Re: Svara that Goldbach's Conjecture is Unprovable

From: Michael N. Christoff (mchristoff_at_sympatico.caREMOVETHIS)
Date: 07/12/04 

Date: Mon, 12 Jul 2004 13:20:14 -0400

"Craig Feinstein" <cafeinst@msn.com> wrote in message
news:b671fc3e.0407120601.7f74f7c6@posting.google.com...
> The argument is as follows. Its very similar to a previous post that I
> made which argues that Twin Primes is unprovable. I'll let you judge
> whether this is a rigorous proof or not. I'm not claiming that it is
> or that it is not - It's just for fun:
>
> Let set S be the set of all q such that 0<q<2n and q is prime.
> Let set T be the set of all q such that 0<q<2n and 2n-q is prime.
>
> Then Goldbach's Conjecture is equivalent to the intersection of sets S
> and T being nonempty for each natural number n.
>
> Now, the condition which defines set S, that q is prime, does not give
> any information about the factors of 2n-q, so it does not give any
> information about whether or not 2n-q is prime.
>
> And the condition which defines set T, that 2n-q is prime, does not
> give any information about the factors of q, so it does not give any
> information about whether or not q is prime.
>
> Therefore, the two conditions which define the two sets are
> independent from one another, meaning that the only way to determine
> whether the intersection of the two sets, S and T, is nonempty is to
> directly calculate the elements in the intersection of S and T, i.e.,
> test if there is a q such that q and 2n-q are both prime. But
> this would take an infinite amount of time, since one would have to
> test this for each natural number n. Therefore, Goldbach's Conjecture
> is unprovable. QED
>
> For those of you who don't buy the argument, let me explain this in
> another way: One can define set S as the days in the year 2003 in
> which twins were born in hospital A. And define set T as the days in
> the year 2003 in which twins were born in hospital B. (A and B are two
> different hospitals in different parts of the world.) The conditions
> which define sets S and T have nothing to do with one another, i.e.,
> they are independent from one another. Therefore, the only way to
> determine whether the intersection of sets S and T is nonempty is to
> directly calculate the elements in the intersection of sets S and T.
> This is obviously feasible to do since the number of days in the year
> 2003 is finite, but if one had to do it for each year until the end of
> time and prove that there is always a day in each year when both
> hospitals give birth to twins, one would have to wait an eternity to
> do such.
>
> Anyway, I welcome comments.
>

Hi Craig. This sounds similar to your proof idea for p!=np. In that proof,
if I recall correctly, you claimed (without adequate proof) that a single
solution to the problem existed (in that case the problem was subset-sum),
and then claimed that this single solution required at least a
super-polynomial amount of computation to solve (hence subset-sum must be in
NP-P so P != NP). In this proposition you state:

"the only way to determine whether the intersection of the two sets, S and
T, is nonempty is to directly calculate the elements in the intersection of
S and T"

You still need to prove that this truly is 'the only way'.

l8r, Mike N. Christoff

Relevant Pages

  • Svara that Goldbachs Conjecture is Unprovable
    ... Then Goldbach's Conjecture is equivalent to the intersection of sets S ... and T being nonempty for each natural number n. ... the year 2003 in which twins were born in hospital B. (A and B are two ... hospitals give birth to twins, one would have to wait an eternity to ...
    (sci.math)
  • Re: Svara that Goldbachs Conjecture is Unprovable
    ... >> sets S and T being nonempty for each natural number n. ... >> determine whether the intersection of the two sets, S and T, is ... >> Goldbach's Conjecture is unprovable. ... >> twins, one would have to wait an eternity to do such. ...
    (sci.math)
  • Independent sets problem
    ... Let A be a nonempty finite set. ... If the intersection of the system I is nonempty, ... the sets in I1 are pairwise disjoint, ...
    (sci.math.research)
  • Re: Basic real analysis results in a physics students utter confusion.
    ... if a family of connected subsets of a space forms ... nonempty intersection, then its union is connected. ... closed under the edge relation of the graph, ...
    (sci.math)
  • Re: cover and subcover (topology)
    ... >>The collection is nonempty, since it contains the full cover. ... consider the intersection of all subcovers in C. ... The intersection could even be empty. ... Prev by Date: ...
    (sci.math)