New Pigeonhole principle - Unknowability of there being infinitely many counter examples of Goldbach Conjecture.

[khongdo...@gmail.com]

To sci.logic,

One of the most difficult issues I've been facing with in sci.logic is to explain
sufficiently clear why it's true that it's not possible (impossible) to know or
verify whether or not there are infinitely many counter examples of Goldbach
Conjecture.

But I've found a way and in this thread I'll elaborate that way. I's a new kind of
Pigeonhole principle, albeit it's about unknowability - not knowability - like the
canonical Pigeonhole principle.

Contrary to what some might have suspected, I'd like to be understood not just talked
to, so in this thread I'll only talk to ones who would readily understand the
terminologies and few logic behind my explanation: I'm sorry in advance I wouldn't
explain to those who don't understand. In particular, cranks and trolls are _not_ welcomed.

Imagine the collection K of uncountably many successor-function bars each of which is
infinitely long in one direction (left-to-right). Each bar would have infinitely many
holes per the following characterizations:

C1: Each hole has a little room beneath that one and only one pigeon can dwell. But
right behind each hole on the surface there's a finite marking of many sticks and
the first hole on the successor chain has zero marking (i.e. zero stick if one
tries to count). And each hole's marking would be unique: there are no two holes
with the same marking - with the same number of sticks.

The length of the marking is the number of sticks in the sequence.

C2: There are infinitely many pigeons one can 1-1 map the set of all pigeons to the
set of holes in any successor-function bar. Let P be the set of all these pigeons
in this discussion.

C3: Each pigeon would have one finite sequence of DNA with a certain number of
DNA-ladders: there's exactly one pigeon with no sequence (hence no DNA-ladder)
and no two pigeons would have the same DNA sequence. The length of the sequence
is the number of ladders in the sequence.

C4: In _each bar_ of the collection K, one can if one wish arrange the pigeons in the
below manner:

(1) The first hole, the second hole, and the third hole, will each have one
pigeon stand next to the hole, and the three pigeons would have the
DNA-sequences of the following length: zero, one, two, respectfully.

(2) Each hole thereafter with the length of the marking being a non-prime number
will have the corresponding pigeon of the _same length_ (of its DNA
sequence).

(3) The conditions (1) and (2) above would hold true to each bar in the
collection K of successor-function bars: the same pigeons would each stand
next to the same hole of each bar - not dwelling in the compartment (room)
beneath!

(4) For each hole, of which the length l1 of the marking is a prime number,
there's exactly one pigeon of which the length l2 of its DNA sequence is
a prime number: but l1 and l2 are not necessarily the same. In fact, it's
given here as part of the characterization of K that for any two different
bars, there are at least two distinct pigeons with distinct odd prime number
DNA sequence lengths dwell in the compartments beneath the corresponding two
holes: the the length of each of the two hole wouldn't be equal to the
length of the DNA sequence of the corresponding dwelling pigeon.

Iow, an (odd) prime pigeon would dwell in a an odd prime hole-compartment but
the two (in length ) might be of different (odd) prime numbers - in any two
distinct successor-function bars of K!

Given the characterization of the collection K as above, then *obviously* if one
claims that out of the uncountably many bars of K, he/she has chosen by Choice one
bar as his/her *the standard* bar denoted by "N_bar", then the following statement
(*) is true of this "the standard" N_bar:

(*) It's impossible to know, to verify, for any given hole h that is of even-length
of the marking, there are two pigeons dwelling in the the compartments and the
sum of their DNA-sequence lengths is equal to the length of the marking of the
hole h.

If you understand and see that (*) is true in this precise analogy (of the new New
Pigeonhole principle) as above, you'd understand that undecide(N |= cGC) where:

- N is the purported "the standard" language structure for the language oif
arithmetic.
- cGC <-> "There are infinitely many counter examples of Goldbach Conjecture".
- undecide(S) <=> "It's impossible to know, to verify, the meta statement S is true".

[khongdo...@gmail.com]

For clarity, let me re-phrase (*):

(*) It's impossible to know, to verify, if it's so true, for any given hole h that is
of even-length in its marking, there are two pigeons dwelling in two compartments

[Peter Percival]

khongdo...@gmail.com wrote:
> To sci.logic,
>
> One of the most difficult issues I've been facing with in sci.logic is to explain
> sufficiently clear why it's true that it's not possible (impossible) to know or
> verify whether or not there are infinitely many counter examples of Goldbach
> Conjecture.
>
> But I've found a way and in this thread I'll elaborate that way. I's a new kind of
> Pigeonhole principle, albeit it's about unknowability - not knowability - like the
> canonical Pigeonhole principle.
>
> Contrary to what some might have suspected, I'd like to be understood not just talked
> to, so in this thread I'll only talk to ones who would readily understand the
> terminologies and few logic behind my explanation: I'm sorry in advance I wouldn't
> explain to those who don't understand.

There is no point in explaining to anyone else!

> In particular, cranks and trolls are _not_ welcomed.

So if someone finds fault in what you write, you won't correct the fault
you'll just call them a crank and a troll.

> Imagine the collection K of uncountably many successor-function bars each of which is

I know what a successor function is, but what is a bar in this context.

--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan

[Mike Terry]

<khongdo...@gmail.com> wrote in message
news:b1fcaff4-55c1-408f...@googlegroups.com...

Just to be clear, the 1st hole has no sticks, the second hole has 1 stick,
the third hole has two sticks etc.? So a successor bar looks a bit like
this :
[VIEW WITH FIXED FONT!]


+----------------------------------------------------------------------
| (1st hole) (2nd hole) (3rd hole) (4th hole)
bar: | +--------+ +--------+ +--------+ +--------+
| | | | I | | II | | III | ......
| +--------+ +--------+ +--------+ +--------+
|
| (has 0 sticks) (1 sticks) (2 sticks) (3 sticks)

+----------------------------------------------------------------------

(So we could say more clearly and with fewer words that the holes are
labelled 0,1,2,3...)

>
> The length of the marking is the number of sticks in the sequence.
>
> C2: There are infinitely many pigeons one can 1-1 map the set of all
pigeons to the
> set of holes in any successor-function bar. Let P be the set of all
these pigeons
> in this discussion.
>
> C3: Each pigeon would have one finite sequence of DNA with a certain
number of
> DNA-ladders: there's exactly one pigeon with no sequence (hence no
DNA-ladder)
> and no two pigeons would have the same DNA sequence. The length of the
sequence
> is the number of ladders in the sequence.

Just to be clear, similar to the question above, is there a pigeon with 1
DNA-ladder? What about 2 DNA ladders? And similarly 3, 4, 5, ... DNA
ladders? [I.e. we could just say more clearly and with fewer words that the
pigeons are labelled 0,1,2,3....)

>
> C4: In _each bar_ of the collection K, one can if one wish arrange the
pigeons in the
> below manner:

Just to be clear, do all the pigeons stand next to a hole? And do all the
holes have a pigeon standing next to it? (I think from below the second
question is a "yes", and I expect the first is a "yes" too?)

[khongdo...@gmail.com]

Right. Except the real intention is to use the numerals which are solely based
on the symbols of the language of arithmetic. Also since this is an analogy, I
intended to use the English words more than technical terms, but this is just
a minor judgement call. (But thanks for the suggestion).

> > The length of the marking is the number of sticks in the sequence.
> >
> > C2: There are infinitely many pigeons one can 1-1 map the set of all
> pigeons to the
> > set of holes in any successor-function bar. Let P be the set of all
> these pigeons
> > in this discussion.
> >
> > C3: Each pigeon would have one finite sequence of DNA with a certain
> number of
> > DNA-ladders: there's exactly one pigeon with no sequence (hence no
> DNA-ladder)
> > and no two pigeons would have the same DNA sequence. The length of the
> sequence
> > is the number of ladders in the sequence.
>
> Just to be clear, similar to the question above, is there a pigeon with 1
> DNA-ladder? What about 2 DNA ladders? And similarly 3, 4, 5, ... DNA
> ladders? [I.e. we could just say more clearly and with fewer words that the

> pigeons are labelled 0,1,2,3....).

Thanks for the correction: "ladder" was misused here. What I meant there is: a
DNA sequence itself is a ladder, while the number of DNA rungs is the length of
the sequence (of the ladder). Each pigeons has exactly one DNA sequence (ladder).
In fact one DNA sequence can be represented by a numeral too, just like in the
case of one hole being labelled by one numeral.

It's just a pigeon whose DNA sequence numeral represents a prime might be in a hole
of which the label represents a different prime: but which we don't know how far
apart these two different primes be, or which one of the 2 primes is less than the
other!

