View Full Version : erroneous proofs and mathematical fun!

Smash_Returns

03-11-2009, 08:37 PM

(e=the base of the natural log; i = sqrt(-1))

e^(pi*i) = -1 (true statement)

(square both sides)

e^(2*pi*i) = 1

(take the natural log of both sides)

ln e^(2*pi*i) = ln 1

(simplify the natural log)

2*pi*i*ln e = ln 1

(further simplify; ln e = 1, ln 1 = 0)

2*pi*i = 0

Q.E.D.

Where is the error in the above proof?

sKratch

03-11-2009, 09:10 PM

The argument of the natural log is some crap to a power. The log is not raised to a power. Therefore you can't bring down the exponent.

Smash_Returns

03-11-2009, 09:19 PM

a log brings down a power. e is raised to 2*pi*i. The natural log allows us to bring down the power. That is the whole POINT of a logarithm.

That is not a problem.

Endymion

03-11-2009, 11:28 PM

a log brings down a power. e is raised to 2*pi*i. The natural log allows us to bring down the power. That is the whole POINT of a logarithm.

That is not a problem.

[edit for easier answer]:

you missed the branch cut.

Budzy

03-11-2009, 11:30 PM

you can only bring down a power if it's real. complex powers don't come down like that.

l2math.

My complex powers come down hard on anyone who displeases me.

sKratch

03-12-2009, 06:24 AM

Endy's right; logs and exponentials are significantly different in the complex plane.

Virtuoso

03-12-2009, 01:14 PM

Here's a short (and easy) one:

Assume that a is a positive number.

a - 1 < a | *(-a)

-a^2 + a < -a^2 | +a^2

a < 0

Haha, so lame...

Smash_Returns

03-13-2009, 08:39 AM

Endymion is correct. When we bring down the 2*pi*i, because the power is complex, the modular base changes.

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