[unreality]

This is a good one

e^iπ = -1

[square both sides]

e^2iπ = 1

[take the natural logarithm of both sides]

2iπ = ln(1) = 0

But something is wrong! 2*pi*i does not equal 0, it equals 2pi*i, ie, ~6.28i. What went wrong?

Post answers or your own mathematical fallacies

[Guest]

this had me going for a bit

the final step is where it looses. cant take ln of i. would have to evaluate in expanded form ln[cos(2*pi) + i*sin(2*pi)] = ln[1] = 0

[Guest]

Okay the answer is as follows:

2 pi needs to be expressed as 0 to create an equivalent expression due to the fact that you are working in polar coordinates with imaginary numbers.

You would end up with 0*i = 0 which is true.

[unreality]

actually - as I understand it ( ) - nobody has gotten it yet ;D

[Guest]

actually - as I understand it ( ) - nobody has gotten it yet ;D

I'm pretty sure Scrappy is right, though his answer is a bit hard to follow. In polar coordinates, an angle of 2*pi radians is the same as 0 radians, and the 2*pi must be reduced to 0 to simplify the expression, just as 4*pi and 6*pi, and any 2*n*pi, where n is an integer...no matter how many times you go around the circle, you still end up at 0. When you enter ln(e^(2*n*pi*i)) for any n into a calculator, it gives you an answer of 0 because it automatically reduces the expression.

edit: btw, welcome to the Den, Scrappy!

Edited January 5, 2009 by lazboy

[unreality]

yes, so the trig functions reduce it on their own - my calculator doesn't do what you're saying So there is at least two fallacies then

the problem occurs when you square it. When you square something, the information is lost. Ie:

5^2 = 25 = -5^2

for that reason, when you take the square root of 25, you're not sure whether it's positive or negative 5, so you write ±5

well in this case, when you square -1, if you leave it as (-1)² instead of +1, then it works:

e^iπ = -1

square

e^2iπ = (-1)²

take ln of both sides

2iπ = ln((-1)²) = 2*ln(-1)

2iπ = 2*ln(-1)

drop the 2s

iπ = ln(-1)

Which is a true statement of course

edit-typo

Edited January 6, 2009 by unreality

[unreality]

I know a couple more good paradoxes, but I'm waiting to see if someone else posts one first

[Guest]

You actually have a fallacy in the very first statement. I don't know how to do the funky formatting though, so I'm going to use ^ to represent exponents.

Instead of e^(in) = -1, you should be using e^i(2n+1). After all, if n is an integer, then n can be any integer, including the even ones. The way to represent odd integers isn't n, it's 2n+1. This would modify the proof from there on out, but I can't do the formatting, and don't want to do the math, so someone else can

[unreality]

you're mistaking it, I think. the π stands for pi, 3.14159... It looks kind of like an 'n' letter but I was trying to use the pi symbol

[Guest]

You actually have a fallacy in the very first statement. I don't know how to do the funky formatting though, so I'm going to use ^ to represent exponents.

Instead of e^(in) = -1, you should be using e^i(2n+1). After all, if n is an integer, then n can be any integer, including the even ones. The way to represent odd integers isn't n, it's 2n+1. This would modify the proof from there on out, but I can't do the formatting, and don't want to do the math, so someone else can

As unreality said, the n isn't an n its a π (or pi). If you change the font, then its a bit more legible: π π π π

e^(iπ) = -1 (or e^(iπ) + 1 = 0) is a famous equation*, and, for me at least, is a pretty surprising one linking so many fundamental numbers.

I first read the equation as e ^ ln (or natural log) and thought unreality had made a different blunder, meaning to put ln(e) = 1.

*eulers equation, i think

Edited January 7, 2009 by armcie

[unreality]

it's called Euler's Identity, and is a special case of Euler's Equation (I guess it's actually called Euler's Formula not Euler's Equation, whatever )

To me, it's just as surprising as a formula for i^x I discovered by accident a few months ago:

i^x = cos(x * pi/2) + i*sin(x * pi/2)

that's off the top of my head btw, but I'm pretty sure it's correct. Of course, the special cases give i^2 = -1, i^3 = -i, i^4 = 1, etc, likewise i^(1/2), the square root of i, gives sqrt(2)/2 + i*sqrt(2)/2

edit: more surprising is that i^i is a real number lol

Edited January 7, 2009 by unreality