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From: davis@hplabs.UUCP (Jim Davis)
Newsgroups: net.math
Subject: Re(2): -1 = 1
Message-ID: <1853@hplabs.UUCP>
Date: Wed, 21-Sep-83 01:51:30 EDT
Article-I.D.: hplabs.1853
Posted: Wed Sep 21 01:51:30 1983
Date-Received: Fri, 16-Sep-83 11:29:22 EDT
References: <734@bronze.UUCP>
Organization: Hewlett Packard Labs, Palo Alto CA
Lines: 37


		Steve Summit writes:

>	The proof that a = b was pretty obvious.  Here's one I discovered
>	by accident and still can't quite figure out:
>
>	1 = 1                   reflexive property of equality
>
>	2/2 = 2/2               another name for 1
>
>	  2/2    2/2
>	-1   = -1               raise -1 to both sides
>
>	  1              2       b/c      th         b
>	-1 = sqrt( ( -1 )  )    a    ==  c  root of a
>
>					     2
>	-1 = sqrt( 1 )          reduce ( -1 )	
>
>	-1 = 1                  reduce sqrt( 1 )
>
>	The mistake is probably in the fourth step, but no book I've seen
>	places restrictions on a, b, or c in that identity.

	Yes, the error in the argument is in step 4.
If (x = y) then it is the case that (sqrt(x) = +or- sqrt(y)).
If you examine any reasonable text of mathematics you will find
                        1/n             1/n
that (x = y) implies ( x   =  omega(k) y   ) where each of the n omegas
                    th
is chosen from the n   roots of unity.

-- 
					Jim Davis (James W Davis)
					...!ucbvax!hplabs!davis
					davis.HP-Labs@UDel-Relay
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