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

 One Equals Two (Posted on 2003-08-22)
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.)
 Puzzle Solution | Comment 8 of 10 |

The number of times that x's are added is a variable. Had the number of times been a constant, then the method would have worked.

For example, let us consider g(x) = (x+x+....+x)(m times), where m is a constant.

Then, g'(x) = (1+1+....+1)(m times) = m

This is indeed true as d/dx (mx) = m

However in the present case, the number of times is a variable so that the number of times (x) also needs to be differentiated w.r.t  x.

Thus, taking f(x) = (x+x+ ....+x) (x times), by means of product rule, we obtain:

f'(x) = (1+1+ ....+1) (x times) + (x+x+...+x) (1 time)
= 1*x + x*1 = 2x.

This is indeed true, since d/dx (x^2) = 2x.

 Posted by K Sengupta on 2007-09-26 06:34:59

 Search: Search body:
Forums (0)