Deeper than primes - Continuation

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There are some traditional mathematicians that can't recognize traditional notions (even if there is a clear comparison between standard and non-standard notations of some standard notion (and in this case the notion of base 2 is represented as <0,1>)) unless it is expressed by traditional notations.
 
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doronshadmi said:
There are some traditional mathematicians that can't recognize traditional notions unless it is expressed by traditional notations.
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What of it?
 
realpaladin said:
"Mutually independent" means they need each other to be independent of, therefore, they can not be independent.

Get it?
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So what, there is no hierarchy of dependency between them, and this is the notion of being mutual independent.

The traditional mathematician in this thread claims that P(S) rises from S, but in http://www.internationalskeptics.com/forums/showpost.php?p=8137722&postcount=890 I clearly show that Both P(S) and S (where S is a placeholder of a set) share the same type of form, such that there is an element of that form in P(S), but not in S.

In other words, given S, it is incomplete because not all the possible forms of a given type are its members.

Once again:

Lets improve the last part of http://www.internationalskeptics.com/forums/showpost.php?p=8137722&postcount=890 (also be aware of
a) There is no additional distinct element (there is nothing) between two distinct elements, which are constructed according to the given form.
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which follows traditional maths' point of view):

Since (every P(S) of distinct elements of a given form has a power set of distinct elements of that given form) AND (for every given set under placeholder S there is a distinct element of that form, which is an element of P(S) but it is not an element of that set) AND (the largest power set of distinct elements of that form does not exist) then no given set (under placeholder S) with distinct elements of that given form is complete (no matter if k is the cardinality of a finite or non-finite non-empty set under placeholder S).


Get it?
 
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doronshadmi said:
So what, there is no hierarchy of dependency between them.

Get it?
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Yeah! It's not like adverbs and nouns are ever related. They are just words. It doesn't matter how you use them. They have no relationship to each other...at least in Doron's world.
 
doronshadmi said:
There are some traditional mathematicians that can't recognize traditional notions (even if there is a clear comparison between standard and non-standard notations of some standard notion (and in this case the notion of base 2 is represented as <0,2>)) unless it is expressed by traditional notations.
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Ah! Lost ground on that whole traditional notation lie, so now you shift to the change the subject technique. You leap from exponentiated tuples to powers of 2 and now you attempt to seamlessly equate all that to the previously unmentioned binary notation.

The starting point was still not traditional mathematical notation, so you still lied there. Powers of 2 and binary are not synonymous, so there's another lie. And there's still that gap between power sets and your silly notation. I predict another lie is in the works to attempt to obscure that Doron-failure, too.
 
Each time when the traditional mathematician in this thread can't comprehend a given notion, even if it is translated to the standard notion, he calls it a lie.

Say no more.
 
doronshadmi said:
Each time when the traditional mathematician in this thread can't comprehend a given notion, even if it is translated to the standard notion, he calls it a lie.

Say no more.
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You are in a rut, Doron, with this same sequence of obfuscation again and again. You need a new shtick. This one's tired.
 
Since the posters here did not get the previous version of my argument, let's improve it according to their replies, by doing it in baby steps.

Here it is:

Let's follow the traditional mathematical notion in order to show that given any non-empty set of distinct elements with a given form, no possible set (where S is a placeholder of a given set of that form) is complete, because:

a) There is no additional distinct element (there is nothing) between two distinct elements, which are constructed according to the given form.

b) Given the powerset P(S) of distinct elements of a given form, S is a placeholder for any possible set of distinct elements of that form.

c) The largest powerset P(S) of distinct elements of that form does not exist.


1) <0,1> is the common notation of the forms that is equivalent to base 2 (where the possible notations under this base are 0 and 1, notated as <0,1>).

2) ^ is the power operation.

3) k is the (finite or infinite) cardinality of any given set of the considered distinct forms under placeholder S.

4) <0,1>^k(where k=1 to ∞) expression (which is equivalent to 2^k expression) is used for both S (where S is a placeholder of a given set of the considered form) and P(S), where P(S) is the powerset of any possible set under place holder S, such that the cardinality of P(S) is determined by 2^k(where k=1 to ∞) and the cardinality of any set under placeholder S is determined by k(where k=1 to ∞).

5) P(S) and any possible set with cardinality k (under placeholder S) have the same type of elements, notated by <0,1> symbols.

6) The order of the presentation of P(S) and the sets under placeholder S, is insignificant, and all we care is the shared type of forms among P(S) (which its cardinality is determined by 2^k(where k=1 to ∞)) and the sets under placeholder S (which their cardinality is determined by k(where k=1 to ∞)).

