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Topic: Matheology § 253
Replies: 30   Last Post: Apr 22, 2013 2:44 PM

 Messages: [ Previous | Next ]
 Virgil Posts: 8,833 Registered: 1/6/11
Re: Matheology � 253
Posted: Apr 22, 2013 2:44 PM

In article
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 21 Apr., 22:25, Virgil <vir...@ligriv.com> wrote:
>

> > > Explain how N elements (n) can be distributed, with repetition, among
> > > M sets (s_k) such that there are all elements n represented at least
> > > once, all s_k are used too, but not all elements n are in in one set
> > > s_k.

> >
> > WM himself has shown us how to do it:
> >
> > FISON_1 = s_1 = { 1 }
> > FISON_2 = s_2 _ { 1 2 } = { 2 1 }
> > FISON_3 = s_3 = { 1 2 3 } = { 3 2 1 }
> > ...
> >
> > So FISONs do it very neatly!

>
> No, you have not shown the other condition, namely that all naturals
> are in those FISONs.

> >
> > > Is it possible to avoid the condition:
> > > exist  j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k"?

> >
> > Yes! Trivially! it is, in fact, impossible for any 4 naturals
> > j, k, m, n not to avoid it.

>
> That is correct.

> >
> >
> >

> > > How?
> >
> > How not?
> >
> > See above.

>
> You have not shown the other condition, namely that all natural are in
> the FISONs.

But that is so obvious that even WM should have been able to figure it
out on his own,

Every natural is, by definition, the determiner of that FISON of which
it is the largest member, so that every natural is automatically in at
least one FISON.
> >
> > Also. from a previous post:
> >
> > WM's claim:
> >

> > > If *not* all naturals in one s, then
> > > exist j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k.

If WM's claim above were true, WM should be able to find specific vaLues
for j, k, m and n such that
m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k holds true.

That he has not done so suggests that he cannot do so because there are
no such numbers satisfying his claimed condition.
>
> that is obvious in mathematics if two conditions have to hold:
> 1) All naturals in FISONs.
> 2) Not all naturals in one and the same FISONs.

If it were so obvious, even someone so inept at proofs as WM should be
able to demonstrate it, but as he cannot, or at least has not, it would
almost certainly be a false claim.

But it is well known that the natural numbers satisfy the trichotomy
properties of a linearly ordered set which disproves WWM's claim.

And, at least everywhere outside of Wolkenmuekenheim,
1) All naturals are in FISONs.
2) Not all naturals are in one and the same FISONs.
3) NOT exist j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k.
--

Date Subject Author
4/18/13 mueckenh@rz.fh-augsburg.de
4/18/13 fom
4/18/13 Virgil
4/18/13 fom
4/19/13 mueckenh@rz.fh-augsburg.de
4/19/13 Virgil
4/19/13 mueckenh@rz.fh-augsburg.de
4/19/13 fom
4/19/13 mueckenh@rz.fh-augsburg.de
4/19/13 fom
4/19/13 mueckenh@rz.fh-augsburg.de
4/19/13 fom
4/19/13 Virgil
4/19/13 fom
4/19/13 Bergholt Stuttley Johnson
4/20/13 mueckenh@rz.fh-augsburg.de
4/20/13 fom
4/20/13 Virgil
4/21/13 mueckenh@rz.fh-augsburg.de
4/21/13 Virgil
4/21/13 mueckenh@rz.fh-augsburg.de
4/21/13 Virgil
4/22/13 mueckenh@rz.fh-augsburg.de
4/22/13 Virgil
4/19/13 Virgil
4/19/13 Virgil
4/19/13 Scott Berg
4/19/13 fom
4/19/13 Michael Klemm
4/19/13 Virgil
4/19/13 fom