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Disproving that One is Equal to Two


Date: 11/13/95 at 10:20:15
From: Anonymous
Subject: Calculus (Derivatives)

   I've been struggling with finding the error in my logic for 
quite some time now, but can't seem to quite get where I messed 
up.

If anyone could point out the error I would appreciate it.
 
   1. Assumption: if a(x) = b(x)+c(x)+d(x)+ ... z(x)
      a'(x) = b'(x) + c'(x) + d'(x)+ ...z'(x)  
      (I'm not 100% sure of this but I think its valid.)
 
   1                                =     1        = 1^2 (1 squared)
   2 + 2                            =     2 x 2    = 2^2 (2 squared)
   3 + 3 + 3                        =     3 x 3    = 3^2 (3 squared)
   4 + 4 + 4 + 4                    =     4 x 4    = 4^2 (4 squared)
   X + X + X + ... + X (X times)    =     X x X    = X^2 (X squared)
 
Now take the derivative of both sides of the last equations
   1 + 1 + 1 + ... + 1 (X times)    =     2X
                                 X  =     2X
                                 1  =     2
 
X isn't 0, I'm not dividing by 0, not taking any plus roots on one 
side and minus roots on the other?  What is the flaw?
 
Tony Chamberlain
achamber@cs.weber.edu


Date: 11/13/95 at 17:41:10
From: Doctor Ethan
Subject: Re: Calculus (Derivatives)

Hello,

   Well your example is very neat and the flaw is pretty subtle,
but I think I can convince you.

1.  Your assumption is correct but we need to clarify it a 
little more before we can really use it well.

When you say, " if a(x) = b(x)+c(x)+d(x)+ ... z(x)" you are 
assuming that this is true for some range of x values

so a(1) = b(1) + ... + z(1)

and a(1.347) = b(1.347) + ... + z(1.347)

and all the numbers in between for example.

So the problem with your equation is that it is not a well defined 
function.  It is only defined at the one integer value.

What you are saying is for any x add x to itself x times.

Either you have to use x^2 to mean that or a chart, as you have 
done. But no point on that chart is differentiable because that 
expression is only good at that point.

I hope that this has made some sense. If it hasn't, then write 
back and I will try to help more.

-Doctor Ethan,  The Geometry Forum

    
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