Re: |{A,A}|=1; C:={A,B}; A=B; |C|=?
From: RASINGER (helmut_rasinger_at_hotmail.com)
Date: 06/05/04
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Date: Sat, 5 Jun 2004 19:58:26 +0200
THANKS A LOT!
Indeed I want to implement this on a computer later, but not in the common
way, too much redundancy and other problems.
> Let C = { x,y } where x,y are two elements which may or may not be sets.
> |C| = 1 iff x = y ('iff' is 'if and only if', same as <-> )
This means in a proof refering the number of elements, I will have to check
the set's elements pairwise if (x=y)
Without doing the check I must use: "|C|<3"
The whole problem arises because we do not ALWAYS explicitly distinct
between elementvariables or elementconstants.
Isn't it contradictionary to say: "x is an element which is not a set". For
me its like to say "4 is not a number, but maybe an index"
In fact I think "element" was used to express in a comfortable way "is a set
with one and only one distinction from empty_set like {x}, but is allowed to
be subdistincted like {{{y},{z}}}"
> > Do mathematicians have an expression for "Elements which have no
Elements"?
In the sense of unsubdistinctable elements. I will call them
elementconstants.
> |{}{}{}| is syntatically incorrect.
Conclusion: elementseparator_set:={,} is obligatory.
This is because my personal and default element_separator_set:={} iff
distinctable.
PS Maybe you like the theoretical formula I found (I dont see practical
use):
(GeometricSeriesApplication)
BinaryDigitAtPostition(k)OfPosInt(n)=((-1)^(n!/(n-2^k)!(2^k)!)-1)/2
It would be a formula to change the representation of a given positive
integer n from any base into binary_base.
I just realized that
1/(x^0-x^(-1)) =
= 1/(x^0-x(-2)) + 1/(x^1-x(-1)) =
= 1/(x^0-x(-3)) + 1/(x^1-x(-2)) + 1/(x^2-x(-1)) =
= ...
THANKS AGAIN FOR YOUR HELP
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