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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.)
Solution flaws | Comment 2 of 10 |
There are two major flaws:
1. The function is defined only for integers, so is not at all continuous and therefore has no derivative.

2. When differentiating the left side, it was done for a fixed value of x (as a fixed number of terms, but that number is dependent on x). But by definition differentiation has to vary depending on the change in x.

Related to the latter is that the description of the function on the left side contains within itself mere English words to take the place of a mathematical function, and thereby hopes to avoid the necessary mathematical treatment. That is,

x + x +...+x (x times)

really is saying
x * x
which everyone agrees is x²

then as both the left hand and the right hand x are changing (not just the left hand one), the correct formula to apply is the product rule. To differentiate x * x you get x * 1 + 1 * x, which is indeed 2x, rather than x.
  Posted by Charlie on 2003-08-22 08:43:50
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