Cont: Deeper than primes - Continuation 2

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phiwum said:
but it is nonsense to think that there are two different numbers 2.
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Unless redundancy is not ignored, for example:

By the ,so called, "established mathematics" [A] and [A,A] are addressed by the distinct pairs {(A,1),(A,2)} exactly because each pair defines different redundancy degree.

A = {1,2,3,...} (the set of all natural numbers).

If redundancy degree > 1, in case of, for example [A,A] there can be more than 1 degrees of freedom of the mapping between the As copies, as addressed in http://www.internationalskeptics.com/forums/showpost.php?p=12158706&postcount=2864.

A = A is simply the particular case of the pair {(A,1)}, so nothing is fundamental here.
 
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jsfisher said:
See? 2 may not be a member of the set containing 2.
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Wrong, [2,2] is different than [2] exactly as the members of {(2,2),(2,1)} are different of each other by their redundancy degrees.

In all the considered multisets above (whether the redundancy degree = or > 1) the copies of 2 are members of a given multiset.
 
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doronshadmi said:
Wrong, [2,2] is different than [2]
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Why do you continue to respond to arguments no one has made?

Of course the multiset [2, 2] is different from the multiset [2]. No one has suggested otherwise, but you seem compelled to dwell on it.
 
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doronshadmi said:
jsfisher, why do you take only a part of my reply to your wrong argument in http://www.internationalskeptics.com/forums/showpost.php?p=12163196&postcount=2922 about 2 as if it is not a member of a given multiset ?
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My statement about 2 may not being a member of {2} was a bit of mockery. It is a clearly bogus claim you'd made in the past. In its larger form, you had claimed although 2 was in the set of even numbers, and in the set of prime numbers, and in the set of integers, and in so many other sets it either wasn't the same 2 in each case or it couldn't be in all of them simultaneously.

You continue this conceptual failure with your understanding of multisets, but that aside for a moment, the multisets [2, 2], [2, 2, 2], and [2] are all different, yet you continue to fly off on a tangent about the rest of us saying otherwise.

Now, as for the reason those multisets are all different, it is simply a matter of definition for multisets. By definition, distinct members may appear more than once in a multiset. It has nothing to do with the tortured vocabulary invention you keep trying to force on us.

I have found it best to limit your posts up to just the first bit of wrongness, so in responding, I clipped the quote at such a point.


ETA: By the way, my post you linked to made no reference to multisets.
 
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jsfisher said:
See? 2 may not be a member of the set containing 2.
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My phrasing was ambiguous and probably misleading for some, especially those for whom English was not their first language. I apologize.

The meaning I meant to convey, the meaning as originally expressed by Doronshadmi when he made the claim was: 2 is not necessarily a member of {2}. It could be a different 2, or maybe 2 was busy off doing something else.
 
jsfisher said:
My statement about 2 may not being a member of {2} was a bit of mockery.
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In other words, you wrote a reply that has nothing to do with http://www.internationalskeptics.com/forums/showpost.php?p=12163184&postcount=2921, or in other words, another red herring of yours.

jsfisher said:
You continue this conceptual failure with your understanding of multisets, but that aside for a moment, the multisets [2, 2], [2, 2, 2], and [2] are all different
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Exactly, they have different redundancy degrees among members' copies.

jsfisher said:
By definition, distinct members may appear more than once in a multiset.
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And this is exactly the reason of why they have more than one degree of freedom, in case that at least two copies of infinite sets are mapped, as addressed in http://www.internationalskeptics.com/forums/showpost.php?p=12158706&postcount=2864.

jsfisher said:
It has nothing to do with the tortured vocabulary invention you keep trying to force on us.
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Worng jsfisher, you are the one that forces the notion of A=A, which is the particular case of {(A,1)}, as if it the only possibility in order to do Math.
 
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jsfisher said:
The meaning I meant to convey, the meaning as originally expressed by Doronshadmi when he made the claim was: 2 is not necessarily a member of {2}. It could be a different 2, or maybe 2 was busy off doing something else.
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No, 2 is an element. If this element is a member of some set, then this case may be written as {2} , {1,2,3} , {2, 4, 6, ...} , {{2}} etc. ad infinitum.

The simple fact in this case is that nothing forces 2 to be a member of some non-empty set, since 2 can also be taken as its own mathematical universe even if it is taken in terms of a "pure" set like {{},{{}}} (Von Neumann constucation) or {{{}}} (Zeremalo costruction).

Moreover, if 2 is not taken in terms of a "pure" set, then 2 is not a member of any framework that is defined in terms of "pure" sets.

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So, jsfisher your red herring is not going to save you from your own artificial restrictions.
 
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Little 10 Toes said:
Doronshadmi, why do you keep making up your own terms and why aren't you using the correct terms?
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I do not use the conventional terms, and by doing so I discover the different degrees of freedom of the mapping between at least two copies of a given infinite set under a given multiset, as done here.

