Deeper than primes

[doronshadmi]

So squares are round?

No.

There is an "unlimited life" beyond the matrix.

 

[doronshadmi]

You're the one who seems to be applying a step-by-step approach, adding members to a set one by one. When the set is defined, the members are in it, they don't have to be looked at one at a time to decide if they are in or out.

I do not look only at each case separately as the step-by-step reasoning does ( please see the results of only step-by-step reasoning on jsfisher's mind in http://www.internationalskeptics.com/forums/showpost.php?p=6921871&postcount=14391 ).

I also use a parallel reasoning.


EDIT:

Please try to get

http://www.internationalskeptics.com/forums/showpost.php?p=6921898&postcount=14397

http://www.internationalskeptics.com/forums/showpost.php?p=6921901&postcount=14398

http://www.internationalskeptics.com/forums/showpost.php?p=6921905&postcount=14399

 

[epix]

A set is an ever increasing mathematical object.

Your only step-by-step reasoning does not help to get it.

There is a finite set |S|= 4 that increments itself diagonally and step by step in order to "prove" that A={1, 2, 3, 4} is incomplete:

1st step: 1 <=> 0001
2nd step: 2 <=> 0010
3rd step: 3 <=> 0011
4th step: 4 <=> 0100

(S|InvDiag {0,0,1,0} = {1,1,0,1}) => (1101bin = 13dec) => (A={1, 2, 3, 4} is incomplete)
Note: That's good. Who cares about bad luck 13 in the set, right?

 

[doronshadmi]

There is a finite set |S|= 4 that increments itself diagonally and step by step in order to "prove" that A={1, 2, 3, 4} is incomplete:

1st step: 1 <=> 0001
2nd step: 2 <=> 0010
3rd step: 3 <=> 0011
4th step: 4 <=> 0100

(S|InvDiag {0,0,1,0} = {1,1,0,1}) => (1101bin = 13dec) => (A={1, 2, 3, 4} is incomplete)
Note: That's good. Who cares about bad luck 13 in the set, right?

Very .

But you do not have to define the mapping only between the decimal notation and its binary notation.

After all

1st step: 1 <=> 2
2nd step: 2 <=> 1
3rd step: 3 <=> 4
4th step: 4 <=> 3

is perfectly legal

and so is

1st step: 1 <=> 0010
2nd step: 2 <=> 0001
3rd step: 3 <=> 0100
4th step: 4 <=> 0100

because the the diagonal method works in any given order.

 

[epix]

It is a "permanently out of the (X^2) box" reasoning.

EDIT:

Please try to understand the permanent question mark (which is a present continuous state) at the top of the following diagram:

[qimg]http://farm6.static.flickr.com/5214/5480734023_6482feb9e5_b.jpg[/qimg]

The question mark is hardly permanent; it is replaced by 1 due to the apex of Pascal Triangle.

2⁰ = 1 = 1
2¹ = 2 = 1 + 1
2² = 4 = 1 + 2 + 1
2³ = 8 = 1 + 3 + 3 + 1
2⁴ =16 = 1 + 4 + 6 + 4 + 1
2⁵ =32 = 1 + 5 +10 +10 +5 + 1

and so on.

2ⁿ can be binomially distributed, if you happen to forget.

 

[jsfisher]

...where the set of all powersets does not exist.

Why do you keep bringing up this irrelevant trivia? Nobody has claimed otherwise.

 

[jsfisher]

It is not about (S (X^2)) < (P(S) (2^X)).

Of course not. What you wrote, (S (X^2)) < (P(S) (2^X)), is gibberish. However, it is all about |S| < |P(S)|.

It is about the bijection between any given natural number and any given P(S) member

No, that's wrong, too. There of course is a bijection between any given natural number and any given element of a set (power set or otherwise). That's one of those trivial things that isn't worth dwelling on.

Since you have trouble keeping up, Doron, I'll demonstrate. Let A be a natural number, and let B be an element of some set. A bijection between A and B would be, wait for it.....wait for it....wait for it, A <-> B.

However, your statement is completely wrong if you meant there was a bijection between the natural numbers (not just one element) and the power set of any infinite set (not just one element).

...
there is a bijection between natural numbers and any P(S), where the set of all powersets does not exist.

Ok, this is yet a different statement of your thesis. It's still wrong. Let's take for example S = {A}. It's power set P(S) = {{},A}. Show us, Doron, a bijection between the natural numbers and P(S). Here, I'll start you out with this table. Just fill in the left-hand side to complete the bijection:

  N       P(S)
------+--------

  ?  <->  {}
  ?  <->  {A}

 

[The Man]

Any given collection is under the laws of probability.

What “laws of probability” are you referring to? The probability that any natural number is included in the set of all natural number is 1 the probability that it is not is 0. “under the laws of probability” you’ve still got nothing to ‘increase’ in the set of all natural numbers.

