Question

Can an algebra have an odd number of elements?

Answer 1

Yes

Answer 2

Gimme an example!!1

Answer 3

Do you really mean "an algebra" like I think you do, basically a vector space in which you can multiply vectors?

Answer 4

Oh ok, so an algebra A is a set of subsets of a set S with 3 conditions:

\[S \in A\] if \(X \in A\) then the complement of tha set is in A, \(S\backslash X \in A\)
and finally, the union of two sets is also in A, \(X \in A \) and \(Y \in A\) the \(X \cup Y \in A\).

Answer 5

This is an interesting question, do you have any thoughts on it so far, anything that could be helpful?

Answer 6

Yeah, the binomial theorem sums to powers of 2

Answer 7

It's my own question, just sorta trying to poke at what this stuff is, started learning measure theory so I can understand Lebesgue integrals cause I'm not the integral master yet

Answer 8

so there are indeed 2^N subsets of a set of size 2, that's true

Answer 9

er, set of size N**, rather

Answer 10

however that's not the full story, we don't necessarily need to take the power set of S to be the algebra A

Answer 11

The reason I suspect the evenness is for a different reason, it's cause the set and its complement come in pairs

Answer 12

but this union condition is throwing me off a bit maybe... just maybe there's some trick idk haha

Answer 13

Well, you're in good company! I'm not an integral master either lol So I'm going to go out on a limb: Have you taken an abstract algebra course, with groups and rings, that sort of stuff?

(It's a very good question, by the way)

Answer 14

I've not taken a course, although I have used rings and fields a lot and abstract algebra concepts a ton. I have a bangup knowledge of them but I know a lot of it.

Answer 15

That's very true! It'll turn out that any algebra is a sub-algebra of the full power set

Well, I'll share my intuition: These should probably be called "boolean algebras" in full generality; they're really algebras over the finite field with 2 elements.
Which probably doesn't mean much, but I have a feeling it's going to imply that we need an even number of elements. Certainly if S has an odd number of elements, it's true:

Answer 16

I also know what a group is specifically, there are 4 main things it must obey.

Identity
Inverses
Closed
Associative

And then Abelian = Commutative is a nice added 5th condition to have haha

Answer 17

So, if S is odd, then the complement of any subset X < S is going to be different from X; that implies that elements of our algebra A have to come in pairs, X, and S\X, and have even size

Answer 18

So I'll let you take a second to think about that (I really mean, at the end, that the algebra A has to have even size, since all of its elements come in pairs), let me know what you think

Answer 19

Ok I think I see what you're saying now.

Each set represents one way to cut the group into two parts from the condition that it and its complement are in the algebra. There's no way to "cut an even number of times" because of this.

Answer 20

\(\mathbb{Z}_p\) where \(p\ne 2\)

Answer 21

Actually, the more I think about it...No matter what the set X is, its complement is distinct from X: That means that elements of this kind of algebra _always_ come in pairs

Answer 22

are groups not algebras?

Answer 23

Yes, I agree! @Kainui ... Except I'm not 100% sure about the "cut an even number of times" part, but I have a feeling you're on the right track

Answer 24

So I think we have a well-defined way to cut the algebra into two halves, A1 and A2. Start by picking any set, and put it in A1. Put its complement in A2, noting that its complement is always different from itself. Do this until you've used everything up- There's probably a nice proof by induction in there :D

Answer 25

like if you cut bread into 3 pieces, that's 2 cuts. But each of those ends up turning into a single cut cause the other cut 'heals' since the complement of each individual cut ends upthere. Ok, sounds completely ridiculous what I'm saying haha.

Answer 26

@zzr0ck3r I'm not sure, I'm thinking about it now. I think an algebra might end up being a requirement for the set in order to be a group due to closure

Answer 27

haha, well, I'm not quite there yet, but at least you've got a good analogy!

Answer 28

I'm fairly sure that not only will a finite algebra have even cardinality, it'll be of size 2^N for some N. I don't really have an "elementary" reason to believe that's true, but I'm fairly sure that every algebra is isomorphic to the power set of S, for some set S

Answer 29

So it's definitely good to ask questions like this! I have to admit though, that I found the sigma algebra portion of real analysis to be the best part of the course, and understood approximately 3% of the course; it didn't really get me _that_ far. But I was able to do well in the course without understanding anything, so it was a happy ending! :P

Answer 30

Yeah, it seems useful and interesting for sure haha. I never took a course that went over it so I've had to self teach myself topology and all this junk. There are some pretty amazing resources on the web though.

Answer 31

I had a strong "I love math" background, and a lot of experience doing _other_ kinds of math; so I took to just mimicking the arguments you see in the course fairly well (evidently).
But of course, your mileage may vary I'd encourage you to keep playing with these things though! I wouldn't worry too much about not having a super strong grasp on these algebras, because they don't really take center stage for very long

Answer 32

Well, it seems simple enough, but as of now I'm really just looking to apply it to fixing up my knowledge of quantum mechanics since you are looking at probabilities of discrete and continuous sets. Bound states, where the total energy is lower than the potential has a discrete spectrum while the scattering state, where the total energy is higher than the potential energy, ends up having a discrete spectrum. So in order to look at probabilities of finding particles in states you have to use some measure theory to make predictions. So ultimately that's what's lead me here. I'd like to know better what it is I'm dealing with haha

Answer 33

Of course the answer to the OP is yes (the algebra \(\{\varnothing\}\) on \(\varnothing\)) but otherwise, the \(2^N\) answers makes sense in finite setting, and I don't know what is "even" for an infinite number of sets.. Maybe there's an abstract definition.. :)

Answer 34

Fair enough, the empty set and infinite sets are probably about as good as it's gonna get haha