> > C4: In _each bar_ of the collection K, one can if one wish arrange the
> pigeons in the
> > below manner:
>
> Just to be clear, do all the pigeons stand next to a hole? And do all the
> holes have a pigeon standing next to it? (I think from below the second
> question is a "yes", and I expect the first is a "yes" too?)

For better clarification:

- There's only one set of infinitely many pigeons.

- There's also only one infinite bar with infinitely many holes, each of which
has a marking (or label) next to it, and also has one compartment underneath.

- For each hole there corresponds exactly one pigeon, which either stands right
next to the hole or dwells in the hole's compartment. Likewise, for each pigeon
there correspond one hole, where the pigeon stand next to or dwells in its
compartment.

- The collection K mentioned previously is the collection of uncountably many
combinations of pigeons and holes of the only bar, the combinations having been
stipulated up to this point.

Hope this has helped to clarify.

[graham...@gmail.com]

MIA CULPA! I spent 6 months solving GoldBach Conjecture

and neither CLAY nor HARPER COLLINS put up their promised $1000000


p(n in PR') = 2 / LOG(n) (n is ODD)

Works same as PRIMES!


http://unsolvedproblems.org/index_files/Solutions.htm

[Peter Percival]

So 0, S0, SS0, ... ?

[Rupert]

On Sunday, October 30, 2016 at 6:41:15 PM UTC+1, khongdo...@gmail.com wrote:
> To sci.logic,
>
> One of the most difficult issues I've been facing with in sci.logic is to explain
> sufficiently clear why it's true that it's not possible (impossible) to know or
> verify whether or not there are infinitely many counter examples of Goldbach
> Conjecture.
>
> But I've found a way and in this thread I'll elaborate that way. I's a new kind of
> Pigeonhole principle, albeit it's about unknowability - not knowability - like the
> canonical Pigeonhole principle.
>
> Contrary to what some might have suspected, I'd like to be understood not just talked
> to, so in this thread I'll only talk to ones who would readily understand the
> terminologies and few logic behind my explanation: I'm sorry in advance I wouldn't
> explain to those who don't understand. In particular, cranks and trolls are _not_ welcomed.
>
> Imagine the collection K of uncountably many successor-function bars each of which is
> infinitely long in one direction (left-to-right).

What is a successor-function bar?

[Dan Christensen]

You might find this useful: http://www.dcproof.com/pigeonhole.htm

It is my formal development of a non-numeric version of the Pigeonhole Principle (in the DC Proof format).


Dan

Download my DC Proof 2.0 software at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[Peter Percival]

Dan Christensen wrote:
> You might find this useful: http://www.dcproof.com/pigeonhole.htm
>
> It is my formal development of a non-numeric version of the Pigeonhole Principle (in the DC Proof format).

The last thing Nam want's to do is learn from others.

[khongdo...@gmail.com]

Ok. So it's not bar (bar is too small); it's a (construction) beam:

http://www.aaanimalcontrol.com/pigeonroost.htm

It's an abstract infinite beam from left to right having infinitely many holes each
of which has a marking (a sequence of symbols) and a compartment right inside where
a pigeon might dwell.

[khongdo...@gmail.com]

A pigeon's DNA sequence (of symbols) is an intrinsic identity of the pigeon, from
which some truth would be based: such as whether or not it's DNA sequence represent
the smallest prime.

A beam-hole's marking is a formalized address of the hole: formalized because it uses
symbols of the language of arithmetic; _and_ from left to right the addresses would
represent the natural numbers with the usual canonical order.

Each hole will have one pigeon either standing next to the hole address (label) or
dwelling inside the hole (compartment) but not both. Each pigeon will have a hole
where it will either stand next to the address label or dwell inside the hole's compartment (but not both).

A combination of the beam's holes and the pigeons is one subjective mapping between
the beam's holes and the pigeons.

The key and rather subtle information:

(a) In a mapping, the order of the pigeons DNA sequences (IDs) from left to right
might not be natural. For instance the pigeon with the ID SSSSS0 (5) might be
on the right of the the pigeon with the ID SSSSSSSSSSS0 (11).

(b) Because of (a), and of the characterizations of K (the uncountable collection of
all the above mappings/combinations), knowing/"summing" the prime-number IDs of
two different pigeons dwelling in two hole-compartments would not necessarily
equal to (identify) the identity of the even-number pigeon standing by a hole
somewhere to the right of the 2 dwelling prime-number pigeons.

Note also that the DNA identity of a multiple-number pigeon is identical to the
address-label of the hole it is standing next too. (Ditto for the 1-number pigeon
and its hole). It's an odd-prime pigeon that has an identity that might not match
with the address of the hole (of which the compartment the pigeon is dwelling).

The long and short of the story is that the definition of even numbers would not
compact enough ( _sufficient_ ) information about the prime numbers so that if the
Goldbach conjecture is true it's not possible (impossible) to verify/know the
Conjecture is so, per HP principle.

[khongdo...@gmail.com]

Proof theoretically, given a formula F of the form 'AxEy...' where Ax quantifies over
infinitely many non-trivial multiples and Ex quantifies over the odd primes, the
truth of F might be inaccessible.

[khongdo...@gmail.com]

On Monday, 31 October 2016 07:44:59 UTC-6, Dan Christensen wrote:
> You might find this useful: http://www.dcproof.com/pigeonhole.htm
>
> It is my formal development of a non-numeric version of the Pigeonhole Principle (in the DC Proof format).

It's not the same kind of Pigeonhole Principle: that's the canonical kind about
know-ability - "Then at least two pigeons will be in the same hole".

The kind I'm employing here is about unknowability: impossible to know.

[khongdo...@gmail.com]

Another succinct way to capture this alluded impossibility is via the below statement
of fact.

It's impossible to formalize the existences of all the odd primes with the language
of arithmetic, in the sense that if a _general_ odd prime p is represented by a
specific numeral X, it's _impossible_ to write down the numeral that would represent
the next prime larger than p.

[khongdo...@gmail.com]

Should have been "bijective mapping".

[Peter Percival]

khongdo...@gmail.com wrote:

> Proof theoretically, given a formula F of the form 'AxEy...' where Ax quantifies over
> infinitely many non-trivial multiples and Ex quantifies over the odd primes, the
> truth of F might be inaccessible.

Are there any quantifiers in the ... part? What, Pproof theoretically,
does "inaccessible" mean?

[Mike Terry]

<khongdo...@gmail.com> wrote in message
news:6c17b4be-b011-4773...@googlegroups.com...

No problem, I'll take the holes to have markings 0, S0, SS0, SSS0 etc..
Hmm, since I get bored typing out e.g. SSSSSSSSS0 with 9 S operations, in
what follows I'll shorten this using notation <9> to mean the same thing!

OK, so we have pigeons labelled as <0>, <1>, <2>, <3>,... etc. as well

ok, I'm with you so far - it is how I envisioned it more or less.

> > >
> > > Given the characterization of the collection K as above, then
*obviously*
> > if one
> > > claims that out of the uncountably many bars of K, he/she has chosen
by
> > Choice one
> > > bar as his/her *the standard* bar denoted by "N_bar", then the
following
> > statement
> > > (*) is true of this "the standard" N_bar:
> > >
> > > (*) It's impossible to know, to verify, for any given hole h that is
of
> > even-length
> > > of the marking, there are two pigeons dwelling in the the
compartments
> > and the
> > > sum of their DNA-sequence lengths is equal to the length of the
> > marking of the
> > > hole h.

But here I am confused. The problem is your phrase "chosen by Choice", and
also it is not a good plan to call the "chosen" bar "the standard N_bar",
because that has connotations that it is analagous somehow to what
mathematicians refer to as "the standard (intended) PA-structure N", which
it is not. (More below on this!)

To avoid arguments about meanings of words, here is my own wording of
something I would agree with, which hopefully captures what you are trying
to say in another way:

If k represents a specific successor bar from K, then dependent on the
choice of k, it could be that the claim above regarding pigeon labels (DNA
rungs/ladders) is true, or that it is false. So with no further information
about k, you might say "it is Impossible to Know the truth of the claim!"
but I would think that wording would be uninteresting to a mathematician
because the source of the "Impossibility to Know" is just that k is
unspecified so it is not even a specific "claim".

It is no different from the following argument:
1. Let x be a Natural number
2. Then it's Impossible to Know whether x > 100, because there are examples
of (1) where x is chosen so that x > 100, and examples where it's chosen
such that not x > 100.

This doesn't seem very profound to me, but if this is all you mean by your
phrase Impossible to Know [for the Pigeon/DNA claim] then I might go along
with that (at least in the context of the Pigeon/DNA claim).