EDIT:

Please try to understand the permanent question mark (which is a present continuous state) at the top of the following diagram:

[qimg]http://farm6.static.flickr.com/5214/5480734023_6482feb9e5_b.jpg[/qimg]

Please try to understand that “the permanent question mark” only represent your “present continuous” and apparently preferred “state” because you never try to learn anything and just choose instead to make up whatever self-contradictory, coloring book and connect the dots nonsense you want.

 

[The Man]

jsfisher, the permanent question mark at the top of the considerd diagram is timeless and therefore has nothing to do with before (past) and after (future).

The Man and you still do not grasp the accurate meaning of present continuous.

Doron you still do not grasp the accurate meaning of increasing.

 

[epix]

Please try to understand the permanent question mark (which is a present continuous state) at the top of the following diagram:

[qimg]http://farm6.static.flickr.com/5214/5480734023_6482feb9e5_b.jpg[/qimg]

Your problem with the question mark exists, coz your graphic rendering shown on the diagram is topological nonsense. One way to partition a circle with a binary scheme looks like this.

So there is no way the question mark would even come into a consideration.

Now, test your context-independent thinking: Is the pic below related to the word "binary," "decimal," or "chicken?"

 

[doronshadmi]

Of course not. What you wrote, (S (X^2)) < (P(S) (2^X)), is gibberish. However, it is all about |S| < |P(S)|.

They are different notations of the same thing, and as a clearly show, there is a bijection bewteen the set of natural numbers and any given set of different objects, whether the amount of the objects is finite or not.

No, that's wrong, too. There of course is a bijection between any given natural number and any given element of a set (power set or otherwise). That's one of those trivial things that isn't worth dwelling on.

Since you have trouble keeping up, Doron, I'll demonstrate. Let A be a natural number, and let B be an element of some set. A bijection between A and B would be, wait for it.....wait for it....wait for it, A <-> B.

However, your statement is completely wrong if you meant there was a bijection between the natural numbers (not just one element) and the power set of any infinite set (not just one element).

Wrong. I explicitly demonstrate a bijection between the set of natural number and any given set of different objects, whether the amount of the objects is finite or not.

Your one round "closed under matrix X^2" reasoning simply can't get it

Ok, this is yet a different statement of your thesis. It's still wrong. Let's take for example S = {A}. It's power set P(S) = {{},A}. Show us, Doron, a bijection between the natural numbers and P(S). Here, I'll start you out with this table. Just fill in the left-hand side to complete the bijection:

  N       P(S)
------+--------

  ?  <->  {}
  ?  <->  {A}

  N       P(S)
------+--------

  1  <->  {}
  2  <->  {A}

 

[doronshadmi]

What “laws of probability” are you referring to?

The ? is permanent exactly because being binominal is the probability of itself.

In the case of <0,1> form the probability that ?=0 or 1 is 1:2 .

Please try to understand that “the permanent question mark” only represent your “present continuous” and apparently preferred “state” because you never try to learn anything and just choose instead to make up whatever self-contradictory, coloring book and connect the dots nonsense you want.

The Man you maybe have the chance, this time really really try to understand what is shown AND written in [qimg]http://farm6.static.flickr.com/5214/5480734023_6482feb9e5_b.jpg[/qimg]

 

[doronshadmi]

Doron you still do not grasp the accurate meaning of increasing.

The Man you still do not grasp the accurate meaning of increasing as a present continuous.

 

[doronshadmi]

The question mark is hardly permanent; it is replaced by 1 due to the apex of Pascal Triangle.



2⁰ = 1 = 1
2¹ = 2 = 1 + 1
2² = 4 = 1 + 2 + 1
2³ = 8 = 1 + 3 + 3 + 1
2⁴ =16 = 1 + 4 + 6 + 4 + 1
2⁵ =32 = 1 + 5 +10 +10 +5 + 1

and so on.

2ⁿ can be binomially distributed, if you happen to forget.

The ? is permanent exactly because being binominal is the probability of itself.

In the case of <0,1> form the probability that ?=0 or 1 is 1:2 .

 

[doronshadmi]

Your problem with the question mark exists, coz your graphic rendering shown on the diagram is topological nonsense.

If it is nonsense for you, you do not have to say anything beyond nonsense.

 

[jsfisher]

We can add bijection to the list of things you don't understand.

  N       P(S)
------+--------

  1  <->  {}
  2  <->  {A}

You seem to have omitted some of the natural numbers from what you claim to be a bijection--3 for example. It is not a bijection unless you account for every natural number, not just a couple.

 

[jsfisher]

The Man you still do not grasp the accurate meaning of increasing as a present continuous.

