@0celo7 it...should?

@heather yes

@alarge Have you been reading my Wayland updates?

@0celo7 hopefully your definition of easy is a little higher than mine

Why can't mathematicians use the physical conventions for factors of $i$ in representation theory?

@heather No, it should be really easy. If I tell you, you will think so too.

@0celo7 i'll keep thinking then

@heather oh, small typo

$A\cup B\supset (A\cap B)\cup (A\Delta B)$

is it obvious now?

what did you change?

oh, nvm

i think so

tell me

What are you two doing?

(set theoretic) boolean algebra

whatever you do to two sets you cannot make a set that isn't within or equal to the combination of both sets.

@heather exactly

easy, right?

@0celo7, yeah. How do you write it formally?

@0celo7 Hm? What's boolean about this?

Well, not quite. If you take a complement you'll end up outside of the sets

oh, true.

but we're not.

@BernardoMeurer A ring of sets is a Boolean ring with multiplication $\cap$ and addition $\Delta$

Let A and B be sets. A $\cup$ B is closed under the operations $\cap$, $\cup$, and $\Delta$ so any combination of $A$ and $B$ using those operations must produce a set that is a subset of $A\cup B$.

@0celo7 What do you mean by $\Delta$?

something like that, maybe?

@heather Sure.

@BernardoMeurer Symmetric difference $A\Delta B=(A-B)\cup (B-A)$

cool.

@0celo7 Sweet

Okay, cool

@0celo7 Once you reach the point, show her why a boolean algebra can never form a vector space, it's a nice thing to think about

Then that's a terrible terminology, because an algebra is a vector space!

@0celo7 Yeah, that's why I thought it would form a vector space :/

hmm... so now that I've proved $\supset$ if I prove $\subset$ then it must be $=$ right @0celo7?

I feel like we've had this conversation before.

@heather yes.

@0celo7 We have

okay, let me take a stab at that then.

@0celo7 To form a vector space it must form a field, and a Boolean Algebra does not form a field (Is this correct?)

Not correct

Other way around, a field is necessarily a vector space

And there's no multiplication in a general vector space

That's what an algebra is

@ACuriousMind BIG BOY

@0celo7 This is a not quite correct thing to say. One can multiply by scalars.

But I know what you mean

There's no multiplication of vectors

@heather First take a point in $A$, and show $A\subset (A\cap B)\cup (A\Delta B)$. Then take a point in $B$ and show $B\subset (A\cap B)\cup (A\Delta B)$. Conclude $A\cup B\subset (A\cap B)\cup (A\Delta B)$

@BernardoMeurer Yes. I haven't had the time to look into setting up stuff myself so I haven't replied

hmm, okay, I needed that hint. thank you.

Also I did look into Wayland again and I think Nvidia's prop drivers don't do things the way the standards dictate, so many of the implementations have decided to ignore Nvidia and recommend using nouveau

which I'm not going to do

Draw a good Venn diagram.

but I understand the sentiment

@ACuriousMind I finally understand angular momentum.

@0celo7 It's about time.

@DanielSank No physics book explains it like this

@0celo7 Then don't use a physics book.

@BernardoMeurer I'm sorry, but if you think ZFC is wrong I don't think I'd invite you for coffee :P

@BernardoMeurer have coffee with me instead.

@ACuriousMind I can show you a nice argument if you'd like, using turing machines, and I'd be happy to get it refuted because I'm a bit worried if it's correct

@DanielSank <3