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1 = 0 Fallacy

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Date: 06/22/2001 at 08:12:29
From: D. Boeijen
Subject: 1 = 0 fallacy

Hi there,

Reading the Dr. Math pages - and especially the ones on 1 = 0
fallacies - I remembered a 'proof' we ran up against during high
school (VWO in the Netherlands). It makes use of integral calculus.

We learned the following rule for 'partial integrating':

Int(f(x)*g(x))dx = f(x)*G(x) - Int(f'(x)*G(x))dx

with G(x): the primitive function of g(x)
and f'(x): the derivative of f(x)

Now watch the following 'proof':

Int(1/x^2 * 2x)dx  =  1/x^2 * x^2 - Int(-2/x^3 * x^2)dx (step 1)

{just take f(x) = 1/x^2 and g(x) = 2x}

this yields:

Int(2/x)dx  =  1 - Int(-2/x)dx  = 1 + Int(2/x)dx        (step 2)

substracting Int(2/x)dx on both sides yields:

0 = 1                                                   (step 3)

Quite remarkable, I think!

We found two arguments that possibly explain the fallacy:

1) 1/x^2 * x^2 = 1 is an invalid step

2) we work with unbounded integrals. If we put lowerbound a and
upperbound b to the integral, we get for step 2:

Int(2/x)dx (a,b)  =  1 (a,b) + Int(2/x)dx (a,b)

which yields:

Int(2/x)dx (a,b)  =  1(b) - 1(a) + Int(2/x)dx (a,b)

because 1(b) = 1(a) = 1 we get:

Int(2/x)dx (a,b)  =  Int(2/x)dx (a,b)

which of course is true.

Which of these two arguments tackles the fallacy?

I hope you can give me an answer.

Thanks and greetings,
D. Boeijen
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Date: 06/22/2001 at 12:47:07
From: Doctor Peterson
Subject: Re: 1 = 0 fallacy

Hi!

Your second explanation is essentially right. I would say it is an
"indefinite" integral, rather than "unbounded." When you work with
indefinite integrals, you always have to keep in mind that an
arbitrary constant can be added to the result, since differentiation
of a constant yields zero. So what you really have is

Int(2/x)dx + C1  =  1 + Int(2/x)dx + C2

with a constant added to each side. This simplifies to

C1 = 1 + C2

which of course doesn't say much, since C1 and C2 could be anything.
That eliminates the problem entirely.

Your method of turning the integrals into definite integrals amounts
to the same thing; evaluating the constant at the limits makes it
disappear, so you can ignore it.

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

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