No, he does. And whether it goes by the name 'present continuous' or 'present progressive', it still refers to something that is happening. Again, it is you, Doron, who doesn't understand, but are trying to project that failure onto others.

 

[jsfisher]

They are different notations of the same thing, and as a clearly show, there is a bijection bewteen the set of natural numbers and any given set of different objects, whether the amount of the objects is finite or not.

No. What you did was (1) scramble together some symbols then claim they have meaning and (2) show that you don't understand the term, bijection.

 

[doronshadmi]

You seem to have omitted some of the natural numbers from what you claim to be a bijection--3 for example. It is not a bijection unless you account for every natural number, not just a couple.

jsfisher, it is not P(S) unless you account for every P(S) member.

Actually you can do it infinity many times and still you will not get a complete collection of different objects in both sides of the 1-to-1 mapping.

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

By using a cross-contexts reasoning, let us use Cantor's construction method in order to define a 1-to-1 correspondence between natural numbers and any set of different objects.

First, let us demonstrate it on some finite case, for example, the powerset of set {1,2,3}.

P({1,2,3}) = ({},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}

By using Cantor's construction method with the members of {1,2,3} and the same amount of members taken from P({1,2,3}), one enables to explicitly define each P({1,2,3}) member and put it with 1-to-1 correspondence with a natural number, such that there is no natural number that is not mapped with some P({1,2,3}) member.

Cantor's construction method constructs an explicit P({1,2,3}) member as follows:

1) The defined explicit P({1,2,3}) member is the result of a 1-to-1 correspondence between {1,2,3} members and the same amount of members taken from P({1,2,3}).

2) The explicit constructed P({1,2,3}) member is the result of some 1-to-1 correspondence, such that it includes a {1,2,3} member only if this member does not exist as one of the members of the P({1,2,3}) member that is in a 1-to-1 correspondence with it.

By using this construction method several times, one enables to define a 1-to-1 correspondence between natural numbers and P({1,2,3}) members, as follows:

Round 1 particular case:
-----------------------
1 ↔ {1} |
2 ↔ {2} | Provides {} ↔ 1
3 ↔ {3} |
-----------------------

Round 2 particular case:
-----------------------
1 ↔ { } |
2 ↔ {2} | Provides {1} ↔ 2
3 ↔ {3} |
-----------------------

Round 3 particular case:
-----------------------
1 ↔ {1} |
2 ↔ { } | Provides {2} ↔ 3
3 ↔ {3} |
-----------------------

Round 4 particular case:
-----------------------
1 ↔ {1} |
2 ↔ {2} | Provides {3} ↔ 4
3 ↔ { } |
-----------------------

Round 5 particular case:
-----------------------
1 ↔ { } |
2 ↔ {3} | Provides {1,2} ↔ 5
3 ↔ {1,3} |
-----------------------

Round 6 particular case:
-----------------------
1 ↔ { } |
2 ↔ {2} | Provides {1,3} ↔ 6
3 ↔ {1,2} |
-----------------------

Round 7 particular case:
-----------------------
1 ↔ {1} |
2 ↔ {3} | Provides {2,3} ↔ 7
3 ↔ {1,2} |
-----------------------

Round 8 particular case:
-----------------------
1 ↔ { } |
2 ↔ {1} | Provides {1,2,3} ↔ 8
3 ↔ {2} |
-----------------------

The same 1-to-1 correspondence construction method works also in the case of infinitely many objects, where in this case P({}, ... , {1,2,3,...}) is the power set of {1,2,3,...}, and also in this case we are able to define a 1-to-1 correspondence between 1 to {}, 2 to {1,2,3,...}, and any natural number > 2 with any P({}, ... , {1,2,3,...}) member between {} and {1,2,3,...}.

It is done by using Cantor's construction method |P({}, ... , {1,2,3,...})| times.

Since the set of all powersets does not exist (because ...P(P({1,2,3}}}... size is unsatisfied) and since we are generally able (by using Cantor's construction method) to construct P(S) members by mapping S members with |S| P(S) members |P(S)| times, and also to define a 1-to-1 correspondence between the natural numbers and any P(S) degree, then also the size of the set of natural numbers is unsatisfied.

In other words, we discover that the concept of collection of different objects is essentially incomplete, or in other words, the whole is greater than the amount of any collection of different objects, where the whole is the cross-contexts existence among this infinity many context-dependent objects, which their amount is unsatisfied (they can't be summed into a whole).

 

[jsfisher]

jsfisher, it is not P(S) unless you account for every P(S) member.

Which, of course, I did. S was the set, {A}, and so P(S) was the set, {{},{A}}. P(S) was a set of two members, {} and {A}.

The failure was yours when you claimed to have a bijection between this simple power set, P(S), and the natural numbers. Your mapping wasn't a bijection at all, but it did expose your most recent claims as the load of rotted dingo entrails they actually are.