Find the FlawDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/