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Hi ppl! Sorry for nongamerelated topic, but imo it''s interesting question, and maybe the most kind people here will answer me It''s known, that a relation R on a set S{(x,y),(a,b),...} is equivalence relation, if the 3 INDEPENDENT conditions are true: 1) Reflectivity: (x,x) are in R (the same as xRx = true) 2) Symmetry: xRy implies yRx 3) Transitivity: xRy and yRz imply xRz Now i''ll proove that "2" and "3" => "1", so that "1" is not neccesary! PROOF: Let xRy = true; according "2": xRy => yRx according z=x in "3", and substituting the result of "2" (yRx), we obtain: xRy and yRx => xRx !!! And xRx is "1" condition! So "2" and "3" => "1" END But it''s known, that these 3 conditions are independent, so where is mistake?

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Thank You very much for Your answer, but it seems that i have incorrectly asked the question... so i'll try to write everything in more detailes:

i don't say that "2" => "1", but rather: if "2"=true, and "3"= true => "1" is not neccesary IN THE DEFINITION of equivalence relation (because it is fulfilled by "default", if "2" and "3" are true), for the R to be equivalence relation...

Given:
Cartesian product of Set A{x, y, z, a, b, c, ...} AxA, consider relation R, which is:
i) Symmetric: xRy => yRx
ii) Transitive: xRy & yRz => xRz
Proove: I will proove that R is reflective:
We have got the condition of transitivity xRy & yRz => xRz, now instead of applying it to some z element, let's apply it to x element (here one may say that it is wrong, but it doesn't contradict with any math rule... or does? then why?) so using "3" we have:
(xRy & yRx) [[<-this is true according "2"]] => xRx.

SO: if the GIVEN RELATION is SYMMETRIC AND TRANSITIVE, it is EQUIVALENCE RELATION!!!

[edited by - truk on September 9, 2002 11:53:41 AM]

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Alvaro is exactly right in what he says: it relies on the existence of the y.

From your last post:
quote:

(xRy & yRx) [[<-this is true according "2"]] => xRx


"2" says that if xRy then yRx. It could be the case that for a given x in A, there is no such y (not equal to x) in A such that xRy. Thus "2" doesn''t apply for that particular x - it''s assumption simply doesn''t hold. Thus you cannot rely on "2" in your proof of "1" in this case. It is still very possible that xRx for this x, and so "1" can still hold, and R can be an equivalence relation.

Miles

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quote:
Original post by truk
We have got the condition of transitivity xRy & yRz => xRz, now instead of applying it to some z element, let''s apply it to x element (here one may say that it is wrong, but it doesn''t contradict with any math rule... or does? then why?)



"Now let''s apply it to x instead of z"... That only makes sense if xRz, doesn''t it?

Just because xRz doesn''t mean xRq for some q( ~R )z.

That''s my tuppence ''apenny worth anyway.
I would be interested in seeing a counter-example relation though...

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THANKS A LOT!!! It''s really that xRy may not be fulfilled for any y, but though the relation will be equivalence one! Again thanks, and my appologizes to alvaro, for i didn''t understand Your answer

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truk,

I need to tell you that apologizing for posting a non-game related topic does not make it okay to post a non-game topic. I''m closing the topic now.

I''m assuming you''re a newbie, since you registered today. I suggest you spend a few minutes reviewing the forum FAQ, located here:

Forum FAQ

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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