Sunday, December 21, 2008

Imaginary paradox

Following my proof of the obvious, I thought it might be fun to prove the impossible. Math paradoxes!

Well, of course you all know about the proof of 1=0? It's included below for completeness.
Let a = b
a*a = b*a
a*a - b*b = b*a - b*b
(a+b)(a-b) = b(a-b)
a+b = b
a = 0
Let a = 1
1 = 0
The experienced algebraist quickly spots the error. One of the steps in the proof is an illegal move.

The following proof also makes an illegal move somewhere, but it breaks a rule that most people haven't ever heard of.

We start with Euler's formula:
eix = cos(x) + i*sin(x)
Here, "i" is used to denote the square root of -1. Euler's formula is far from a mere mathematical curiosity. I do not exaggerate when I say we use it all the time in physics. Euler's formula is the main reason that imaginary numbers are of any use at all. But I digress.

Using Euler's formula, we know...
e = -1
(e)i = (-1)i
(-1)i = eiπi = e
But then...
ei3π = -1
(ei3π)i = (-1)i
(-1)i = ei3πi = e-3π
Therefore, e = e-3π. But clearly this is wrong. Therefore, our proof is wrong. But where?

5 comments:

foole said...

How sad is it that my first thought for your first proof was "hey, this is like nullity[0]." The guy proposing nullity actually using a similar "proof" to a group of British schoolchildren (taking mere stupidity and making it truly awful).

[0]: http://www.bbc.co.uk/berkshire/content/articles/2006/12/06/divide_zero_feature.shtml

Anonymous said...

I'm not sure I know what is wrong with it, but I'm wondering whether this move is legitimate: Moving from

e^iπ = -1

To:

(e^iπ)^i = (-1)^i

I worry about it because it is not clear what you are multiplying both sides with. With non-imaginary numbers, you can usually put exponents on both sides of an equation because it is short hand for multiplying both sides with the same number.

If I'm write about this move being illegitimate, then the similar move in the second part is also illegitimate.


(I'm also worried that my answer is wrong, because my diagnosis doesn't show the mistake to be breaking a rule most people never heard of.)

miller said...

In general, if we have a function f(z) which returns a single value for every number z, then a=b implies f(a)=f(b). The function f(z) does not need to be a multiplication.

On second glance, I think the paradox is perhaps a bit more subtle than I initially made it out to be. I think there are at least two places where the error could be, depending on what conventions you're using.

Yoo said...

I seem to have a vague recollection that general exponentiation with powers of imaginary numbers is not single-valued ...

And under one reasonable single-valued definition of such exponentiation, I'm pretty sure that (e^(i3π))^i is not equal to e^(i3πi). For that matter (x^y)^z would not generally be equal to x^(yz).

miller said...

Yep, you've got it. Either you've got to consider the full set of values of complex exponentiation, or you have to define a "principle" value in much the same way that we usually define the "principle" square root to be the positive root rather than the negative one.