But all this is irelevent, because the scenario with pigeons and
successor-bars is not analagous to the claim about cGC! (more below)

> > >
> > > If you understand and see that (*) is true in this precise analogy (of
the
> > new New
> > > Pigeonhole principle) as above, you'd understand that undecide(N |=
cGC)
> > where:
> > >
> > > - N is the purported "the standard" language structure for the
language
> > oif
> > > arithmetic.
> > > - cGC <-> "There are infinitely many counter examples of Goldbach
> > Conjecture".
> > > - undecide(S) <=> "It's impossible to know, to verify, the meta
statement
> > S is true".

Well, I'm afraid even if I agreed with the Pigeon/DNA claim (as I discussed
above) I wouldn't agree with your claim about cGC. In fact, I'm not clear
exactly how you think they correspond! So you should spell this out a
bit... What follows is my guess at how you're thinking:


You are claiming a correspondance between K (the collection of the various
successor-bars) and various possible language structures of PA, right? So I
take it that a given successor-bar (with assigned pigeons) corresponds to
some specific language structure of PA? Ultimately I think you're aiming to
conclude that the different language structures / successor bars give
different answers to a cGC-like question, and "we don't know which language
structure / successor bar is the "intended" one, so we can't answer the
question....

And the holes in the successor bar going from left to right (in my diagram
above showing what a successor-bar looks like) represent successive
applications of the S (successor) function. Right? (Otherwise the name
"successor-bar" is very poorly chosen! :))

Now at this point I see you could intend one of two interpretations for your
description!

1) The universe (elements to be quantified over) for the PA language
structure corresponding to a specific successor-bar is the set of HOLES
themselves, which I think we've agreed are labelled left to right by the
language numerals <0>, <1>, <2>,.... So I'll refer to the n'th hole as
H(n) = hole_<n>.

2) The universe is the set of Pigeons, which I think we've agreed are also
labelled by the language numerals <0>, <1>, <2>,.... So I'll refer to the
n'th Pigeon as P(n) - fully understanding of course that the P(n) isn't
necessarily in hole H(n). (For some values of n, like all even n, this is
true, but for others not. Also it would not be true that S(P(n)) = P(n+1)
for all n.)

Which of (1) or (2) do you intend?

If you intend (1), then the pigeons are irrelevant to the interpretation of
cGC, because it only talks about holes H(n), never the pigeons P(n). So in
this case, all your uncountable successor bars (after ignoring the pigeons)
would be completely identical and you would have just ONE representation of
the standard (intended) model for PA, containing elements {H(0), H(1), H(2),
...} i.e. elements are H(n) for (standard) natural numbers n, and there is
even the natural correspondence for language terms where
<0> --> H(0)
<1> --> H(1)
<2> --> H(2)
etc.

So perhaps you intended (2), and so the universe for the language structure
is the set of pigeons {P(n): n in N}, N being the (standard) natural
numbers. Now we no longer have the neat natural correspondence we had for
(1) above, because <n> in the language structure does not necessarily
correspond to P(n). E.g. here is the correspondence for one particular
successor bar + pigeon assignment that I've just made up!
<0> --> P(0)
<1> --> P(1)
<2> --> P(2)
<3> --> P(5)
<4> --> P(4)
<5> --> P(3)
<6> --> P(6)
<7> --> P(7)
...
(Here we see pigeons labelled 3 and 5 are not in their corresponding holes -
just to prove I have followed what you're saying about which pigeons are
moved around! :)) Also here we see the successor of P(2) is P(5).

BUT HERE IS THE PROBLEM FOR YOU: Although you would have an uncountable
collection of different PA language structures (if my interpretation of what
you're meaning is correct), THEY ARE ALL ISOMORPHIC TO EACH OTHER, AND IN
PARTICULAR THEY ARE ALL ISOMORPHIC TO THE STANDARD (INTENDED) LANGUAGE
STRUCTURE OF PA.

In other words they are all isomorphic to the natural numbers N together
with the usual definition of zero, successor, addition, multiplication and
so on. More to the point: EXACTLY THE SAME LANGUAGE STATEMENTS ARE TRUE IN
THE LANGUAGE STRUCTURES CORRESPONDING TO EACH AND EVERY SUCCESSOR-BAR!

Coming back to cGC, which I take to be a statement in the language of PA, it
is either true in all the above language structures (corresponding to ALL of
the successor-bars), or false in ALL of them. Your claim that "obviously"
it is "Impossible to Know" the truth of your statement concerning
Successor-Bars, Pigeons and Holes was simply based on the rather mundane
observation that for some Successor-Bars the statement would be true, whilst
for other Successor-Bars it would be false. This has to be contrasted with
the corresponding PA-language structre/cGC statement, which is true in all
of the corresponding structures or false in all of them! So this argument
for "Impossible to Know" for cGC simply does not work.


What has gone wrong here? The answer is that your interpretation of cGC in
terms of successor-bars and pigeons is Wrong. When you talk about the
interpretation of a language statement in a language structure, what matters
is the interpretation of the statement using the interpretations of
functions, constants, predicates etc. AS DEFINED WITHIN THE LANGUAGE
STRUCTURE ITSELF. The labels applied to elements of the universal set
matter NOT ONE IOTA in this respect! In the case of PA, it's the
interpretation of the successor function and addition/multiplication
functions that give the individuals in the structure their meaning, not any
labels attached externally to the individuals.

Here I reproduce from above your claim about pigeons/holes:

> > > (*) It's impossible to know, to verify, for any given hole h that is
of
> > even-length
> > > of the marking, there are two pigeons dwelling in the the
compartments
> > and the
> > > sum of their DNA-sequence lengths is equal to the length of the
> > marking of the
> > > hole h.

You are trying to take this as corresponding to an arithmatic equivalent
statement:

It's impossible to know, to verify, for any given even number h,
that there are two prime numbers adding to h.

In this implied correspondence, you identify the even number h by its hole
markings, which correspond to the language numeral <h> (and so ultimately to
the number h in N), but then you ask about whether there are (prime) PIGEONS
with labels adding to h, BUT THESE LABELS DO NOT HAVE ANY INTERPRETATION AS
PA LANGUAGE ELEMENTS in the structure, so the whole comparison is completely
meaningless. (And so the correspondence and your conclusing fails.)

The CORRECT correspondance would be asking consistently about the language
numerals <n> and their interpretation in the language structures for each of
the successor-bars. Doing it this way (consistently) the corresponding
question for successor-bars should be purely referring to Hole numbers,
because as noted above

<n> = H(n) = Hole_<n> [i.e. the hole with n sticks, i.e. the n'th
hole! in the bar]

This is true of EVERY bar, regardless of any pigeon-assignment, so all such
arithmentic questions have the same answer for every bar. All this is just
an obvious consequence of the language structures for every bar all being
isomorphic, so OF COURSE such arithmetic questions come out with the same
answers in all bars, and the whole pigeon thing is an irrelevance! :)

Regards,
Mike.

[khongdo...@gmail.com]

I understand the boredom, but that's a bad temptation here. You wouldn't be the first
one though. There are important reasons why here the holes are labeled as precise
terms of the language of arithmetic instead of <0>, <1>, <2>, ... which are based
on informal intuitions about the natural numbers. And I've already explained one of
the reasons: the definition of even numbers doesn't yield sufficient information for
us to conclude the Goldbach conjecture is true should it be so.

If you want to understand my pigeon analogy, please try not to paraphrase it without
thorough understanding what the analogy is trying to convey.

That's what you think, and why you're wrong. The so called "the standard (intended)
PA-structure N" is exactly a choice of the this collection K in this analogy, once
you isomorphically replace "pigeon" by "natural numbers", "hole-label" by "numeral"
(or "term of L(PA)"), etc...

>
> To avoid arguments about meanings of words, here is my own wording of
> something I would agree with, which hopefully captures what you are trying
> to say in another way:
>
> If k represents a specific successor bar from K, then dependent on the
> choice of k, it could be that the claim above regarding pigeon labels (DNA
> rungs/ladders) is true, or that it is false. So with no further information
> about k, you might say "it is Impossible to Know the truth of the claim!"
> but I would think that wording would be uninteresting to a mathematician
> because the source of the "Impossibility to Know" is just that k is
> unspecified so it is not even a specific "claim".

You've completely misunderstood the fine subtle points of the analogy. At this point
it's not helpful to chat further until you agree that my statement (*) as written is
true. If you still don't understand something about the analogy please let me know
and I'll try to elaborate again. But until you agree with (*) - or clearly point
out why it - as written by me - is wrong, it wouldn't help the arguing at all.

>
> It is no different from the following argument:
> 1. Let x be a Natural number
> 2. Then it's Impossible to Know whether x > 100, because there are examples
> of (1) where x is chosen so that x > 100, and examples where it's chosen
> such that not x > 100.
>
> This doesn't seem very profound to me, but if this is all you mean by your
> phrase Impossible to Know [for the Pigeon/DNA claim] then I might go along
> with that (at least in the context of the Pigeon/DNA claim).

