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From: shauns@tekcad.UUCP
Newsgroups: net.math
Subject: Re: 2=1!? - (nf)
Message-ID: <28@tekcad.UUCP>
Date: Sat, 10-Sep-83 04:26:15 EDT
Article-I.D.: tekcad.28
Posted: Sat Sep 10 04:26:15 1983
Date-Received: Mon, 12-Sep-83 19:22:17 EDT
Sender: ricks@tekcad.UUCP
Organization: Tektronix, Beaverton OR
Lines: 26

#R:ihuxe:-33200:tekcad:10500001:000:987
tekcad!shauns    Sep  9 22:00:00 1983

  OH YEAH, sure....
  
1)start with a fundamental relationship(1=1)              1 = 1
2)add "c" to both sides                               c + 1 = c + 1
3)assign value 2 to "c" and substitute on right side  c + 1 = 3
4)multiply both sides by (a-b) 
    "a" and "b" are the two                      (c+1)(a-b) = 3(a-b)
    numbers to be proven equal
5)multiply out terms                        ac + a - bc - b = 3a - 3b
6)place "a" terms on left                  
  place "b" terms on right                          ac - 2a = bc - 2b
7)factor                                             a(c-2) = b(c-2)
8)cancel common factors                                   a = b

Wrong! cancelling common factors means that one DIVIDES by the common factor.

Since c has been defined as 2, the denominator is zero, and division by zero
is, of course, undefined, so we really know nothing about a and b.

It's a cute trick, though... can suck somebody right in with that innocent
looking step (3).