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) = x

Differentiating both sides with respect to x, we get:

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

(Since the derivative of x 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.)
Some Thoughts Additional Consideration | Comment 9 of 10 |
(In reply to Puzzle Solution by K Sengupta)

The method as given in the problem would also have failed if the number of times were a variable but the quantity to be summed were a constant.

Thus, if h(x) = (n+n+.....+n) (x times), where n is a constant.

Then apparently, h'(x) = (0+0+....+0) (x times) = 0
which is incorrect as d/dx (nx) = n

However, remembering the requirement of differentiation upon the number of times, we obtain:

h'(x)
= (0+0+....+0) (x times)+ (n+n+....+n) ( 1 time)  
= 0*x + n*1
= n, which is indeed true.

Edited on September 26, 2007, 12:52 pm
  Posted by K Sengupta on 2007-09-26 06:36:05

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 (10)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information