Proof that math is wrong

[Marty]

Proof

Where is the mistake?

 

[nater]

uhh what do I do with the I's

 

[acidvoodoo]

dunno 😕

 

[Anubis]

its because of teh imaginary numbers

 

[DoNotDisturb]

that doesn't prove that math is wrong, that proves that for that problem it is an exception.

if u were to add: 1 + 1 using the traditional real number system, would it not equal 2? yes, and thats math.

if 5x = 15, would not x = 3? yes, it would, thus, that proof is not valid for everything.

 

[Kev]

I dressed myself.

 

[tarheelmm]

un huh

 

[brunswickite]

my life is now complete

 

[WinkOsmosis]

x=0

 

[Marty]

Whoops, there was a small typo. It's gone now.

 

[MonkOnXanax]

Originally posted by: Marty
Whoops, there was a small typo. It's gone now.

hahaha, i was wondering how that pi cancelled out.

 

[RaynorWolfcastle]

not sure why yet, but I'm pretty sure that the problem is going from step 2 to 3... I'd look into it, but I'm lazy

 

[aux]

Short answer: you are using (line 2 -> line 3) something like e^(a*b) = (e^a)^b, which is true for real a and b only (and in your case a = 2*pi*i, which is a complex number).

This has something to do with the funny way the logarithm function (as an inverse of the power function) is defined in a complex domain, but the argument would be too technical for ATOT 🙂

 

[DoNotDisturb]

anyone here try working on the navier stokes equation? =)

 

[RaynorWolfcastle]

Originally posted by: aux
Short answer: you are using (line 2 -> line 3) something like e^(a*b) = (e^a)^b, which is true for real a and b only (and in your case a = 2*pi*i, which is a complex number).

This has something to do with the funny way the logarithm function (as an inverse of the power function) is defined in a complex domain, but the argument would be too technical for ATOT 🙂

heh, time to break out the ol' Complex vars textbook 😉

 

[Snapster]

Proof that math is wrong.....using math, oxymoron or what. 😛

 

[DoNotDisturb]

paradox.

 

[bleeb]

imaginary numbers.. it becomes zero? instead of 2?

0.9999.... != 1

 

[DoNotDisturb]

Originally posted by: bleeb
imaginary numbers.. it becomes zero? instead of 2?

0.9999.... != 1

lol dont start with that again!

 

[Marty]

... so why does exponetiation not hold for complex numbers?

 

[Triumph]

Why would e^2*pi*i = 1? e is a number, 2.71828182828182818 or close to it. 2*pi is roughly 6.28. I don't think e^2*pi*i is = 1.

 

[Marty]

Proof that e^(i*2*pi) = 1

 

[spidey07]

old trick...NEXT!

-edit- someone's trying to get extra credit. 🙂

 

[TuxDave]

I think the flaw comes in when it comes from the step that (1)^(x/2pi) = 1. In fact, I shall disprove it by counter example. =)

Proof why (1)^(x/2pi) != 1 all the time

I choose x to be pi.... therefore 1^1/2 = +- 1

-Dave-

<edit> Cuz x is a variable, you can't conclude much about 1^x for any generic case.

 

[Grminalac]

Hmm, wouldn't
cos(x)=([1] x/2pi + [1] -x/2pi);
cos(x)=(x/2pi+ -x/2pi);
cos(x)=(0/2pi);
cos(x)=0