Here is a finite example (k is finite):

By <0,1>^1 (which is equivalent to 2^1)
P(S)=
{0,1}
where the sets (with cardinality k) under placeholder
S=
(
{0} → 1
or
{1} → 0
)

By <0,1>^2 (which is equivalent to 2^2)
P(S)=
{00,01,10,11}
where some possible sets (with cardinality k) under placeholder
S=
(
{
10,
11
} → 00
or
{
10,
00
} → 01
or
{
00,
11
} → 10
or
{
01,
10
} → 11
)

...

Nothing is changed even if k=∞ (in this case the cardinality of P(S) is 2^∞ and the cardinality of any possible set under placeholder S is ∞) because also in this case, given a set under placeholder S with distinct elements of the considered form, there is a distinct element of that form, which is an element of P(S) but it is not an element of the considered set under placeholder S.

Since (the distinct elements of the considered form are elements of powerset P(S)) AND (for every considered set -with cardinality k- under placeholder S, there is a distinct element of that form, which is an element of P(S) but it is not an element of that set) AND (the largest powerset of distinct elements of that form does not exist) then no given set (under placeholder S) with distinct elements of that given form is complete (no matter if any set (with the considered form) under placeholder S, has finite or infinite cardinality k(where k=1 to ∞)).

This post can be simplified by further editing that does not change the result.
 
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realpaladin said:
http://www.internationalskeptics.com/forums/showthread.php?t=232772
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Irrelevant.

Next time please be honest and simply say that you have to clue of http://www.internationalskeptics.com/forums/showpost.php?p=8140608&postcount=933 subject even if it is given by baby steps.

You simply do not get that there are at least 2^∞ sets (where each set has cardinality ∞) under placeholder S, where each one of them has ∞ members of <0,1> form and no one of them is complete, because there is always a given element of <0,1> form, which is a member of P(S), but it is not a member of anyone of the 2^∞ sets (where, again, each one of them has ∞ members of <0,1> form, and now we realize that no one of them is complete).

This fact is not changed, whether k (of the expression 2^k) is finite or infinite.
 
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doronshadmi said:
Since the posters here did not get the previous version of my argument, let's improve it according to their replies, by doing it in baby steps.

Here it is:

Let's follow the traditional mathematical notion in order to show that given any non-empty set of distinct elements with a given form, no possible set (where S is a placeholder of a given set of that form) is complete, because:
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Wrong. The set can be complete. My set S is the set of positive single digit numbers when written out in English have only 4 letters in them. My set S is (four, five, nine). It is complete. Since you claim that it is incomplete, please provide any missing elements of this set.
 
doronshadmi said:
Let's follow the traditional mathematical notion in order to show that given any non-empty set of distinct elements with a given form, no possible set (where S is a placeholder of a given set of that form) is complete,
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So many extraneous words. Stripping the preceding of the Doron-ese gibberish and word salad, we are left with: To be shown: no set is complete.

...because:

a) There is no additional distinct element (there is nothing) between two distinct elements, which are constructed according to the given form.
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Since "between" is not a set concept, this is irrelevant.

b) Given the powerset P(S) of distinct elements of a given form, S is a placeholder for any possible set of distinct elements of that form.
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This is gibberish coupled with Doron's inverted processing. Given a set, S, he could then consider its power set, but not the other way around.

By virtue of being meaningless, this one is also irrelevant.

c) The largest powerset P(S) of distinct elements of that form does not exist.
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It is impossible to say if Doron intended this to be a premise or a deduction. Either way, it has no basis and gets chucked into the trash heap with the first two.

1) <0,1> is the common notation of the forms that is equivalent to base 2 (where the possible notations under this base are 0 and 1, notated as <0,1>).
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Common to whom? Be that as it may, this is a mighty tortured way to introduce binary numbers.

2) ^ is the power operation.

3) k is the (finite or infinite) cardinality of any given set of the considered distinct forms under placeholder S.
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Just saying "Let k = |S|" would have been too easy?

4) <0,1>^k(where k=1 to ∞)...
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So, we now abandon the previous meaning assigned to k.

...expression (which is equivalent to 2^k expression)
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No, they are not equivalent. The former is being suggested as binary notation (I say suggested because, as always, Doron never completes his thoughts), and the best that can be inferred for <0,1>^4, as an example, is some non-specific 4-character string of 0's and 1's; 2^4, on the other hand, is 16.

...is used for both S (where S is a placeholder of a given set of the considered form) and P(S), where P(S) is the powerset of any possible set under place holder S, such that the cardinality of P(S) is determined by 2^k(where k=1 to ∞) and the cardinality of any set under placeholder S is determined by k(where k=1 to ∞).
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Well, this certainly jumps around, doesn't it? It also assumes quite a bit. To this point in the presentation, the relationship between |S| and |P(S)| has not been established. And given the inverted approach Doron insists on abusing, he'll be hard-pressed to establish that |S| is the base-2 log of |P(S)| nor that the arithmetic is meaningful for non-finite sets.