Again, please try not to go too far with your own examples. Let's see if we could see
agree with my (*) first.

>
> But all this is irelevent, because the scenario with pigeons and
> successor-bars is not analagous to the claim about cGC! (more below)

Again, you're way too far ahead than necessary ... Let's agree or disagree on (*)
first. If you think (*) is wrong, why?

Well see ... This is one of the subtleties that people have missed: you're wrong
here since you can't establish the (non-logical) isomorphism between any two
structures in the collection K!

>
> Coming back to cGC, which I take to be a statement in the language of PA, it
> is either true in all the above language structures (corresponding to ALL of
> the successor-bars), or false in ALL of them. Your claim that "obviously"
> it is "Impossible to Know" the truth of your statement concerning
> Successor-Bars, Pigeons and Holes was simply based on the rather mundane
> observation that for some Successor-Bars the statement would be true, whilst
> for other Successor-Bars it would be false. This has to be contrasted with
> the corresponding PA-language structre/cGC statement, which is true in all
> of the corresponding structures or false in all of them! So this argument
> for "Impossible to Know" for cGC simply does not work.

You completely misunderstood and characterized the analogy: I wasn't talking
being true or being false. Rather I was talking about "It's impossible to know, to verify [...]"!

> What has gone wrong here? The answer is that your interpretation of cGC in
> terms of successor-bars and pigeons is Wrong. When you talk about the
> interpretation of a language statement in a language structure, what matters
> is the interpretation of the statement using the interpretations of
> functions, constants, predicates etc. AS DEFINED WITHIN THE LANGUAGE
> STRUCTURE ITSELF. The labels applied to elements of the universal set
> matter NOT ONE IOTA in this respect!

You're wrong of course. In this context, "The labels applied to elements of the
universal set" are called terms of the language denoting the individuals - the
natural numbers - and of course THAT MATTERS INFINITELY MANY IOTAs!

> In the case of PA, it's the
> interpretation of the successor function and addition/multiplication
> functions that give the individuals in the structure their meaning, not any
> labels attached externally to the individuals.

>
> Here I reproduce from above your claim about pigeons/holes:

Please don't reproduce anything here, when you don't understand much of it.
(Please see more below).

> > > > (*) It's impossible to know, to verify, for any given hole h that is
> of
> > > even-length
> > > > of the marking, there are two pigeons dwelling in the the
> compartments
> > > and the
> > > > sum of their DNA-sequence lengths is equal to the length of the
> > > marking of the
> > > > hole h.

Well there was an overlook (typo): it should have been "for a general hole h".

>
> You are trying to take this as corresponding to an arithmatic equivalent
> statement:
>
> It's impossible to know, to verify, for any given even number h,
> that there are two prime numbers adding to h.

Again, should have been "for a general even number h".

>
> In this implied correspondence, you identify the even number h by its hole
> markings, which correspond to the language numeral <h> (and so ultimately to
> the number h in N), but then you ask about whether there are (prime) PIGEONS
> with labels adding to h, BUT THESE LABELS DO NOT HAVE ANY INTERPRETATION AS
> PA LANGUAGE ELEMENTS in the structure, so the whole comparison is completely
> meaningless. (And so the correspondence and your conclusing fails.)

Wrong, the labels of the holes are actually numerals (terms) of the language
of arithmetic, denoting "PA LANGUAGE ELEMENTS in the structure" (your phrase).

>
> The CORRECT correspondance would be asking consistently about the language
> numerals <n> and their interpretation in the language structures for each of
> the successor-bars. Doing it this way (consistently) the corresponding
> question for successor-bars should be purely referring to Hole numbers,
> because as noted above
>
> <n> = H(n) = Hole_<n> [i.e. the hole with n sticks, i.e. the n'th
> hole! in the bar]
>
> This is true of EVERY bar, regardless of any pigeon-assignment, so all such
> arithmentic questions have the same answer for every bar. All this is just
> an obvious consequence of the language structures for every bar all being
> isomorphic, so OF COURSE such arithmetic questions come out with the same
> answers in all bars, and the whole pigeon thing is an irrelevance! :)

You've misunderstood the analogy and its subtleties big time! Sorry.

[khongdo...@gmail.com]

It's best if you and I agree or disagree that (*) is true, first. Anything else
we could dissect later.

If you agree: good, we can examine the analogy further.

If you don't agree: why not? Can you, in this case, figure out 2 primes p1, p2,
corresponding to a _general_ even e such that e = p1+p2?

[khongdo...@gmail.com]

I mean, if you could figure out p1, p2 for a _general_ even e, I'd concede my pigeon
analogy is bogus. But not until then.

[graham...@gmail.com]

GoldBach Conjecture is ruse!

Its no different to:


EVERY EVEN NUMBER IS THE SUM OF 2 ODDS WITHOUT A 7 IN THEM

WHY DOES IT HOLD?



BECAUSE THE PRIME DISTRIBUTION FUNCTION

pr(n e PRIMES) = 1 / LOG(n)


ALL GOLDBACH DID IS 1 SIMPLE TRICK! 1 SIMPLE TRICK!




pr(n e NUMBERSET) = 1 / LOG(n)

NUMBERSET = { 3,4,6,7,8,11,15,22,23,26,30... }


SAME DISTRIBUTION AS PRIMES! BUT JUST ANY RANDOM NUMBERS IN THAT DISTRIBUTION!





NOW TAKE THE ODD NUMBERS ONLY!


TRY CLICKING REFRESH A FEW TIMES AND SEE HOW YOU GO!


www.tinyurl.com/goldbachconjecture



MODIFIED PRIME DISTRIBUTION FUNCTION

Pr( n e SET ) = 2 / log(n) where n is Odd



EXAMPLE DISTRIBUTION

3 5 7 9 11 15 17 21 23 27 31 35 39 43 45 47 59 65 67 73 77 83 85 87 91 93

3+3=6

3+5=8

3+7=10
5+5=10

3+9=12
5+7=12

3+11=14
5+9=14
7+7=14

5+11=16
7+9=16

3+15=18
7+11=18
9+9=18

3+17=20
5+15=20
9+11=20





NOTHING TO DO WITH PRIMES WHAT-SO-EVEN!

[khongdo...@gmail.com]

What this pigeon analogy is really saying is that all you know about any two
different combinations (of holes and pigeons) in the uncountably many combinations
in K is that they (the two combinations) are just different: having two unspecified
primes.

Iow, you'd have two *different* successor functions sharing the _same_ symbol 'S',
sharing the _same_ domain (universe), sharing the _same_ numbers, relations that can
be inductively (but positively) defined or established.

This undifferentiable difference is the foundation weakness of the concept of the
natural numbers as technically connoted by the expressibility of the language of
arithmetic L(0,<,S,+,*).

In the 1st place, you can't establish a non-logical isomorphism between two
different language structures in K even though you can choose any of them to be
the concept of the natural numbers. (Of course you can choose none but in which
case you can't talk about "models" of PA, and Completness as a statement would
be nonsensical in its inception, since there would be no language structure
candidate for PA's models even if it's syntactically consistent).

In the 2nd place, the collection K reveals that - in expressibility - the language of
arithmetic L(0,<,S,+,*) is incomplete, insufficient, in enforcing the expected
coupling between the successor function denoted by 'S' to the rest of the binary
relation and functions, denoted by '<', '+', '*', respectively. Let me elaborate
further.

In _all_ language structures in K, zero, one, two, and all the multiple numbers
(or "individuals", or "pigeons") have their identities lined up naturally with the
order of their positions in the successor chain: this _natural order_ is can be
established inductively (positively). So in so far as proving any formula inductively
(positively) to be true w.r.t to K, it's immaterial which element of K we could use
as the underlying language structure, standard or not.

Otoh, for (odd) prime numbers, the expressibility of the language of arithmetic
L(0,<,S,+,*) is agnostic of a way to ensure the intrinsic natural order of the primes
be honored. Consequently in general we can't tell where the sum or the product of the
two primes be.

The key consequence of the two observations above is that if we're given only
information about a general even number, you can't pick out the "right" element
of K to be the "expected" standard language structure: since the reduction of the
successor function S, w.r.t. even numbers, could be decoupled from the reductions
of S w.r.t. the relation symbolized by '<', and w.r.t. functions symbolized by '+',
'*'. And these decoupling would ensure an impossibility to know occur.

(I know my English isn't my best ability. But I could try to explain further if you
still have questions on this.)

[Peter Percival]

khongdo...@gmail.com wrote:

> (I know my English isn't my best ability. But I could try to explain further if you
> still have questions on this.)

