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### Proof that 2 Equals 1 Using Derivatives

```
Date: 05/10/2000 at 04:56:22
Subject: How can this be?

How can this be?

kx = x + x + ... + x  (k-times)   .......................[1]

xx = x + x + ... + x  (x-times)   .......................[2]

x^2 = x + x + ... + x  (x-times)   .......................[3]

dx(x^2) = 2x  (diff. wrt x)   ................................[4]

dx(x + x + ... + x) = 1 + 1 + ... + 1  (x-times)   ...........[5]

so

2x = 1 + 1 + ... + 1  (x-times) {from eq. [4],[5]}   ....[6]

so we have

2x = x   ................................................[7]

so

2 = 1  (x <> 0)   ......................................[8]

Thank you!
```

```
Date: 05/10/2000 at 09:51:35
From: Doctor Rick
Subject: Re: How can this be?

Hi, Akram.

In taking the difference, you forgot that not only has each term
changed its value, but also the NUMBER of terms has changed. Let's put
in some numbers to make this clear. Let x = 3 and delta(x) = 1. Then:

x^2 = 3 + 3 + 3

(x+delta(x))^2 = 4 + 4 + 4 + 4

delta(x^2) = 1 + 1 + 1 + 4

We still don't have the 2x that you expected; we've got 7 instead of
6. Why is this? You forgot something else. 2x is the DERIVATIVE of x^2
- the limit of delta(x^2)/delta(x) as delta(x) approaches zero. But
the function as we have defined it (as a sum of x terms) has meaning
only for integer values of x, so delta(x) can't be less than 1. The
derivative is not defined. All we can define is a DIFFERENCE, as I
have done (with delta(x) = 1, the smallest possible value), and this
is not equal to 2x.

Proofs, Classic Fallacies", linked on our main FAQ page:

http://mathforum.org/dr.math/faq/

At the bottom there is a link to "derivatives," an item in our
archives that is directly related to your problem.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 08/15/2001 at 02:11:02
From: Akshay Dayal
Subject: Explanation of proof needed

In the explanation you gave to show why the proof was wrong,

x^2 = 3 + 3 + 3

(x+delta(x))^2 = 4 + 4 + 4 + 4

delta(x^2) = 1 + 1 + 1 + 4

what is delta(x^2)?

There's a delta(x) = 1, and x = 3, but I thought delta(x) was a term
of its own, not a function of x.

Thanks for the help.
```

```
Date: 08/15/2001 at 09:31:55
From: Doctor Rick
Subject: Re: explanation of proof needed

Hi, Akshay.

I took some shortcuts in my explanation, trying to correct the writer's
notation without changing it too much. I'll go through it in a different
way for you.

We're interested in finding the derivative of the function

f(x) = x^2

using the definition of x^2 as a sum of x copies of x. The claim was that
you can differentiate

f(x) = x + x + x + ... + x (x times)

by taking the derivative of each term and adding:

df(x)/dx = 1 + 1 + 1 + ... + 1 (x times)
= x

In my explanation of why this is wrong, I can't really talk about
derivatives, because the function has been defined only for integer values
of x. Therefore I wrote in terms of finite differences: delta(x) is a
finite (integer) change in x, and delta(x^2) is the change in x^2 due to
this change in x. Normally we would use the Greek capital delta, and drop
the parentheses around the x. You're right, it's not a function. Delta(x^2)
is defined formally as follows:

delta(x^2) = f(x+delta(x)) - f(x)

You might prefer it if I talk in terms of independent variable x and
dependent variable y:

[1]  y = f(x) = x + ... + x                     (x times)

Let's remain general rather than choosing delta(x) = 1. For any value of x
and any change in x, delta(x), we can evaluate f(x+delta(x)), which will
differ from y = f(x) by an amount delta(y):

[2]  y + delta(y) = (x+delta(x)) + ... + (x + delta(x)) (x+delta(x) times)

Subtract [1] from [2] to get delta(y):

delta(y) = delta(x) + ... + delta(x)          (x times)
+ (x+delta(x)) + ... + (x+delta(x))  (delta(x) times)

It's that second line that the writer ignored. Applying the "definition" of
(integer) multiplication, we get

delta(y) = x*delta(x) + delta(x)*(x+delta(x))
= 2x*delta(x) + delta(x)^2

Then

delta(y)/delta(x) = 2x + delta(x)

If our function were defined on all real numbers rather than just integers,
we would find the derivative by taking the limit of delta(y)/delta(x) as
delta(x) approaches zero. The delta(x) term would go away, and we'd get the
correct derivative. As it stands, you can see why (in my example in the
original explanation) I got a difference of 7 instead of 6: there's an
extra term  delta(x)^2 = 1.

Another correspondent suggested that we "define" multiplication for
non-integers like this (in my own notation):

x*y = x + ... + x ([y] times)
+ x*(y-[y])

where [y] is the greatest integer less than y. It's not very helpful,
because it only defines multiplication by a number greater than 1 in terms
of multiplication by a number less than 1 (namely, y-[y]). However, it does
allow us to take delta(x) to zero. If you work through it, you'll find that
the derivative works correctly.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus
High School Calculus

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