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

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mueckenh@rz.fh-augsburg.de

Posts: 15,072
Registered: 1/29/05
Re: Matheology § 253
Posted: Apr 22, 2013 3:00 AM
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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.
>
> 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.


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.

Regards, WM



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