How does the above square with the

in this thread I'll only talk to ones who would readily understand
the terminologies and few logic behind my explanation: I'm sorry
in advance I wouldn't explain to those who don't understand.

of your first post?

[khongdo...@gmail.com]

On Wednesday, 2 November 2016 12:05:29 UTC-6, Peter Percival wrote:
> khongdo...@gmail.com wrote:
>
> > (I know my English isn't my best ability. But I could try to explain further if you
> > still have questions on this.)
>
> How does the above square with the
>
> in this thread I'll only talk to ones who would readily understand
> the terminologies and few logic behind my explanation: I'm sorry
> in advance I wouldn't explain to those who don't understand.
>
> of your first post?

This thread isn't about "squaring", so I don't have to spend time on this question.

[khongdo...@gmail.com]

Fwiw, I also don't seem to have a habit of answering a questioning-automaton, a kind
of (parrot/robot-like) entity that would incessantly spit out questions but would
never in-take an answer-content.

[Peter Percival]

Maybe my use of the word "square" is misleading, so I'll try to ask
differently. Here is a quote from you:

I could try to explain further if you
still have questions on this

and here's another quote from you:

I wouldn't explain to those who don't
understand

Will you explain, or won't you? (Note that there is no need to explain
to those who do understand.)

[Peter Percival]

If you won't answer, there is no "answer-content" to be taken in.

[khongdo...@gmail.com]

On Wednesday, 2 November 2016 14:44:55 UTC-6, Peter Percival wrote:
> khongdo...@gmail.com wrote:
> > On Wednesday, 2 November 2016 14:04:28 UTC-6, khongdo...@gmail.com wrote:
> >> On Wednesday, 2 November 2016 12:05:29 UTC-6, Peter Percival wrote:
> >>> khongdo...@gmail.com wrote:
> >>>
> >>>> (I know my English isn't my best ability. But I could try to explain further if you
> >>>> still have questions on this.)
> >>>
> >>> How does the above square with the
> >>>
> >>> in this thread I'll only talk to ones who would readily understand
> >>> the terminologies and few logic behind my explanation: I'm sorry
> >>> in advance I wouldn't explain to those who don't understand.
> >>>
> >>> of your first post?
> >>
> >> This thread isn't about "squaring", so I don't have to spend time on this question.
> >
> > Fwiw, I also don't seem to have a habit of answering a questioning-automaton, a kind
> > of (parrot/robot-like) entity that would incessantly spit out questions but would
> > never in-take an answer-content.
>
> If you won't answer, there is no "answer-content" to be taken in.

Which, for a (parrot/robot-like) questioning-automaton, would be a fine situation
by me.

[Peter Percival]

What is anyone who reads your posts to make of "I wouldn't explain to
those who don't understand"? Surely it is when people don't understand
that explanantions are needed. You may not wish to respond to automata,
but I'm asking about _people_ who don't understand.

[khongdo...@gmail.com]

Didn't I already use the word "entities" here?

[khongdo...@gmail.com]

"entity" anyway.

[Peter Percival]

Yes, but reminding me that you did so doesn't help me to understand why
you will only explain to people who understand, and therefore are not in
need of explanations.

[khongdo...@gmail.com]

Btw, if I din't mention I don't dialog with trolls on technical matters, consider
I've mentioned so. You're a troll (among other being a ...). Hope I don't have to
explain that.

[Peter Percival]

I'm not asking a technical question. I'm asking about the non-technical
opening remarks of your first post.

> , consider
> I've mentioned so. You're a troll

Why do you think that?

> (among other being a ...

A what?

[khongdo...@gmail.com]

Good trolling.

[Virgil]

In article <8410eb3b-9422-4e8a...@googlegroups.com>,

Since you are the one asking many questions but not answering any, you
AR the model of a modern major (parrot/robot-like) questioning-automaton
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

[Jim Burns]

On 11/2/2016 5:15 PM, Peter Percival wrote:
> khongdo...@gmail.com wrote:
>> On Wednesday, 2 November 2016 14:44:55 UTC-6,
>> Peter Percival wrote:
>>> khongdo...@gmail.com wrote:
>>>> On Wednesday, 2 November 2016 14:04:28 UTC-6,
>>>> khongdo...@gmail.com wrote:
>>>>> On Wednesday, 2 November 2016 12:05:29 UTC-6,
>>>>> Peter Percival wrote:
>>>>>> khongdo...@gmail.com wrote:

>>>>>>> (I know my English isn't my best ability.
>>>>>>> But I could try to explain further if you still have
>>>>>>> questions on this.)
>>>>>>
>>>>>> How does the above square with the
>>>>>> in this thread I'll only talk to ones who would
>>>>>> readily understand the terminologies and few logic

>>>>>> behind my explanation: I'm sorryin advance I

>>>>>> wouldn't explain to those who don't understand.
>>>>>> of your first post?
>>>>>
>>>>> This thread isn't about "squaring", so I don't have to
>>>>> spend time on this question.

The use of "squaring" here is closer in meaning to
"straightening" or "making square" perhaps metaphorically
the way one might make a wall square.

<https://en.wiktionary.org/wiki/square#Verb>
Definitions 2 and 3 seem appropriate.

>>>> Fwiw, I also don't seem to have a habit of answering a
>>>> questioning-automaton, a kind of (parrot/robot-like)
>>>> entity that would incessantly spit out questions but
>>>> would never in-take an answer-content.
>>>
>>> If you won't answer, there is no "answer-content"
>>> to be taken in.
>>
>> Which, for a (parrot/robot-like) questioning-automaton,
>> would be a fine situation by me.
>
> What is anyone who reads your posts to make of
> "I wouldn't explain to those who don't understand"?
> Surely it is when people don't understand that
> explanantions are needed. You may not wish to
> respond to automata, but I'm asking about _people_
> who don't understand.

Perhaps Nam is waiting for some who can explain
_to Nam_ what Nam means. Clearly, neither automata nor
people who do not already understand him would be able
to do that.

What could be more a reasonable explanation of Nam's
declaration than this?

[Virgil]

In article <nvdj6c$ok1$1...@news.albasani.net>,

Peter Percival <peterxp...@hotmail.com> wrote:

> khongdo...@gmail.com wrote:
> > On Wednesday, 2 November 2016 12:05:29 UTC-6, Peter Percival wrote:
> >> khongdo...@gmail.com wrote:

> Here is a quote from you:
>
> I could try to explain further if you
> still have questions on this
>
> and here's another quote from you:
>
> I wouldn't explain to those who don't
> understand
>
> Will you explain, or won't you? (Note that there is no need to explain
> to those who do understand.)

In order to make sure that those whom he does not wish to
understand won't understand, he mostly refuses to explain at all!

[Jim Burns]

On 11/2/2016 6:01 PM,

khongdo...@gmail.com wrote:
> On Wednesday, 2 November 2016 15:55:12 UTC-6,
> Peter Percival wrote:

>> Yes, but reminding me that you did so doesn't help me to
>> understand why you will only explain to people who
>> understand, and therefore are not in need of explanations.
>
> Btw, if I din't mention I don't dialog with trolls on
> technical matters, consider I've mentioned so. You're a
> troll (among other being a ...). Hope I don't have to
> explain that.

No need for an explanation. You're clearly accusing
Peter of being a hyper-intelligent shade of blue.

What is not clear is why you are accusing him of that.

Peter? Do you have something to tell us?

[khongdo...@gmail.com]

On Wednesday, 2 November 2016 16:35:24 UTC-6, Jim Burns wrote:
> On 11/2/2016 6:01 PM,
> khongdo...@gmail.com wrote:
> > On Wednesday, 2 November 2016 15:55:12 UTC-6,
> > Peter Percival wrote:
>
> >> Yes, but reminding me that you did so doesn't help me to
> >> understand why you will only explain to people who
> >> understand, and therefore are not in need of explanations.

A dishonest distortion of what Nam had said, of course. No surprise from the troll
PP of course.

> >
> > Btw, if I din't mention I don't dialog with trolls on
> > technical matters, consider I've mentioned so. You're a
> > troll (among other being a ...). Hope I don't have to
> > explain that.
>
> No need for an explanation. You're clearly accusing
> Peter of being a hyper-intelligent shade of blue.
>
> What is not clear is why you are accusing him of that.
>
> Peter? Do you have something to tell us?

Why do you, Jim Burns, think PP in this thread (at least) is a sincere poster
whom should be taken seriously?

[khongdo...@gmail.com]

"who should be taken seriously?" it was meant.

[graham...@gmail.com]

THESE NUMBERS WORK TOO!



