin who pointed out that the way you understand the meaning of "completeness" greatly differs from the way it is understood by the rest of the world.

Your misleading <0,1> argument holds the following set incomplete

Not only that the size of the set is defined and is used by axioms/rules of the game, but there is a physical restriction given by the size of the triangle that cannot accommodate other balls so they would touch the felt. So that's an example of completeness that you have such a hard time to comprehend and so that's why you came up with your own definition of it.

The consequence of this limited capability to comprehend the basic notions showed up in the gender example. Your <0,1> argument holds the following finite set incomplete.

G = {male, female}

When I posted a photo of a man and a woman asking you to identify the third kind of gender, you asked me, "Where is the child?"

Obviously, the source of your inability to comprehend the idea of completeness must wreak havoc with your mind in other aspects, as it did when you confused the category male/female with the category adults/children . It follows that there can be no special (mathematical) argument that you would be able to comprehend and critically examine your phantasmagoric constructs.

The initial idea of an axiom was about a theorem that cannot be further proven. (Of course, according to your argument, the set of theorems particular to a unique problem is always incomplete and therefore you never arrive at axiom(s), and the truth of any math proposition therefore cannot be proven, excluding your <0,1> -- obviously.) Euclid called axioms "common notions." But if a common notion cannot be understood, then the logic of it cannot be explained. And that's what is happening in your case -- which is really phenomenal.

 

[doronshadmi]

EDIT:

Excellent. So you have once again contradicted yourself. Your list is indexed. Doron, really, can't you get anything right?

jsfisher, can't you get that any collection of distinct objects is incomplete if it is translatable to <0,1> form ?

And, again, you add index-based constraints. You are requiring the number of bits in each item equal the number of list items.

It does not change the fact that the S diagonal has less <0,1> bits than the of P(S) diagonal, whether S collection is finite or not.

Can you get things right?

 

[doronshadmi]

I think that it was zooterkin who pointed out that the way you understand the meaning of "completeness" greatly differs from the way it is understood by the rest of the world.

Your misleading <0,1> argument holds the following set incomplete

[qimg]http://www.masterrack.com/images/product_us8ball.jpg[/qimg]

Not only that the size of the set is defined and is used by axioms/rules of the game, but there is a physical restriction given by the size of the triangle that cannot accommodate other balls so they would touch the felt. So that's an example of completeness that you have such a hard time to comprehend and so that's why you came up with your own definition of it.

Your triangular form is a partial case of <0,1> form, for example:

1
11
111
1111
11111
...

and yet 00000..., which is the inverse of the diagonal of your triangular, is not in the range of the triangular, or in other words, the triangular form is incomplete.

 

[jsfisher]

jsfisher, can't you get that any collection of distinct objects is incomplete if it is translatable to <0,1> form ?

Even given your bizarre misinterpretation of "incomplete", yes, I understand the nonsense you are trying to peddle. I also understand fully that you didn't address the point I raised but instead responded with a non sequitur. Not surprising, though.

It does not change the fact that the inverse of diagonal 0000... of S has less 1 bits than the inverse of diagonal 0000... of P(S) , whether S collection is finite or not.

We've been over this already, Doron. You blew it. You denied your list was indexed. You denied its correspondence with the natural numbers. The diagonal method proof only works if there is a one-to-one correspondence with the natural numbers, and it is now apparent that your construction assumes a one-to-one correspondence.

Yes, the master of the blaming everyone else for having hidden assumptions is denuded with yet another of his own hidden assumptions.

Your bogus proof dissolves under your hidden assumption and consequential contradiction.

Can you get things right?

History would indicate that yes I can and usually do. The same test for you, however, gives the opposite indication.

 

[doronshadmi]

We've been over this already, Doron. You blew it. You denied your list was indexed. You denied its correspondence with the natural numbers. The diagonal method proof only works if there is a one-to-one correspondence with the natural numbers, and it is now apparent that your construction assumes a one-to-one correspondence.

It has nothing to do with one-to-one correspondence with the natural numbers.

----------------

Let us do it simpler:

EDIT:

Excellent. So you have once again contradicted yourself. Your list is indexed. Doron, really, can't you get anything right?

jsfisher, can't you get that any collection of distinct objects is incomplete if it is translatable to <0,1> form ?

And, again, you add index-based constraints. You are requiring the number of bits in each item equal the number of list items.

It does not change the fact that the S diagonal has less <0,1> bits than the of P(S) diagonal, whether S collection is finite or not.

EDIT:

Furthermore, the inverse of the P(S) diagonal (whether P(S) collection is finite or not) is not in the range of P(S).

Can you get things right?

--------------------------

 

[jsfisher]

It has nothing to do with one-to-one correspondence with the natural numbers.

Are you now saying it isn't indexed? That would be contradicting your contradiction, you know.

...
jsfisher, can't you get that any collection of distinct objects is incomplete if it is translatable to <0,1> form ?

We've been over this already, Doron. Is your reading comprehension issues compounded by short-term memory failure?

It does not change the fact that the S diagonal has less <0,1> bits than the of P(S) diagonal, whether S collection is finite or not.

It emphasizes that your assumption about power sets was bogus. It emphasizes that your assumption about proof was bogus. All this goes hand-in-hand with your definition of incompleteness, also bogus.

You are batting 1000, there, Doron.

Can you get things right?

Yep, and I provide a sterling example you should emulate.