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### 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.

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

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

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Imaginary/Complex Numbers
College Number Theory
High School Imaginary/Complex Numbers
High School Number Theory
High School Transcendental Numbers

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