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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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Solution 1st post, correct solution (i think) | Comment 7 of 13 |
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
this is flawed, and since (a-b) = 0 (as a and b are equal) it would read correctly as
a=b
a^2-ab=a^2-b^2
a(a-b)=(a+b)(a-b)
a(0)=a+b(0)
a(0)=a+a(0)
a(0)=2a(0)
1(0)=2(0)
0=0
i think that is correct ?
  Posted by Aim Jayy on 2002-12-25 22:11:06
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