Pr( n e P' ) = 2 / log(n) where n is Odd



EXAMPLE DISTRIBUTION

3 5 7 9 11 13 19 29 31 33 35 39 41 43 51 53 55 57 59 61 65 69 71 77 79 83 85 87 91

3+3=6

3+5=8

3+7=10
5+5=10

3+9=12
5+7=12

3+11=14
5+9=14
7+7=14

3+13=16
5+11=16
7+9=16

5+13=18
7+11=18
9+9=18

7+13=20
9+11=20

3+19=22
9+13=22
11+11=22

5+19=24
11+13=24

7+19=26
13+13=26

9+19=28

11+19=30

3+29=32
13+19=32

3+31=34
5+29=34

3+33=36
5+31=36
7+29=36

[Virgil]

In article <418991a0-e2cb-472e...@googlegroups.com>,

khongdo...@gmail.com wrote:

> On Wednesday, 2 November 2016 16:35:24 UTC-6, Jim Burns wrote:
> > On 11/2/2016 6:01 PM,
> > khongdo...@gmail.com wrote:
> > > On Wednesday, 2 November 2016 15:55:12 UTC-6, Peter Percival
> > > wrote:
> >
> > >> Yes, but reminding me that you did so doesn't help me to
> > >> understand why you will only explain to people who understand,
> > >> and therefore are not in need of explanations.
>
> A dishonest distortion of what Nam had said, of course.
>No surprise from the troll PP of course.

Takes one to know one!

[Virgil]

In article <c59577f8-e194-4242...@googlegroups.com>,

Try proof reading before posting, rather than only after

[khongdo...@gmail.com]

On Wednesday, 2 November 2016 22:10:17 UTC-6, Virgil wrote:
> In article <418991a0-e2cb-472e...@googlegroups.com>,
> khongdo...@gmail.com wrote:
>
> > On Wednesday, 2 November 2016 16:35:24 UTC-6, Jim Burns wrote:
> > > On 11/2/2016 6:01 PM,
> > > khongdo...@gmail.com wrote:
> > > > On Wednesday, 2 November 2016 15:55:12 UTC-6, Peter Percival
> > > > wrote:
> > >
> > > >> Yes, but reminding me that you did so doesn't help me to
> > > >> understand why you will only explain to people who understand,
> > > >> and therefore are not in need of explanations.
> >
> > A dishonest distortion of what Nam had said, of course.
> >No surprise from the troll PP of course.
>
> Takes one to know one!

Idiotic ranting from the hypocrite crank Virgil.

[Rupert]

On Monday, October 31, 2016 at 3:54:30 PM UTC+1, khongdo...@gmail.com wrote:
> On Monday, 31 October 2016 02:40:55 UTC-6, Rupert wrote:
> > On Sunday, October 30, 2016 at 6:41:15 PM UTC+1, khongdo...@gmail.com wrote:
> > > To sci.logic,
> > >
> > > One of the most difficult issues I've been facing with in sci.logic is to explain
> > > sufficiently clear why it's true that it's not possible (impossible) to know or
> > > verify whether or not there are infinitely many counter examples of Goldbach
> > > Conjecture.
> > >
> > > But I've found a way and in this thread I'll elaborate that way. I's a new kind of
> > > Pigeonhole principle, albeit it's about unknowability - not knowability - like the
> > > canonical Pigeonhole principle.
> > >
> > > Contrary to what some might have suspected, I'd like to be understood not just talked
> > > to, so in this thread I'll only talk to ones who would readily understand the
> > > terminologies and few logic behind my explanation: I'm sorry in advance I wouldn't
> > > explain to those who don't understand. In particular, cranks and trolls are _not_ welcomed.
> > >
> > > Imagine the collection K of uncountably many successor-function bars each of which is
> > > infinitely long in one direction (left-to-right).
> >
> > What is a successor-function bar?
>
> Ok. So it's not bar (bar is too small); it's a (construction) beam:
>
> http://www.aaanimalcontrol.com/pigeonroost.htm
>
> It's an abstract infinite beam from left to right having infinitely many holes each
> of which has a marking (a sequence of symbols) and a compartment right inside where
> a pigeon might dwell.

You mean it's a total ordering of type omega of the set of closed terms of the first-order language of arithmetic? (This is the best I can do by way of turning what you said into an actual mathematical definition.)

[khongdo...@gmail.com]

Yes, from what I could gather.

[khongdo...@gmail.com]

Basically the set of the hole-markings is the set of all the numerals in the natural order, to turn it to mathematical definition.

[Rupert]

But there are uncountably many different bars? I thought the idea was meant to be that the collection K was the collection of all possible total orderings of order type omega of the set of numerals. (So, in particular, most of the time the ordering would be different to the natural order.) I misunderstood?

[khongdo...@gmail.com]

I think you misunderstood something: why did you say "there are uncountably many
different bars"?

[khongdo...@gmail.com]

About the analogy I had had some meta statement (*) before but with a typo, but which
has been corrected. The corrected version is (*'):

(*') It's impossible to know, to verify, if it's so true, that for a general hole h
of even-length in its marking, there are two compartment-dwelling pigeons with
the sum of their DNA-sequence lengths being equal to the length of the marking
of h.

It's best that we should understand (*') of the analogy first, before going to the
analogous mathematical case of the underlying "standard" language structure.

Would you understand and agree (*') is true?

[Peter Percival]

khongdo...@gmail.com wrote:

> I think you misunderstood something: why did you say "there are uncountably many
> different bars"?

In your very first post in this thread: "out of the uncountably many
bars". Maybe it's an English problem? If I say "out of the ginger
sporrans..." I do imply that there are ginger sporrans. So when you
write "out of the uncountably many bars" you do indeed imply that there
are uncountably many bars.

[Peter Percival]

Wouldn't it be better to deal your claim that there are uncountably many
bars and that there aren't uncountably many bars first? Simplest would
be for you to say which of these claims you now wish to withdraw:

* there are uncountably many bars

* there aren't uncountably many bars.

It is not bad to admit to a mistake.

[khongdo...@gmail.com]

Idiotic ranting utterance from Peter Percival the hater.

[khongdo...@gmail.com]

On Thursday, 3 November 2016 07:39:54 UTC-6, Peter Percival wrote:
> khongdo...@gmail.com wrote:
>
> > I think you misunderstood something: why did you say "there are uncountably many
> > different bars"?
>
> In your very first post in this thread: "out of the uncountably many
> bars". Maybe it's an English problem? If I say "out of the ginger
> sporrans..." I do imply that there are ginger sporrans. So when you
> write "out of the uncountably many bars" you do indeed imply that there
> are uncountably many bars.

Looks like PP is stupid enough not to recognize there have been more posts in this
thread way ... beyond the "very first post in this thread".

[Peter Percival]

You have claimed that there both are and are not uncountably many bars.
Which is it?

[Rupert]

Well, the collection K is uncountable, isn't it?

[Rupert]

No, I don't understand why that would be.

[khongdo...@gmail.com]

Right: "uncountably many combinations of pigeons and holes of the only bar".

A clarification for what K of the analogy is was given in:

https://groups.google.com/d/msg/sci.logic/ov5-dvdmBrs/vP61vxaqCQAJ

<quote>

For better clarification:

- There's only one set of infinitely many pigeons.

- There's also only one infinite bar with infinitely many holes, each of which
has a marking (or label) next to it, and also has one compartment underneath.

- For each hole there corresponds exactly one pigeon, which either stands right
next to the hole or dwells in the hole's compartment. Likewise, for each pigeon
there correspond one hole, where the pigeon stand next to or dwells in its
compartment.

- The collection K mentioned previously is the collection of uncountably many
combinations of pigeons and holes of the only bar, the combinations having been
stipulated up to this point.

Hope this has helped to clarify.

</quote>

[khongdo...@gmail.com]

So, for a *general* hole h of even-length in its marking, you could find two

compartment-dwelling pigeons with the sum of their DNA-sequence lengths being

equal to the length of the marking of h?

Wow ... that looks incredible: that would be virtually proving the Goldbach
Conjecture is true! (I didn't think anyone could do that).

[khongdo...@gmail.com]

Then again, you might have thought there were uncountably many different bars ...

Anyway hope it's clear now and hope you now would agree the impossibility stated in
(*') is genuinely true. Right?

[Ross A. Finlayson]

There are only countably many primes,
and only countably many interchanges,
thus, as for example one could encode
the pairs of their locations, as one
would encode N x N, for each n -> p_n.
(It is so then for each of those, and
so on, for countably many interchanges
in a permutation.) That this gets into
a countable union of countable sets is
often attributed to Hausdorff.

[khongdo...@gmail.com]

Then again, how many real numbers can you have with countably many decimal expansion
places, and countably many value-choices per each place?

