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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 21, 2013 4:25 PM

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

> On 19 Apr., 22:50, Virgil <vir...@ligriv.com> wrote:
>

> > > There exist m, n, a, b in |N,
> > > such that m is in FISON(a) and not in FISON(b) and n is in FISON(b)
> > > and not in FISON(a).

> >
> > For each natural n in |N, FISON(n) is defined by WM to be the set of all
> > naturals less than or equal to the natural number n.
> > Thus for natural numbers m ands n,
> > n is in FISON(m)  if and only if  n <= m, and
> > similarly n is NOT in FISON(m)  if and only if  m < n.
> > Thus if m is in FISON(a) and not in FISON(b), then b < m <= a
> > and if n is in FISON(b) and not in FISON(a), then a < n <= b
> >
> > Thus in WM's world one must have b < a and a < b simultaneously.

>
> 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!

> 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.
>
> How?

How not?

See above. If WM claims such naturals can exist, let him show us an
example of specific values which do satisfy that false claim.

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.

Once before claimed and also once beforE DEBUNKED.

NOTE that a e s_b iff a <= b and ~( a e s_b) iff a > b, so that
m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k
iff
(m <= j) & (m > k) & (n >j) & ( n <= k)
But if (m <= j) & (n >j) then m < n
And if (m > k) & (n <= k) then n < m
So it follows that m > m and n > n, at least in WM's world.

But OUTSIDE WM's world, NOT all naturals can be in an one FISON, s_n
>
> If only ~(n e s_j) & n e s_k, then also m e s_k is possible, and then
> n and m are in s_k. No evidence that not all naturals in one s.

Maybe no such evidence in Wolkenmuekenheim, but there is enough outside
Wolkenmuekenheim, to satisfy everyone other than WM.

For every i in |N, max(s_i) = i and i+1 is not in s_i.

Which is sufficient evidence OUTSIDE WOLKENMUEKENHEIM to show that not
all naturals are in any one of the s_i,
at least not OUTSIDE WOLKENMUEKENHEIM .
--

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