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 A flawed proof? (Posted on 2002-12-20)
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.

 Subject Author Date Solution To The Problem K Sengupta 2007-03-20 12:24:44 iseeit. dorkdork 2003-08-02 22:09:49 Solution rajesh 2003-06-30 21:11:55 No no michael 2003-04-12 04:31:45 Not really Gamer 2003-04-11 09:52:01 No Subject N__A__T__A 2003-01-03 07:42:18 1st post, correct solution (i think) Aim Jayy 2002-12-25 22:11:06 Solution Jen 2002-12-25 10:51:33 Solution Gautam Joshi 2002-12-23 14:18:27 solution ananth 2002-12-21 17:10:22 re(2): A flawed proof? ( Solution ) Ravi Raja 2002-12-21 07:21:16 re: A flawed proof? ( Solution ) Dulanjana 2002-12-21 05:56:58 A flawed proof? ( Solution ) Ravi Raja 2002-12-20 19:14:19

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