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 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)

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 General Case Comment 10 of 10 |
(In reply to Additional Consideration by K Sengupta)

In general, if:
h(x) = (x^m+x^m+ .....+x^m) (x^n times)
then, in terms of product rule, we obtain:
h'(x)
= [m*x^(m-1)+m*x^(m-1) + .....+ m*x^(m-1)][x^n times] +
[x^m+x^m+ .....+x^m) [n*x^(n -1)times]

= x^(m-1)*X^(n-1)(mx+nx)
= (m+n)*x^(m+n-1)

This is indeed true, since:
d/dx[x^(m+n)]
= (m+n)*x^(m+n-1)

 Posted by K Sengupta on 2007-09-26 06:37:27

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