Re: About pii and integers

From: Keckman (keckman_at_welho.com)
Date: 09/30/04 

Date: Thu, 30 Sep 2004 18:44:38 +0300

On Wed, 29 Sep 2004 17:25:25 -0500, Daniel W. Johnson
<panoptes@iquest.net> wrote:

Ööö?!?!?

> Keckman <keckman@welho.com> wrote:
>
>> On Tue, 28 Sep 2004 14:58:35 -0500, Daniel W. Johnson
>> <panoptes@iquest.net> wrote:
>>
>> > Induction lets you reach a conclusion about every element of N.
>> > It does NOT yield a conclusion about N itself.
>>
>> Where that's comes from? Is it in the Peano's axioms ?
>
> 5. (induction axiom.) If a set S of numbers contains zero and also the
> successor of every number in S, then every number is in S.
>

How can that says, that the way we define set S tells us nothing about set
S?

I thought that the set of naturals is defined by the Peano's axioms and
still you declare
that Peano's axioms don't tell nothing about S? Ööö?? I must say.

Of course induction tells us something only for the set S item's but that
is not the holeness
of the axioms. And btw. in original axioms it is not talked about numbers
just sets and items.

http://www.wordiq.com/definition/Peano_axioms

> Note that it does NOT say
>
> 5. (fake induction axiom.) If a set S of numbers contains zero and also
> the successor of every number in S, then every number is in S, as is the
> set of numbers.

I sometimes write badly and make writeing erros by myself too but i did
not get that.

>
>> Induction somehow even qualify the conception for example Natural
>> numbers.
>> If something tells us something to all of set S items it must tell us
>> something about the set itself.
>
> No. Consider the set of two-element subsets of {a,b,c,d}. Each item in
> the set has two elements. But the set itself has six elements.

Öö??? What this got to do with Peanos axioms? where is the order relation?

You mean {(a,b) (a,c) (a,d) (b,c) (b,d) (d,b) (c,d)} for example ?

No,no. You are counting about how many different ways it is possible to
take two from set that has 4 item.

But the set that has 4 item has 2^4=16 subsets (if we do not inlcude empty
set) This is easy rule
to remember because when selecting those sub set you have two choise: you
eithert take or not: 2
possibilities since 2 powered the number of items). This rule is 2^n rule
is valid if we take empty
set along 2^n-1 if not.

And now - not saing that from those 16 subset there is 16 subset that have
two item - but
that from those 16 subset (they have 1,2,3 or for item) can be modified
subsets that
have two element in 16 other way:

(aa),(ab),(ac),(ad),(ba),(bb),(bc),(bd),(ca),(cb),(cc),(cd),(da),(db),(dc),(dd)

And if you from some reason don't want to include same item twice in same
"sub"set (aa)
then these: (ab),(ac),(ad),(ba),(bc),(bd),(ca),(cb),(cd),(da),(db),(dc) 12
pcs.

But what does this to do whit Peano's axiom? Nothing.

You are really confuseing now something. Maybe you tried to say something
like that
"Allthough induction can tell us something about the amount of items
amount it does not tell us
anything about the hole set" Well. That's sound horrible terrblle dösjfdös
ködöfjskdlöoi d.
It sounds like saying because cars cannot be treated as yellow table i
must eat my chocolate
tomorrow evening. It really does have meaning att all. What induction
tells us about the amount
of items - i think it have not told nothing particularly if you don't
believe a very simple fact.
For every n in N there is n precessor, n is finite. 

-- 
Petri Keckman

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