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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?

See The Solution Submitted by phil    
Rating: 2.9167 (12 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): A flawed proof? ( Solution ) | Comment 3 of 13 |
(In reply to re: A flawed proof? ( Solution ) by Dulanjana)

Yes even I agree to what Dulanjana has commented on the flaw in the 6th line of the given proof, but my dear friend, you have considered a particular case of the problem when
a = b= 0, but I have considered the general problem where a and b can take any value. Anyway, both the reasoning is the same, whether it is the 3rd, 4th or the 6th step. But, if a flaw is found in the 'n'th step, then obviously it follows that at least one (if not all) step after the 'n'th step(inclusive) contains a mistake.
  Posted by Ravi Raja on 2002-12-21 07:21:16

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