1 = -1?!

[Not Good]

My brain hurts after reading this thread. So I present this:

 

[Sevre]

Once I screwed up a maths equation and got 1=2 for the final answer. Quite hilarious at the time.

 

[Overlord_Dave]

Just remember, its only an equation! Its not real!!!! 1 does not actually equal -1!

I'm afraid that the majority of the world's philosophers and mathematicians will disagree. Something proved mathematically is immutably, undeniably and universally true.

There's just a mistake in the working somewhere.

 

[EeveeElectro]

Maths and a scary clown all in one page... I'm outta here.

 

[Galli]

I GET IT!
Of course... So obvious

 

[Vohn_exel]

I have a learning disability in math. I can't do higher then 8th grade algebra and thats on a good day when all the planets are aligned.

I get long division wrong sometimes, for pete's sake.

 

[Helston]

Being a maths person I'm taking up this challenge, but what does the very first line mean? I'm not too good with maths jargon symbols.

EDIT: Does it mean X = 2n * pi, such that n is a natural number?

 

[adventurecat]

e^ix is always equal to 1 when x is only allowed to be equal to 2npi for some n (as stated in the original problem).

The problem is with raising both sides to the power pi/x.
Because x is equal to 2npi we have both sides being raised to the power 1/2n, which is always less than 1. There isn't just one root to a number, so saying that 1^pi/x = 1 is incorrect, there are several answers to it, including 1, -1 and various complex answers if n > 1.
So, when we multiply the two sides by pi/x we are saying that the roots of each side are equal to the roots of the other side, not that every root of each side is equal to every root of the other side.

 

[ILPPendant]

Being a maths person I'm taking up this challenge, but what does the very first line mean? I'm not too good with maths jargon symbols.

Consider the set X which contains all the even natural numbers multiplied by pi.

EDIT:

e^ix is always equal to 1 when x is only allowed to be equal to 2npi for some n (as stated in the original problem).

The problem is with raising both sides to the power pi/x.
Because x is equal to 2npi we have both sides being raised to the power 1/2n, which is always less than 1. There isn't just one root to a number, so saying that 1^pi/x = 1 is incorrect, there are several answers to it, including 1, -1 and various complex answers if n > 1.
So, when we multiply the two sides by pi/x we are saying that the roots of each side are equal to the roots of the other side, not that every root of each side is equal to every root of the other side.

DING DING DING!!! We have a winner!

 

[Helston]

e^ix is always equal to 1 when x is only allowed to be equal to 2npi for some n (as stated in the original problem).

The problem is with raising both sides to the power pi/x.
Because x is equal to 2npi we have both sides being raised to the power 1/2n, which is always less than 1. There isn't just one root to a number, so saying that 1^pi/x = 1 is incorrect, there are several answers to it, including 1, -1 and various complex answers if n > 1.
So, when we take multiply the two sides by pi/x we are saying that the roots of each side are equal to the roots of the other side, not that every root of each side is equal to every root of the other side.

GARRR - was just thinking that, or something along those lines.

 

[Lukeje]

Edit: Wait; that's nonsense. Ignore.

 

[Exocet]

You also made the mistake in the first line:
it should be |N* (all the natural(?) numbers excluding 0) since you then raise to (pi/x) and 2*0*pi=0 and a division by 0 is a sure way to fuck up your grade.

I don't see were the rest went wrong yet,but I'll find it!

 

[megalomania]

I think:

e^ix = cos(2pi)+isin(2pi) is fine but then you use the identity

cos(2pi)+sin(2pi) = 1 is not equal to cos(2pi)+isin(2pi)

The trigonometric identity you used becomes invalid when you have an imaginary sin part.

 

[Haberley]

It's equations like this that make me glad I droppped Maths at A Level...

 

[Vohn_exel]

Yeah well...the Square Root of Nine is Three!

 

[ILPPendant]

You also made the mistake in the first line:
it should be |N* (all the natural(?) numbers excluding 0) since you then raise to (pi/x) and 2*0*pi=0 and a division by 0 is a sure way to fuck up your grade.

I don't see were the rest went wrong yet,but I'll find it!

Ah, this is an awkward point for sure but I've always been taught that zero is not a natural number.

Yeah well...the Square Root of Nine is Three!

Or minus three.

 

[Doug]

First line is wrong. e^ix is not always = 1. You've made the mistake of cos x + sin x = 1, when its cos^2 x + sin^2 x == 1.

Also, there is an i on the sin that you seem to have completely ignored.

EDIT:
Also, e^ix for all x IS NOT = cos 2pi + sin 2pi. Its e^ix = cos x + i sin x, if I remember my maths correctly.

Your challenge is only valid if x is something other than what I've defined it to be in the first two lines.

EDIT: Notice line three states for all x in X, not for all x in R or N.

Well, OK THEN.

e^(i*2*pi*n) = cos (2*pi*n) + i sin (2*pi*n) NOT THE SAME AS e^(i*pi) then.

e^(i*2*pi*n) == 1 (IF n == an integer)
e^(i*pi) == -1 (NOTE: NOT THE SAME AS THE PREVIOUS LINE)

You're comparing 2 different equalations.

 

[JemJar]

Does the fact that an Escapist solved this in under two hours not speak volumes about how much attention people pay to the noticeboard in your Maths Dept?

 

[Vohn_exel]

"Or minus three."

LIES!

 

[ILPPendant]

Well, OK THEN.

e^(i*2*pi*n) = cos (2*pi*n) + i sin (2*pi*n) NOT THE SAME AS e^(i*pi) then.

e^(i*2*pi*n) == 1 (IF n == an integer)
e^(i*pi) == -1 (NOTE: NOT THE SAME AS THE PREVIOUS LINE)

You're comparing 2 different equalations.

Yes, that's pretty much it, well done.

Does the fact that an Escapist solved this in under two hours not speak volumes about how much attention people pay to the noticeboard in your Maths Dept?

I hope for the sake of my future education that that's the real reason.