Find the Flaw

Date: 08/02/2001 at 23:12:48
From: Dan Swanton
Subject: I can't find the flaw in my false proof.

I have carefully read the Dr. Math FAQ and still don't understand 
where the following proof goes wrong.  

e^(i*x) = cos(x) + sin(x)*i
e^(2*pi*i) = 1
e^((2*pi)/i) = 1
e^(2*pi*i) = e^((2*pi)/i)
ln(e^(2*pi*i)) = ln(e^((2*pi)/i))
2*pi*i = (2*pi)/i
i = 1/i
i^2 = 1
i^2 = -1

Perhaps I should attempt to reclaim all points that I've ever lost on 
sign errors.

Date: 08/02/2001 at 23:33:24
From: Doctor Peterson
Subject: Re: I can't find the flaw in my false proof.

Hi, Dan.

The problem is that the log is a multivalued "function" when you work 
with complex numbers. Think about e^(ix); actually e^(ix + 2kpi) has 
the same value (for any integer k), since the sine and cosine are 
periodic. So there are many exponents that give the same value.

Try graphing the complex numbers e^(2 pi i) and e^(-2 pi i) in polar 
form to see what I mean.

Here are some pages from our archives about this:

   e^(pi*i) = -1: A Contradiction?
   http://mathforum.org/dr.math/problems/rwitte.8.17.96.html   

   Functions of Imaginary Numbers
   http://mathforum.org/dr.math/problems/pooh.7.31.96.html   

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

Date: 08/03/2001 at 20:57:07
From: Dan Swanton
Subject: Re: I can't find the flaw in my false proof.

I'm afraid I wasn't very clear, or must be confused.  I think that I 
understand the multivalued nature of e^(i*pi) and was trying to use it 
to prove that 1 = -1.  Why can't I?

Date: 08/03/2001 at 23:18:21
From: Doctor Peterson
Subject: Re: I can't find the flaw in my false proof.

Hi, Dan.

Well, you can't prove it because it's not true...

Maybe what you mean is that you don't understand why the fact that 
e^(ix) has the same value for different x makes your proof invalid. 
Let's build up some background to explain what a valid proof requires.

What I was trying to say is that, because ln(z) is multivalued (has 
more than one value for a given z, and therefore is not really a 
function as we usually think of it), you can't use it in a proof. When 
you prove something algebraically, what you are really doing is 
transforming an equation step by step, through a series of equivalent 
equations, from a known fact to an unknown one. In order for each 
equation to be equivalent to the one before, each transformation must 
be valid and reversible. You are probably familiar with this from 
solving algebra problems, which is similar to proofs. If I have the 
equation

    3x - 1 = 5

then I can add 1 to both sides, because any value of x that satisfies 
the equation also satisfies

    3x = 6

and any value that satisfies the latter also satisfies the former. 
They are equivalent equations.

The same is true in proofs. For example, if I start with

    1 = 1

and then say that

    1 - 1 = 0

I am subtracting one from each side. If the first equation is true, so 
is the second, since equal numbers remain equal when you subtract one. 
Also, if the second equation is true, so is the first one, because my 
transformation is reversible: I can add one to both sides of the 
second equation, and the first will be true.

In false proofs, generally there is one step that does not produce an 
equivalent equation. (Or, in proofs that depend only on implication 
rather than equivalence, the step attempts to reverse an irreversible 
transformation.) For example, dividing by zero is not valid. If I 
start with the fact that

    1 * 0 = 2 * 0

and divide both sides by zero, I get

    1 = 2

The first equation is true, but the second is not. When you divide by 
zero, you are trying to reverse multiplication by zero. If a = b, it 
will be true that 0a = 0b, but not the other way around! You can't say 
that two numbers whose product with zero is zero are the same number, 
because multiplying by zero converts ALL numbers to zero, and 
therefore dividing by zero would convert zero to anything.

As another example, taking the square root (or rather, removing a 
square) does not produce an equivalent equation. If I know that

    (1)^2 = (-1)^2

I can't say that therefore

    1 = -1

because there are in fact TWO numbers whose square is 1. If a = b, 
then a^2 = b^2, but not vice versa, since squaring takes two numbers 
to the same value. Again, we are trying to undo an action (squaring 
this time, multiplication by zero last time) that is not reversible, 
so we are losing possibilities in our equation.

If we really took the square root (that is, the positive square root) 
of each side, we would get

    |1| = |-1|

which is valid.

In your attempted proof, you are taking the log of both sides, saying 
that

    e^(2*pi*i) = e^((2*pi)/i)

implies

    ln(e^(2*pi*i)) = ln(e^((2*pi)/i))

But just as with the square root, since the log has multiple values 
(is not actually a function) for complex numbers, this is not 
necessarily true. If you take different values of the log, they will 
not be equal. In order to make this step valid, you would have to take 
the _principal value_ of the log, just as we took the positive value 
of the square root in order to correct our square root example.

To put it differently, if you avoid the "ln" notation and simply claim 
that

    e^(2*pi*i) = e^((2*pi)/i)

implies

    2*pi*i = (2*pi)/i

then you are assuming that if the power is the same, the exponent must 
be the same, that is, that exponentiation is reversible, as it is for 
real numbers. Since this is not true (e^z can be the same for 
different z), it is not valid to drop "e^" from both sides. The 
resulting equation is not equivalent.

Does that make the problem with your "proof" clearer?

I searched our archives for other discussions of equivalent equations, 
and found these, which deal with solving equations rather than proofs, 
but may shed some light:

   Subtracting One Equation from Another
   http://mathforum.org/dr.math/problems/lee.05.30.01.html   
   
   Why Multiple Roots?
   http://mathforum.org/dr.math/problems/duff.9.9.96.html   

   Equivalent Expressions / Equivalent Equations
   http://mathforum.org/dr.math/problems/brady2.7.27.97.html   

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