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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)

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A flawed proof? ( Solution ) | Comment 1 of 13
Given: a=b. Applying some basic identity transformations, we get:
a=b --------(Step 1)
a^2-ab=a^2-b^2 --------(Step 2)
a(a-b)=(a+b)(a-b) --------(Step 3)
a=a+b --------(Step 4)
a=a+a --------(Step 5)
a=2a --------(Step 6)
1=2 --------(Step 7)

Given a = b implies (a - b) = 0, and as we know that anything divided by zero is undefined, thus the flaw lies in the fourth step which we have obtined from step 3 on division by (a - b), that is, on division by zero, which cannot be done.
That's it.

  Posted by Ravi Raja on 2002-12-20 19:14:19
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