Was Euler wrong? 2*Pi=0?Date: 03/13/2002 at 17:06:43 From: Warren Subject: Was Euler wrong? 2*Pi=0? It is well known that e^(Pi*i) = -1, according to Euler's formula. While I was surfing the Internet last week, however, I stumbled across a website with an interesting proof that shows that 2*Pi = 0 by using Euler's famous equation. As far as I can tell, all of the steps are mathematically sound. I've been puzzling over this problem over the past few days and I can't seem to make much sense of it. Here's the proof: Let x = e^(Pi) 1. x^i = -1 2. (x^i)^i = (-1)^i 3. x^(-1) = (-1)^i 4. [x^(-1)]^i = [(-1)^i]^i 5. x^(-i) = (-1)^(-1) 6. x^(-i) = -1 7. x^(-i) = x^i 8. e^(-Pi*i) = e^(Pi*i) 9. [e^(-Pi*i)]^i = [e^(Pi*i)]^i 10. e^(Pi) = e^(-Pi) 11. ln[e^(Pi)] = ln[e^(-Pi)] 12. Pi = -Pi 13. 2*Pi = 0 The key step is #7, where step #1 is combined with step #6. I've even checked this on my TI-83 calculator: when I enter e^(Pi*i) it returns a -1, and, likewise, when I enter e^(-Pi*i) it returns a -1. If both are equal to -1, this implies that e^(Pi*i) = e^(-Pi*i). Raise both sides to the power of i and you end up with e^(-Pi) = e^(Pi), which makes no sense whatsoever. One value is approximately 23.141 and the other is about 0.043, yet they are equal? From this, you can do some more mathematical manipulation and end up with 2*Pi = 0. If this were true, then that would mean that the circumference of any circle is 0. Obviously, this can't be true. If you can help clarify this situation, or come up with a possible answer as to why this proof is not mathematically sound, I'd be very grateful. Thank you. Date: 03/13/2002 at 23:25:52 From: Doctor Peterson Subject: Re: Was Euler wrong? 2*Pi=0? Hi, Warren. See if this explanation of a very similar "proof" helps: Find the Flaw http://mathforum.org/dr.math/problems/dan.08.02.01.html It is very tricky; even though I wrote that answer, I had trouble with this one. Your step 8 is fine; it still just says that -1 = -1. (In fact, most of the previous steps could be left out.) But whereas in "Find the Flaw" the problem lies in taking the logarithm, here step 10 is already bad before you've done that. That's because complex powers, as well as logs, can have multiple values. That is mentioned at the bottom of this page: Imaginary Exponents and Euler's Equation - Dr. Math FAQ http://mathforum.org/dr.math/faq/faq.euler.equation.html What you've done here is to show, not that -pi = pi, but that raising any number, even -1, to an imaginary power can give multiple values, and therefore is not allowed in a proof. And that's what false proofs like this are really all about: teaching us to be careful when we do the "obvious" in algebra! Here are some pages from the Dr. Math archives that more directly explain the idea that complex powers are multivalued: e^(i*pi) = -1: pi = 0 ? http://mathforum.org/dr.math/problems/koehler10.17.97.html Exponentiation http://mathforum.org/dr.math/problems/pomerance8.16.97.html I'll add a little further discussion of my own. We can write any complex number as r e^(it). Let's calculate this number raised to a complex power: (r e^(it))^(a + bi) = (r e^(it))^a * (r e^(it))^(bi) = r^a e^(iat) r^bi e^(-bt) = r^a e^(iat) e^(ln(r)bi) e^(-bt) = r^a e^(-bt) e^[(at + b ln(r))i] \_________/ \____________/ abs val angle But wait a minute: the angle t is not uniquely defined for a given number. Any angle t + 2k pi could have been used, for any integer k. Let's repeat using any such angle: (r e^(i(t + 2k pi)))^(a + bi) = r^a e^(-b(t+2k pi)) e^[(a(t+2k pi) + b ln(r))i] = r^a e^(-bt) e^(-2kb pi) e^[(at + b ln(r))i] e^(2ka pi i) = r^a e^(-bt) e^[(at + b ln(r))i] e^(-2kb pi) e^(2ka pi i) \_________/ \____________/ \_________/ \____/ abs val angle dilation rotation \___________________________/ \______________________/ principal value varies with k This tells us that the absolute value of a complex power has infinitely many values, whose spacing depends on b, while the angle can take different values dependent on a. In fact, if a is an integer, the angles will all be equivalent, but when it is not an integer, the angle will spiral around while the absolute value changes. Weird, isn't it? But in a way it's not that surprising; we see the same with fractional real exponents, which are likewise multivalued (there are two square roots and three cube roots, for example). Would you expect imaginary numbers to be better behaved than fractions when you use them as exponents? In your case, you have a pure imaginary exponent and a real base: (-1)^i = (e^((1 + 2k)pi i))^i = e^(-(1+2k)pi) So you get infinitely many positive real numbers. Your "proof" just assumes that two of them are equal, namely those for k=0 and -1. - 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- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/