5) P(S) and any possible set with cardinality k (under placeholder S) have the same type of elements, notated by <0,1> symbols.

6) The order of the presentation of P(S) and the sets under placeholder S, is insignificant, and all we care is the shared type of forms among P(S) (which its cardinality is determined by 2^k(where k=1 to ∞)) and the sets under placeholder S (which their cardinality is determined by k(where k=1 to ∞)).
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Gibberish. Fortunately, Doron will follow up with an example to increase the muddle:

Here is a finite example (k is finite):

By <0,1>^1 (which is equivalent to 2^1)
P(S)=
{0,1}
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Last time anyone other than Doron checked, 2^1 = 2, so no, the alleged equivalence isn't there.

Doron still persists in his ass-backwards P(S) leads to S nonsense, too. Apparently, too, we are to infer that <0,1>^n really means the set of all possible n-character strings of 0's and 1's.

where the sets (with cardinality k) under placeholder
S=
(
{0} → 1
or
{1} → 0
)
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This is a rectal extraction. Doron has provided no basis for this. This also contradicts the meaning of <0,1>^n invoked for P(S).

Be that as it may, though, neither possibility for set S leads to the alleged power set, above. So, not only is Doron just scribbling according to some arbitrary, unstated, variable rules, he doesn't even get the right result when he's done.

...<second incompetent example, but one with an even more bizarre process for generating S>...

Nothing is changed even if k=∞
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Yes, nothing changes. The whole thing continues to be fantastically wrong.
 
doronshadmi said:
"Equivalent to ..." is not the same as "the same as ..." but our traditional mathematician here does not know that.
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Perhaps, Doron, you should work on those abysmal reading comprehension skills of yours. No one but you has suggest equivalence means the same as.

Moreover, the equivalence you claimed does not exist. Making things up -- which you so often do -- doesn't make them true.

He also unable to get that P(S) members and the members of the sets under placeholder S, are of the same form.
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That would be the form you tried to demonstrate and failed, miserably.

As a result he totally misses http://www.internationalskeptics.com/forums/showpost.php?p=8140608&postcount=933.

Say no more.
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Not much more needs to be said. Doron makes an incompetent post. Doron attempts to pass gibberish as intelligence. Doron defends it by blaming everyone else.

Nothing new to see here.
 
jsfisher said:
Doron makes an incompetent post. Doron attempts to pass gibberish as intelligence. Doron defends it by blaming everyone else.
Click to expand...

That about sums it up.
 
Some traditional mathematician says: :"That would be the form you tried to demonstrate and failed, miserably."

Argument:

Given a non-empty set with a given form, there is a power set of these forms, such that there is an explicit form that is a member of the power set, but it is not a member of the given set.

As a result the given set is incomplete, because not all the given forms are its members.



Proof by construction:

The form is based on 0;1 symbols.

S={0}

P(S)={0,1} where form 1 is an explicit member of P(S) but it is not a member of S.


S={1}

P(S)={0,1} where form 0 is an explicit member of P(S) but it is not a member of S.


By using Diagonalization among the forms that are based on 0;1 symbols, we always are able construct an explicit form that is a member of the power set, but it is not a member of the given set.

S=
{
10,
11
}

P(S)={00,01,10,11} where form 00 is an explicit member of P(S) that is not a member of S.


S=
{
10,
00
}

P(S)={00,01,10,11} where form 01 is an explicit member of P(S) that is not a member of S.


S=
{
00,
11
}

P(S)={00,01,10,11} where form 10 is an explicit member of P(S) that is not a member of S.


S=
{
01,
10
}

P(S)={00,01,10,11} where form 11 is an explicit member of P(S) that is not a member of S.


...


Things do not change even if the 0;1 forms are infinite, as shown by Cantor.

Since the largest P(S) does not exist, we can conclude that no collection with distinct forms is complete.

Q.E.D
 
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doronshadmi said:
Argument:

Given a non-empty set with a given form, there is a power set of these forms, such that there is an explicit form that is a member of the power set, but it is not a member of the given set.

As a result the given set is incomplete, because not all the given forms are its members.
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So, a set is Doron-incomplete if it doesn't contain things that are not a member of it? Gee, that makes the Doron-incomplete concept fundamentally useless.

Proof by construction:

The form is based on 0;1 symbols.

S={0}

P(S)={0,1} where form 1 is an explicit member of P(S) but it is not a member of S.
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No. If S = {0} then P(S) = {{}, {0}}. Moreover, neither member of P(S) is a member of S.

Doron, until such time as you stop blending your thoughts and uttering confused gibberish, your nonsense hits a brick wall.

By the way, this ass-backwards excursion into binary numbers adds a level of confusion (for you) and adds no clarity (for us).
 
