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Path: utzoo!linus!decvax!tektronix!bennety
From: bennety@tektronix.UUCP (Bennet Yee)
Newsgroups: net.math
Subject: re:A new paradox?
Message-ID: <1375@tektronix.UUCP>
Date: Thu, 22-Sep-83 01:44:34 EDT
Article-I.D.: tektroni.1375
Posted: Thu Sep 22 01:44:34 1983
Date-Received: Fri, 16-Sep-83 23:21:58 EDT
References: <406@5941ux.UUCP>
Organization: Tektronix, Beaverton OR
Lines: 27

From ...!{hocda,ihnp4}!houxm!houxf!5941ux!dje:
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	"Theorem:  In any set of N elements (N >+ 1), all elements in the
		  set are equal."

	"Proof":  By induction on N.

	For N=1, the theorem is clearly true.  If N>1, then we can inductively
	assume that the theorem is true for all sets of N-1 elements.  Select
	a set of N elements x(1), ..., x(N).  The elements x(1), ..., x(N-1)
	constitutes a set of N-1 elements, which by the inductive hypothesis
	are all equal.  Similarly, the elements x(2), ..., x(N) constitues a
	set of N-1 elements, whcih by the inductive hypothesis are all equal.
	Consequently, by the transitive property of equality, all elements
	x(1), x(2), ..., x(N-1), x(N) are all equal.  This completes the
	proof.

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The transitivity argument holds iff there is overlap in your subsets.
For the case of N=2, however, this is not the case.  The two subsets are
disjoint, and transitivity is no longer applicable.


					Bennet Yee
					tektronix!bennety