[Ross A. Finlayson]

Well-order the reals.

[Rupert]

Is this basically another way of saying, given an even number greater than 2, can I find two prime numbers which add up to that even number?

If you give me a *specific* even number greater than 2 I may be able to do that. Right now I do not know an algorithm which is guaranteed to do it in every case but in the future someone may come up with a proof that a certain algorithm works.

[Rupert]

I don't really get what would be meant by "verifying" it for a "general hole".

Obviously, if you give me a specific numeral for an even number greater than 2, then I can computationally check whether that even number is a sum of two primes.

[khongdo...@gmail.com]

What about a general even number greater than 2? Can you do the computational
checking?

And ... what about a general even number greater than 2: can you conclude whether
that even number is a sum of two odds?

Please concede that my (*') is true.

[Rupert]

Right now I do not know whether it is the case that every even number greater than 2 is a sum of two primes, or whether some of them are and some of them aren't. Someone might settle that question in the future, but right now I don't know.

> And ... what about a general even number greater than 2: can you conclude whether
> that even number is a sum of two odds?

Yes, it is easy to prove that every even number greater than 2 is a sum of two odd positive integers.

[Peter Percival]

Obviously. If n is an even number greater than 2, then 1 and n - 1 are
two odd numbers, and n = 1 + (n - 1). That + is computable and (since n
> 0) so is that -.

[khongdo...@gmail.com]

I was asking you about a general even number - ONE number -, just like below.

[Rupert]

Well, am I supposed to assume that I've been told what the number is?

[khongdo...@gmail.com]

I think it has been extremely clear that the number is a general even number that is,
say, greater than 2. Iow, the number is _not_ specific in its numeral form.

[Rupert]

So, if I knew that every even number greater than 2 was a sum of two primes, then I could answer definitely yes, regardless of which even number greater than 2 it is, it is definitely a sum of two primes. And for all I know someone might prove that someday.

As things stand, I don't know whether or not it is the case that every even number greater than 2 is a sum of two primes, so if you said to me "I am thinking of some even number greater than 2, can you tell me whether or not it is a sum of two primes?" I would have to answer "No". In my current state of knowledge, I don't know that. But that's no reason why I might not be able to know it in the future on the basis of some yet-to-be-discovered mathematical proof.

[graham...@gmail.com]

On what grounds?

A chaotic distribution that begins HIT HIT HIT HIT MISS HIT HIT HIT trails off to a MISS

The same chaotic distribution that begins HIT HIT HIT HIT HIT HIT HIT continues with infinite sequence of hits.


There is absolutely NO NEED for an algebraic proof of some pattern in Prime Numbers when the 1/log(n) distribution is far greater than the minimum distribution sqrt(n)



3 5 7 9 11 15 17 19 21 23 25 27 29 33 35 39 45 49 51 59 65 71 73 75 77 81 85 89 93 97 99

3+3=6

3+5=8

3+7=10
5+5=10

3+9=12
5+7=12

3+11=14
5+9=14
7+7=14

5+11=16
7+9=16

3+15=18
7+11=18
9+9=18

3+17=20
5+15=20
9+11=20

3+19=22
5+17=22
7+15=22
11+11=22

3+21=24
5+19=24
7+17=24
9+15=24

3+23=26
5+21=26
7+19=26
9+17=26
11+15=26

3+25=28
5+23=28
7+21=28
9+19=28
11+17=28

3+27=30
5+25=30
7+23=30
9+21=30
11+19=30
15+15=30

3+29=32
5+27=32
7+25=32
9+23=32
11+21=32
15+17=32

5+29=34
7+27=34
9+25=34
11+23=34
15+19=34
17+17=34

3+33=36
7+29=36
9+27=36
11+25=36
15+21=36
17+19=36

[khongdo...@gmail.com]

"Might" of course always connotes a degree of wishful-thinking, and in this case
_that_ "might" is very much detrimental to our mutual understanding on the matter
(even just the analogy).

"Might be possible to prove" is only possible if "might be" is even possible in the
first place: if the corresponding impossibility is genuine, there's no differenece
as to how many times you say "someone might prove that someday". Have you ever waited
for something that has never come because it would never come in the first place?

> As things stand, I don't know whether or not it is the case that every even number greater than 2 is a sum of two primes, so if you said to me "I am thinking of some even number greater than 2, can you tell me whether or not it is a sum of two primes?" I would have to answer "No". In my current state of knowledge, I don't know that. But that's no reason why I might not be able to know it in the future on the basis of some yet-to-be-discovered mathematical proof.

As things stand, if you don't even understand this new pigeon principle is true then
you will never understand it's impossible for you or any mathematics professor to
prove Goldbach conjecture is true - even if it is so true.

[graham...@gmail.com]

prime(p) <-> A(n) n>1 p%n=/=0

Its a Pseudo-Random number generator with a greatly exaggerated history of being sparse in N

With 100 digit number

35958958794378953983879890503483289237823487273489783428924894043340932404202021128474636237474747478

Every 100th number is prime!

That's 10E98 many primes!






There are
10000000000000000000000000000000000000000000000000000000000000000000000000 primes

less than
1000000000000000000000000000000000000000000000000000000000000000000000000000


and you only need 10000000000 random odds for GC to hold


LEFT COLUMN RIGHT COLUMN
10000000000 10000000000

LXR =
1000000000000000000000000000000000000000000000000000000000000000000000000000

PAIRS OF ODD NUMBERS

[Rupert]

Well, I don't know that that's impossible, do I? You haven't given me any good reason to think that it's impossible.

[khongdo...@gmail.com]

I guess you don't. Which is a surprise to me: the meta statement (*') about the
pigeons I'd say even a freshman would understand without many problems. Specifically
which part of (*') that you'd be unable to understand? (Please: (*') isn't the
Goldbach Conjecture; it's an abstract example about pigeons).

> You haven't given me any good reason to think that it's impossible.

I gave sci.logic viewers plenty of reasons to know that it's impossible. That you
or a few particular posters couldn't understand (or refuse to understand?) isn't
something I could really help much further.

Fwiw, I suspect some Ph.D and professor level posters would have no problem of
understanding the truth of (*') in this pigeon analogy.

[Rupert]

It's not a question of not understanding it. I told you what I think about it.

[khongdo...@gmail.com]

It's the issue of you not understanding it.

**********************************************************

---> Nam stated (*'):

(*') It's impossible to know, to verify, if it's so true, that for a general hole h
of even-length in its marking, there are two compartment-dwelling pigeons with
the sum of their DNA-sequence lengths being equal to the length of the marking
of h.

It's best that we should understand (*') of the analogy first, before going to the
analogous mathematical case of the underlying "standard" language structure.

---> Nam asked Rupert (*'):

Would you understand and agree (*') is true?

---> To which Rupert responded:

No, I don't understand why that would be.

**********************************************************

So, it's the issue of you not understanding it - the truth of (*').

[glambedo]

The pigeon is in general's asshole.

[graham...@gmail.com]

I got so bored of NAM going on and on about whether a proof exists or not to GoldBach Conjecture and all of sci.logic using double hypothetical arguments about Gödel's Proof [given GC or ~GC] that nothing made sense to anyone so I went ahead and solved it!


http://unsolvedproblems.org/index_files/Solutions.htm


No one has posted since!


P(n in PR') = 2/log(n) | n is odd


T H A T - I S - I T !


RANDOM ODD NUMBERS LOGARITHMIC DENSE IN N ADD UP TO ANYTHING

[Rupert]

Or, as some might put it, the issue of you not having offered the least reason to think it is true.

[Peter Percival]

khongdo...@gmail.com wrote:

> I gave sci.logic viewers plenty of reasons to know that it's impossible. That you
> or a few particular posters couldn't understand (or refuse to understand?) isn't
> something I could really help much further.
>
> Fwiw, I suspect some Ph.D and professor level posters would have no problem of
> understanding the truth of (*') in this pigeon analogy.

And they're all keeping quiet and have been keeping quiet for years.

[Peter Percival]

That isn't Nam's point. He claims that it is impossible to know if the
Goldbach conjecture has infinitely many counter-examples. Over many
years he has done nothing to argue (never mind prove) his point.

> and all of sci.logic using double hypothetical arguments about Gödel's Proof [given GC or ~GC] that nothing made sense to anyone so I went ahead and solved it!
>
>
> http://unsolvedproblems.org/index_files/Solutions.htm
>
>
> No one has posted since!
>
>
> P(n in PR') = 2/log(n) | n is odd
>
>
> T H A T - I S - I T !
>
>
> RANDOM ODD NUMBERS LOGARITHMIC DENSE IN N ADD UP TO ANYTHING

If you've proved it you should submit it to a reputable journal -
editors and referees need a good laugh just like the rest of us.

