Why are addition and multiplication commutative? Every math

>>9074740

Just use induction

>>9074747
but that requires the successor function doesn't it?

Please take introduction to real analysis before posting on this board
Look up what a field is

>>9074756
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?

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.

>>9074787
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.

>>9074740
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.

>>9074814
What you said doesn't change his proof as you have defined addition to be a different operation as he has.

>>9074740
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.

>>9074787
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 <commonsense.h>

>>9074835
>.h

>>9074830
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?

>>9074838
just by this post, and your previous ones, you aren't going to make it, anon...
read over these notes: http://www.math.ucla.edu/%7Etao/resource/general/131ah.1.03w/week1.pdf

>>9074846
i will read it, but just so you know, if this doesn't answer my question i'm going to Dedekind cut your shit.

>>9074866
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.

>>9074879
which might be commutative/distributive/associative*

>>9074836
yeah, what did you want no extension like some kind of C++ ngrfgt

Because that's how we defined them to be.

>>9074740
You need to read up on axioms

>>9074740
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.

>>9074838
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.

Muptiplication and addition of ordinal numbers is not commutative

>>9074740
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)

>>9076638
exponents are non associative