All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > General > Tricks
One Equals Two (Posted on 2003-08-22) Difficulty: 3 of 5
Given 'x' not equal to 0, let us consider the follwoing relation:

x + x + x + .... +x (added 'x' times) = x²

Differentiating both sides with respect to x, we get:

1 + 1 + 1 + 1 + .... + 1 ('x' times) = 2x

(Since the derivative of x² with respect to 'x' is 2x).

So we now have:

x = 2x

Cancelling 'x' from both sides, we have:

1 = 2

Now the very obvious question follows:

Where is the flaw ?

See The Solution Submitted by Ravi Raja    
Rating: 3.1667 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution I think I've got it... | Comment 1 of 10
The problem lies with your differentiation of both sides. The operation of differentiation is valid only on well defined mathematical functions. "1 + 1 + 1 + ... + 1 ('x' times)" is not a well defined function; it is an algorithm (specifying to add 1 repeatedly), whereas "2x" is a well defined function.

Therefore, the resulting equation is faulty.
  Posted by Your buddy on 2003-08-22 08:40:55
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information