Every math book i read starts off talking about how it assumed this. Why isn't there a proof of some kind?
isn't this a big flaw in math? you say you make a proof of all these things like the fundamental theorem of calculus and the fundamental theorem of algebra, yet you can't prove what you used to prove them with.
instead it's "let _____be commutative/associative/distributive under this set"; why can't we prove this first?
am i being insane? it seems off that math touts rigor then does this.
but that requires the successor function doesn't it?
Ryan Bailey
Please take introduction to real analysis before posting on this board Look up what a field is
Ryan Stewart
a field is a simply a set that is closed under multiplication and addition of elements within said set is it not? what i'm asking is what makes that notion of addition capable of commutativity without the context of a set. does that exist?
Connor Sanchez
X+Y=? Y+X |X=1 |Y=2 (1)+(2)=3 (2)+(1)=3 3=3 QED X+Y=Y+X The proof is just considered trivial.
Chase Green
except that breaks when you do that with systems that have addition defined as (x+y)=(x^y)
x=1 y=2 (1^2)=1 (2^1)=2 1=/=2 conclusion:not closed under addition
question is how the general addition you just used is capable of it and not this. that is, why is shit like exponents and log not commutative, but addition is? this is in no way trivial. hell, even units hold. 1km+2km=3km,2km x 2km = 2km^2, but 1km^2km is undefined. how is any of this shit trivial? if anything it's contrived. addition and multiplication just HAPPEN to work like this.
Christian Powell
They don't have to be, they just are in lots of certain cases. Feel free to make your own system of math where they are not commutative.
Grayson Wilson
What you said doesn't change his proof as you have defined addition to be a different operation as he has.
Adrian Evans
you're having one of those moments where you're not able to realize something obvious. to start, multiplication is defined in terms of addition, so it suffices to show addition is commutative/distributive/associative. to do that, start with the Peano axioms and use induction. if you've ever had any sort of halfway-decent analysis course, you've done this before.
Christian Russell
Your addition is not the same addition as his. You could argue that every single math reference in the world would have to include a definition like his, but everyone takes it for granted unless specified because so many people do it. We basically use #include
Isaiah Hall
>.h
Asher Ward
but then to use induction you need a successor function to prove both the base case and induction hypothesis. for that you have to define the successor function which has addition within it. isn't that circular logic?
i will read it, but just so you know, if this doesn't answer my question i'm going to Dedekind cut your shit.
Lincoln Hill
to start, the successor function does NOT have addition in it. i don't know what idiot told you that. second, addition refers to a very specific set of operations; when talking about it rigorously, one should specify what set they are applying addition on. e.g. naturals, integers, rationals, or reals. for each of those sets, addition is defined differently but intuitively the same. (note that addition can be shown to be commutative on all of those sets.) also, keep in mind that some sets, such as a set of permutation functions, don't have a notion of "addition". that said, we can define a different operation on them, like composition, which might.
Ayden Gonzalez
which might be commutative/distributive/associative*
Landon Myers
yeah, what did you want no extension like some kind of C++ ngrfgt
Gabriel Allen
Because that's how we defined them to be.
Justin Martinez
You need to read up on axioms
Bentley Collins
I think the proof could use induction. Suppose you know (for +) it is true for all a,b less than N. suppose now you try it for a+1,b (a+1)+b =(1+a)+b (comm. with 1,a) =1+(a+b) (assoc.) =1+(b+a) (comm. a,b) =(1+b)+a (assoc.) =(b+1)+a (comm. 1,b) =b+(1+a) (assoc.) =b+(a+1) (comm. 1,a)
Induction pretty much done.
The commutativity of multiplication follows a similar argument and also uses the commutativity of addition as well as distributativity.
Ethan King
The successor function does not require addition in general. You need the Peano axioms, in particular induction and S(n), and some suitable set theory if you want to be rigorous.
David Rivera
Muptiplication and addition of ordinal numbers is not commutative
Nathan Perez
Associativity seems to be the real elephant in the room.
It's easy to think of things that are not commutative (socks and shoes).
Imagining non-associative operators is a nightmare (octonions, sedinions: to name a couple)
