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?
(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.
re(2): A flawed proof? ( Solution )
