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From: debray@sbcs.UUCP (Saumya Debray)
Newsgroups: net.math
Subject: Re: A new paradox?
Message-ID: <475@sbcs.UUCP>
Date: Sat, 17-Sep-83 14:44:10 EDT
Article-I.D.: sbcs.475
Posted: Sat Sep 17 14:44:10 1983
Date-Received: Sun, 18-Sep-83 07:54:23 EDT
References: <406@5941ux.UUCP>
Organization: SUNY at Stony Brook
Lines: 20


	Theorem:  In any set of N elements (N >= 1), all elements
		  in the set are equal.

Ah yes! The problem with the proof, of course, lies in the induction step
going from N=1 to N=2:

	Select a set of N elements x(1), ..., x(N).  The elements
	x(1), ..., x(N-1) constitute a set of N-1 elements, which
	by the inductive hypothesis are all equal.  Similarly, the
	elements x(2), ..., x(N) constitute a set of N-1 elements,
	which by the inductive hypothesis are all equal.  Consequently,
	by the transitive property of equality, all the elements
	x(1), x(2), ..., x(N-1), x(N) are equal.

For the set with N=2, the subsets {x(1), ..., x(N-1)} and {x(2), ..., x(N)}
are disjoint, so the transitivity of equality can't be applied.

Saumya Debray
SUNY at Stony Brook