[khongdo...@gmail.com]

I believe "not having offered the least reason to think" to you (and perhaps to PP
who hates both the message and the messenger), not to some others.

I did explain it in sci.logic to you, et al., which I think an undergraduate would
have no trouble understanding it. That you couldn't see "the least reason" is of
course your own issue.

Let me explain again why (*') is true. You, Nam, and everyone who has some minimum
undergraduate mathematics knowledge could prove that given the definition of K as Nam
has stipulated, the hole-labels would _ONLY DENOTE_ the pigeons: the labels
themselves are _NOT_ the pigeons, hence the sum of any 2 odd-prime-DNA sequences (of
any 2 dwelling pigeons) does _NOT NECESSARILY_ have to equal to any greater-than-4
hole-label. Iow, in general, there's a decoupling between the length of an odd prime
pigeon's DNA's sequence and that of an even hole label: "decoupling" in the sense
the expected natural order between the two would not be honored. Hence, given there
are infinitely many greater-than-4 even hole labels and infinitely many pigeons with
odd odd-prime DNA sequences, the un-necessity would lead to the said impossibility
mentioned in (*').

Would you understand (*') is true now? If not, which specific part of the above
paragraph is still not clear to you, and why - can you elaborate?

[khongdo...@gmail.com]

On Saturday, 5 November 2016 07:59:34 UTC-6, Peter Percival wrote:
> khongdo...@gmail.com wrote:
>
> > I gave sci.logic viewers plenty of reasons to know that it's impossible. That you
> > or a few particular posters couldn't understand (or refuse to understand?) isn't
> > something I could really help much further.
> >
> > Fwiw, I suspect some Ph.D and professor level posters would have no problem of
> > understanding the truth of (*') in this pigeon analogy.
>
> And they're all keeping quiet and have been keeping quiet for years.

Yeah. They'd probably wait until the haters, trolls, racists like PP stop uttering
first. Why should any understanding people contribute much of any saying while
hooligans like PP are about and loud?

[Peter Percival]

Those others who remain silent and never come to your aid?

> I did explain it in sci.logic to you, et al., which I think an undergraduate would
> have no trouble understanding it. That you couldn't see "the least reason" is of
> course your own issue.
>
> Let me explain again why (*') is true. You, Nam, and everyone who has some minimum
> undergraduate mathematics knowledge could prove that given the definition of K as Nam
> has stipulated, the hole-labels would _ONLY DENOTE_ the pigeons: the labels
> themselves are _NOT_ the pigeons, hence the sum of any 2 odd-prime-DNA sequences (of
> any 2 dwelling pigeons) does _NOT NECESSARILY_ have to equal to any greater-than-4
> hole-label. Iow, in general, there's a decoupling between the length of an odd prime
> pigeon's DNA's sequence and that of an even hole label: "decoupling" in the sense
> the expected natural order between the two would not be honored. Hence, given there
> are infinitely many greater-than-4 even hole labels and infinitely many pigeons with
> odd odd-prime DNA sequences, the un-necessity would lead to the said impossibility
> mentioned in (*').
>
> Would you understand (*') is true now? If not, which specific part of the above
> paragraph is still not clear to you, and why - can you elaborate?
>

[khongdo...@gmail.com]

On Saturday, 5 November 2016 14:01:03 UTC-6, Peter Percival wrote:
> khongdo...@gmail.com wrote:
> > On Saturday, 5 November 2016 03:49:28 UTC-6, Rupert wrote:

> >> Or, as some might put it, the issue of you not having offered the least reason to think it is true.
> >
> > I believe "not having offered the least reason to think" to you (and perhaps to PP
> > who hates both the message and the messenger), not to some others.
>
> Those others who remain silent and never come to your aid?

Idiotic trolling of the racist-hater PP: it's Rupert who seems to have needed aid
in understanding (*') is true. Of course the racist-hater PP is beyond help.

[Rupert]

Part of the probem here is that you have not in fact given a definition of the collection K.

Am I to understand that each bar corresponds to a total ordering on the set of numerals of order type omega, and that in fact in each case this is the standard ordering? So that in fact the only difference between the different members of K comes in when you specify some relationship with the pigeons?

[khongdo...@gmail.com]

Huh? Are you serious? Didn't I already inform you as to what K should be in the
below, just a mere 2 days ago:

https://groups.google.com/d/msg/sci.logic/ov5-dvdmBrs/NqGEGdNfAgAJ

?

> Am I to understand that each bar corresponds to a total ordering on the set of numerals of order type omega, and that in fact in each case this is the standard ordering? So that in fact the only difference between the different members of K comes in when you specify some relationship with the pigeons?

Why do you keep saying "each bar": there's only ONE bar! There are however
un-countably many mappings (combinations) between ONE bar with infinitely many
holes, and one set of infinitely many pigeons!

(This is very ... very simple, Rupert! Why do I have to keeping repeating?)

[mitch]

On 10/30/2016 12:41 PM, khongdo...@gmail.com wrote:
> To sci.logic,
>

< snip >

Nam,

In another thread, Mr. Finlayson and I
talked about "numbers must be numbered
before they are counted" as expressing
the sense by which identity in logic
depends on names as opposed to numeric
relations (or, at least, identity in
a model).

Naturally, such a statement had not
be well-received.

Are you saying something similar here?

What you seem to be saying is that the
composite numbers are structurally
fixed; that the divisibility relations
of composite numbers reduce to atoms
relative to divisibility; and that,
with exception for the unique even
prime (and its successor???), prime
numbers may be arbitrarily permuted.

Please forgive me if this is not
what you mean. I am not trying to
put words in your mouth. I am simply
trying to understand your position.

mitch

[Ross A. Finlayson]

If there's a reason why this recognition
of the structure of the numbers is ill-
received, it includes that it makes the
otherwise so-clear and conclusive arguments
of "counting the infinity" not quite so
cut-and-dried, which is a usual metaphor
then of raw materials so refined then to
be stable in their structure for application,
i.e., denoted then let fit as thus defined.
Instead indeed there are notions like that
quantification over (all) fixed things
makes an un-fixed thing, or that the
continuum between zero and one exists
"before" algebra.

So, it's not so "naturally" ill-received
as not very simple. It's "naturally"
exactly they way it is. Here what's
"natural" in numbers is a usual measure
of ease, then elegance, the most for
the least, as it were (simpliciter).


"Identity" as among notions of likeness
or sameness would have that there are
others, for example along the lines of
"identity, equality, tautology" and
convergence, the transitive and non-
transitive, then to the inductive,
and so on. So, that sense so expressed
is particular, and we can formalize it.
In fact then that's establishing the
individuation of things, "identification".

[graham...@gmail.com]

It is a published solution to GoldBach Conjecture by University Of Sydney

There is a 2 year peer review then I WIN THE COVETED $500!



The only reason you all believe in UN-PROVABILITY is because you were CONNED!

There are n/log(n) PRIMES < n

That's 100000000000000000000000000000000000 PRIMES
less than 10000000000000000000000000000000000000

N IS QUITE DENSE IN PRIMES



If you think PRIME1 + PRIME2 = EVEN is a difficult problem that would explain it!



Try it with a RANDOM NUMBER SET instead of PRIMES!

www.tinyurl.com/goldbachconjecture

WORKS 99% OF THE TIME!



And that is CLEARLY PROOF why it works for PRIMES TOO!

[Rupert]

I didn't see that post of yours earlier. In your first post you said there were uncountably many bars, that's why you need to clarify.

Thank you for your clarification.

Yes, I now grant that (*') is true in the sense that you can't know whether or not there are two compartment-dwelling pigeons with the property that the sum of their DNA-lengths is equal to the number corresponding to the numeral of the hole h unless you have access to information about what the relation is between the pigeons and the holes (over and above the fact that we have an element of the set K as you defined it).

[khongdo...@gmail.com]

Thank you. I'll further more relevant comments as soon as I have time. Cheers.

[khongdo...@gmail.com]

Thanks, Mitch, for your post. I'll respond as soon as I could.

[Peter Percival]

Who is this you all? I don't believe that the Goldbach conjecture is
unprovable.

> There are n/log(n) PRIMES < n
>
> That's 100000000000000000000000000000000000 PRIMES
> less than 10000000000000000000000000000000000000
>
> N IS QUITE DENSE IN PRIMES
>
>
>
> If you think PRIME1 + PRIME2 = EVEN is a difficult problem that would explain it!
>
>
>
> Try it with a RANDOM NUMBER SET instead of PRIMES!
>
> www.tinyurl.com/goldbachconjecture
>
> WORKS 99% OF THE TIME!
>
>
>
> And that is CLEARLY PROOF why it works for PRIMES TOO!
>