Generally, Mathematics is much more richer than any particular view, including my non-conventional view of the considered subject.
 
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doronshadmi said:
In other words, you wrote a reply that has nothing to do with http://www.internationalskeptics.com/forums/showpost.php?p=12163184&postcount=2921, or in other words, another red herring of yours.
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If you really think that, then you are truly missing the meaning of your own words. You see, phiwum wrote:

phiwum said:
it is nonsense to think that there are two different numbers 2.
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This is a perfectly correct statement. Mathematics doesn't need a Xerox machine. No matter how it is expressed, a single 2 is sufficient for all its needs.

You then contradicted phiwum's statement with:

doronshadmi said:
Unless redundancy is not ignored
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(By the way, you might want to look up the meaning of that word. Its really meaning is rather amusing in light of how you continue to abuse it.) You continue to labor under a conceptual block that prevents you from accepting that a multiset can have, for example, 2 as a member of it 17 times without the need for any copies.

And this all ties quite nicely to comments you have made in the past about 2 and its possible non-membership in {2}. My post was completely topical and stood as a reminder to phiwum that you think copies are necessary.

So, no, no red herrings were used in the making of my posts.
 
Out of curiosity, Doronshadmi, have you figured out yet you do not need to appeal to multisets to define mappings from N to N? If you wanted to discuss mappings, you could have gone directly there.
 
jsfisher said:
This is a perfectly correct statement. Mathematics doesn't need a Xerox machine. No matter how it is expressed, a single 2 is sufficient for all its needs.
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You jsfisher, and phiwum are simply closed under the notion of A=A that is simply the particular case of {(A,1)}.

jsfisher said:
(By the way, you might want to look up the meaning of that word. Its really meaning is rather amusing in light of how you continue to abuse it.)
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It has more than one meaning, so you have no case.

jsfisher said:
You continue to labor under a conceptual block that prevents you from accepting that a multiset can have, for example, 2 as a member of it 17 times without the need for any copies.
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It has 17 copies, but since your notion is restricted to {(A,1)}, you are not aware of it.

jsfisher said:
And this all ties quite nicely to comments you have made in the past about 2 and its possible non-membership in {2}. My post was completely topical and stood as a reminder to phiwum that you think copies are necessary.
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Also http://www.internationalskeptics.com/forums/showpost.php?p=12163696&postcount=2930 is not closed under your convectional notions.

jsfisher said:
So, no, no red herrings were used in the making of my posts.
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It is a red herring, and in both paths your restricted notions can't catch the red herring nor the fox.

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Generally, Mathematics is much more richer than any particular view, including my non-conventional view of the considered subjects.
 
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jsfisher said:
Out of curiosity, Doronshadmi, have you figured out yet you do not need to appeal to multisets to define mappings from N to N? If you wanted to discuss mappings, you could have gone directly there.
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Out of curiosity, jsfisher, have you figured out that multiset is generalization of the concept of set, such that the mappings from N to N (such that redundancy is ignored) is only the particular case of redundancy = 1 ({(N,1)}). If you wanted to discuss mappings you can't cover the considered subject by {(N,1)} particular case.

Jsfisher, under {(N,2)} ([N,N]) bijection is not the only possible mapping, but you are unaware of it since your notions of the considered subject are restricted to {(N,1)}.
 
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To those who follow the last discussion, jsfisher and phiwum understand, for example [A,A] in terms of {(A,1),(A,1)} and not in terms of {(A,2)} as I do.

In other words, their notion is restricted only to (A,1).
 
doronshadmi said:
To those who follow the last discussion, jsfisher and phiwum understand, for example [A,A] in terms of {(A,1),(A,1)} and not in terms of {(A,2)} as I do.
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Phiwum and I understand multisets in terms of how the concept is defined. You, Doronshadmi, don't bother with such formalities.

By the way, will you as some point tell us all what you had in mind with this (A,1) and (A,2) notation or will that be just another mystery of doronetics?
 
jsfisher said:
Phiwum and I understand multisets in terms of how the concept is defined.
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Your understanding of, for example, [A,A] is understood in terms of {(A,1),(A,1)}.

jsfisher said:
You, Doronshadmi, don't bother with such formalities.
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Wrong, you and Phiwum don't bother to comprehend the formality of [A,A] in terms of {(A,2)}.

jsfisher said:
By the way, will you as some point tell us all what you had in mind with this (A,1) and (A,2) notation or will that be just another mystery of doronetics?
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No mystery. (A,2) has redundancy degree > 1, where (A,1) has redundancy degree = 1.

Under redundancy degree > 1 two copies of infinite sets have more than 1 degrees of freedom of the mapping between them, as very simply addressed in http://www.internationalskeptics.com/forums/showpost.php?p=12158706&postcount=2864.

{(A,2)} is a mystery if observed in terms of {(A,1),(A,1)}.
 
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