"No. If S = {0} then P(S) = {{}, {0}}." is a good example of a mathematician that can't get any notion beyond the traditional notation.

Say no more.
 
jsfisher said:
Not much more needs to be said. Doron makes an incompetent post. Doron attempts to pass gibberish as intelligence. Doron defends it by blaming everyone else.
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Q.E.D.
 
doronshadmi said:
zooterkin said:
And what if

S=
{
00,
01,
10,
11
}


? What element is missing then?
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First please construct its power set, in order to explicitly construct the missing member.
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Ok, I see that you still do not get the generality of 0;1 form.

Please be aware that in this particular case we are using here distinct members under powersets with cardinality ...2^(2^(2^(2^1)))... and the number of 0;1 symbols must grow in order to save their distinction w.r.t each other.

By using it, the powerset of S with cardinality 4 is:
Code: 
0 0 1 1
0,1,0,1
-------
P(S)=
{
0 0 0 0,
1 0 0 0,
0 1 0 0,
1 1 0 0,
0 0 1 0,
1 0 1 0,
0 1 1 0,
1 1 1 0,
0 0 0 1,
1 0 0 1,
0 1 0 1,
1 1 0 1,
0 0 1 1,
1 0 1 1,
0 1 1 1,
1 1 1 1,
}

such that each member is made of 4 0;1 symbols and the cardinality of S is 4, where the cardinality of P(S) is 2^4=16

Now please use Diagonalization among the 4 forms that are based on 0;1 symbols as follows:

S=
{
1001,
0111,
0111,
0001,
}

such that 0000 is an explicit member of P(S) that is not a member of S, etc., as shown in http://www.internationalskeptics.com/forums/showpost.php?p=8144371&postcount=942.

In other words, even if the cardinality of P(S) 2^∞, there is S with ∞ members and both S AND P(S) have members with 0;1 form, where each member has ∞ symbols.

Since (P(S) and S have the same construction of forms, but there are members of this construction in P(S), but not in S) AND (the largest P(S) does not exist) we can conclude that no collection with distinct members is complete (whether the cardinality > 0 is finite or not).
 
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The traditional mathematician here does not understand yet the result of the general use of 0;1 from among cardinality 2^k(where k = 1 to ∞) whether the cardinality grows by ...2^(2^(2^(2^1)))... or not, where by this general use S and P(S) have forms with the same number of 0;1 symbols.

The reason: he can't get any notion beyond the traditional notation.

<snip> is its best.
 
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doronshadmi said:
The traditional mathematician here does not understand yet the result of the general use of 0;1 from among cardinality 2^k(where k = 1 to ∞) whether the cardinality grows by ...2^(2^(2^(2^1)))... or not, where by this general use S and P(S) have forms with the same number of 0;1 symbols.

The reason: he can't get any notion beyond the traditional notation.

<snip> is its best.
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The ''traditional'' notation works. Your made-up notation does not work.
 
jsfisher said:
Doron, why does it surprise you so that a set of 16 elements will have some elements that do not appear in a set of only 4?
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Doron, in your dismissive fit you neglected to address the question posed to you. I've quoted it above in full for your convenience.

In due time, too, we can explore your unfounded belief that 2^|S| behaves as you think for non-finite sets.
 
Sorry, haven't tuned in in quite a while. Just thought I ask: hey Doron, anything published yet? Any results at all? Anything? Or still just wanking on internet forums over the attention some bored folks give you there?
 
He had some sidekicks, then he managed to put them off as well. Now he is with the old stick of:

- If you get it, you get it.
- If you don't get it, you don't get it.
- Only Doron gets it.
 
Let's simplify the argument.

Given a collection of distinct elements that are constructed by 0;1 symbols, such that the size of each element is the same as the size of the collection, there is always an explicit element of 0;1 symbols that is constructed by using Diagonalization, which is not an element of the given collection, no matter if the considered size is finite or infinite.

By following this fact, we can conclude that no such collection of distinct elements is complete.
 
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doronshadmi said:
Let's simplify the argument.

Given a collection of distinct elements that are constructed by 0;1 symbols, such that the size of each element is the same as the size of the collection, there is always an explicit element of 0;1 symbols that is constructed by using Diagonalization, which is not an element of the given collection, no matter if the considered size is finite or infinite.

By following this fact, we can conclude that no such collection of distinct elements is complete.
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So what? I think that Cantor mentioned that.
 
doronshadmi said:
Let's simplify the argument....
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No need for all that binary nonsense nor pretending things to be power sets when they really aren't. It is just a very simple idea you seem hell-bent on presenting in the most convoluted manner possible. The question remains, though: Why does it surprise you so that a set of 16 elements will have some elements that do not appear in a set of only 4?
 
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