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A flawed proof? (Posted on 2002-12-20) Difficulty: 2 of 5
Given: a=b. Applying some basic identity transformations, we get:
       a=b
       a^2-ab=a^2-b^2
       a(a-b)=(a+b)(a-b)
       a=a+b
       a=a+a
       a=2a
       1=2
With such a proof, we can show that 1=2, pi=E, 10000000000000=1, etc.... Can you spot the flaw?

  Submitted by phil    
Rating: 2.9167 (12 votes)
Solution: (Hide)
The flaw, as many have pointed out, is in step 4, where both sides are "simplified", or divided by (a-b). Since we assumed that a = b, this expression is equal to zero. The proof hinges on the ability to divide both sides of the equation by zero. Since such division is illegal, the proof itself is invalid.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolution To The ProblemK Sengupta2007-03-20 12:24:44
iseeit.dorkdork2003-08-02 22:09:49
SolutionSolutionrajesh2003-06-30 21:11:55
No nomichael2003-04-12 04:31:45
Not reallyGamer2003-04-11 09:52:01
No SubjectN__A__T__A2003-01-03 07:42:18
Solution1st post, correct solution (i think)Aim Jayy2002-12-25 22:11:06
SolutionSolutionJen2002-12-25 10:51:33
SolutionGautam Joshi2002-12-23 14:18:27
solutionananth2002-12-21 17:10:22
re(2): A flawed proof? ( Solution )Ravi Raja2002-12-21 07:21:16
re: A flawed proof? ( Solution )Dulanjana2002-12-21 05:56:58
A flawed proof? ( Solution )Ravi Raja2002-12-20 19:14:19
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