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From: ss@rabbit.UUCP
Newsgroups: net.math
Subject: Re: A new paradox?
Message-ID: <1932@rabbit.UUCP>
Date: Thu, 15-Sep-83 09:02:21 EDT
Article-I.D.: rabbit.1932
Posted: Thu Sep 15 09:02:21 1983
Date-Received: Fri, 16-Sep-83 00:23:16 EDT
References: <406@5941ux.UUCP>
Organization: Bell Labs, Murray Hill
Lines: 30


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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)
	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.  This
	completes the proof.

	Where's the fallacy?

	Dave Ellis / Bell Labs, Piscataway NJ
------------------

What are you trying to prove?? All that the above says is that IF x(1) -- x(N)
are in the set of equals THEN all N-1 element sub-sets of this set are in
the set of equals and (here is the nice part) THEREFORE x(1) --- x(N) are in
the set of equals.

It certainly does not prove that ANY N element set has equal elements.

Sharad Singhal