This is a really dumb question...

But what's more likely? 1) P == NP, or 2) Riemann Hypothesis is false?

It's easy to analyze 2) from a statistics perspective, but it's really hard to analyze 1)

Ok, so this is crazy: Apparently you can have multiple people charged with a crime, and if there is sufficient evidence to convict both of them, both can be convicted

vistacriminallaw.com/…

I don't know of a better alternative: You don't want a court where you are trying to play the game of "which person did it", but this is a rather interesting side-effect

Wow, that's weird.

@DJMcMayhem on that note: twitter.com/HLForum/status/1042670700652318720

Yeah, that's somewhat what inspired the question

@flawr Whaaaat?

@El'endiaStarman many consider it unlikely that he has a correct proof

Yeah, I'm pretty skeptical

at the same time it would be very cool to witness that, even if I probably wouldn't understand the tiniest bit:)

I'm curious as to why though. Did people have the same reaction to Andrew Wiles' proof of Fermat's Last Theorem?

(Which, incidentally, did have some errors at first, but they were eventually fixed.)

It was mentioned that he had claimed to have a proof for some other "famous" problem.

@flawr What is an n-sphere?

AFAIK, a regular old real life sphere (like a globe or a ball) is a 2-sphere, right?

right

(just the surface)

I would intuitively expect that to be a 3-sphere (3-dimensional)

basically all points in n+1 dimensions with distance 1 to the origin

@DJMcMayhem well the surface is just 2 dimensional

Is there a standard coordinate system? (Lat/long?)

@flawr Is a 0-sphere defined, or does it start at 1? (Since I assume a 1-sphere is really just a circle)

@DJMcMayhem right, a 1-sphere is a circle, a 0-sphere is just the set {1,-1} with this logic (but I don't think it is usually considered a sphere)

Interesting how every N-sphere has an infinite number of points except for a 0-sphere

@DJMcMayhem depends what field you're using. right now we were talking about real n-spheres

a complex 0-sphere has infinitely many points

o_O

So what is a complex n-sphere?

you just replace each real number with a complex number:)

again you consider C^(n+1) (n+1 dimensions, each entry of a vector here is represented as a complex number)

and consider all points in C^(n+1) with distance 1 to the origin

So, a complex 0-sphere is every complex point with a euclidean distance of 1 to (0, 0)? So it includes (0i, 1), (0i, -1), (i, 0), and (-i, 0)? (and others)

now I think you're mixing two things

it looks like you're referring to points in C^2

OK, I think you've lost me. What is C?

I think you mean the complex points with a distance of 1 to 0 = 0+0i

@DJMcMayhem C=complex numbers={a+bi where a,b are both real numbers}

@DJMcMayhem facepalm. It's {a+bi}, not {ai+b}

the complex 0-sphere (as a subset of C=C^1) includes 0i+1, 0i-1, i+0, -i+0

@DJMcMayhem ah no that is irrelevant, both work

but (0i, 1) would usually be considered as an element of C^2

but I think you meant 0i+1 as an element of C

@flawr Oh wait, I think I get my mistake. (0i, 1) is two dimensions (where each coordinate is complex) and 0i+1 is a single point in complex space

@DJMcMayhem exactly!

OK, that makes more sense. But there's still one thing I'm missing. Why do you refer to C (or C^n) as a set?

And C^n means complex n-dimensional space, not C ** n, right?

@DJMcMayhem exactly

if you have any set S then S^n is usually the cartesian product of S taken n-times with itself

so S^n is the set of n-tuples ("vectors") where each entry is an element of S

That makes way more sense. Thanks for your patience :)

I'm sorry for the confusion:)

I think a lot of my confusion comes from the fact that I think of complex numbers as 2-tuples rather than numbers. So it feels 2-dimensional to me, so it seems like an arbitrary distinction between a complex 0-sphere and a real 1-sphere (and that an n-sphere is in n+1 dimensions also seems arbitrary to me)

But don't worry, I'm not gonna post another wall on that topic :)

@DJMcMayhem Right, it is just convenient for us to represent complex numbers using two real numbers, so it is no suprize that we do imagine them that way.

You know now that I think about it, it's kinda weird that imaginary numbers all sorta have a real part. (not the technical definition of real part) Like, 6i is really 6 (real) * i (imaginary). And even i is really 1 (real) * i (imaginary). But real numbers have nothing imaginary to them. 6 is just 6, 1 is just 1

Although I guess you could argue that real numbers have the imaginary part i^4n

@DJMcMayhem Then again, defining spheres S^n = {x in S^(n+1) with ||x||=1 } could be considered just one "nice" way of describing them, but depending on the perspective you have on them, you do not really think of them as a subset of S^(n+1), but just a set by themselves with some properties (for example form a topological point of view).

Or... 2n?

@DJMcMayhem or 0 :)

n could still be 0 :P

or you could set n=i

(actually works*)

What the hell... Why is i^i real? That's so confusing

@DJMcMayhem exponentials a^b are usually defined as a^b = exp(b*ln(a))

now the logarithm ln(a) of a complex number "a" is slightly tricky, as there are many ways to write "a" as "a = exp(x)"

so you have to "choose" one of these ways (called a "branch" of the complex logarithm)

as you rightly said, i^(4n) = exp(i*pi/2)^(4n) = exp(2*pi*i *n) = 1^n= 1 for all integer values of n

so to find i^i we need to first find some value x such that i = exp(x)

do you have an idea what x could be?

Hold on a sec, I'm thinking

@flawr I don't quite see how i^(4n) = exp(i*pi/2)^(4n). I can see exp(4n * ln(i))

@DJMcMayhem exp(i*x) = cos(x) + i*sin(x), you could take that as a definition

So exp(4n * ln(i)) == exp(4n * pi / 2 * i) == exp(2 pi * i) == cos(2 pi) + i * sin(2 pi) == 1

yup!

@flawr Pi / 2!

@DJMcMayhem 100 points :)

Which incidentally is equal to pi / (2!)

but that is just one possibility, can you find another one?

Oh gosh

@flawr Cheap answer, but 2n * pi + pi / 2 for all integer n :P

bingo:)

@DJMcMayhem wolframalpha agrees: wolframalpha.com/input/?i=i%5Ei

@flawr OK, I'm super close, but I accidentally proved that i^i == i. Where did I go wrong?

i ^ i
exp(i * ln(i))

ln(i) --> e ^ x == i

e^x == cos(x) + i sin(x)
cos(pi/2) + i sin(pi / 2) == i --> ln(i) == pi / 2

exp(i * pi / 2)

cos(pi/2) + i sin(pi / 2) == 0 + i

@DJMcMayhem it is e^(i *x) = cos(x)+i sin(x)

I probably have to go now. I'll come back to this later

Thanks again for all your help :)

@DJMcMayhem cu!

@DJMcMayhem no worries, complex numbers are fun! :)