I still don't get why multiplication is commutative?!

Because when you multiply 5x3, you are either adding five to itself three times (5+5+5) or adding three to itself five times (3+3+3+3+3).

ISA bravo is that you son?

I suggest you get yourself a bag of objects, and place them in three rows of fives and five rows of 3's and count them

>multiplication is commutative because of (insert a consequence of multiplication being commutative)

circular argument senpai
you're saying that it's commutative because 5x3 is the same as 3x5

if you cant wrap your head around this basic fact, just give up. and no, you're not being deep or clever by disputing it.

...

Yeah, you don't need to look at it any harder than this.

Because 5 + 5 +5 = 2 + 3 + 2 + 3 + 2 + 3
And 2 + 2 + 2 = 3 + 3
So 5 + 5 + 5 = 3 + 3 + 3 + 3 + 3

The relationship between addition and multiplication is really interesting.

Oiii that's nice

THIS.
OP, you can consider the definitions and axioms on the matter... but nothing can substitute for a firm foundation in geometry, which aids one's intuition greatly.

...

This. Multiplication comes from area.

Well, multiplication is commutative because the addition is, and the addition is commutative as a principle.
Ex: 2x3= 3+3 = (1+1+1) + (1+1+1) = (1+1)+(1+1)+(1+1) = 2+2+2 = 3x2
Note that the parenteses are only there as a visual guide, it doesn't serve any other perposes

...

The real question is, why aren't powes communitative?
ie x^y=/=y^x
>inb4 sometimes it is

>t. Pucci

I'm sorta glad that Jojo is equal parts stupid and intellectual

This has to be fake. Source?

Jesus christ this thread. you cant just say 'because the area is the same'. if you want it in hat form:

multiplication is defined as a*b = a + a*(b-1)

directly you get a*b = a + a*(b-1) = a + a + a(b-2) = ... = a + a + a + ...

alternatively, a*b = 1*a*b = (1 + 1*(a-1) )*b = b + 1*(a-1)*b = b + (1 + 1*(a-2) )*b = ... = b + b + b + b...

heres a bunch of proofs for it.
proofwiki.org/wiki/Natural_Number_Multiplication_is_Commutative

this might be the best and most simple explanation I've seen.

Now explain to me why this "area" function is rotation invariant.

Because addition is commutative and multiplication is just repeated addition. Why is addition commutative? Proof is left as an excercise for the reader. Hint: (recursion) prove that 0+n=n, prove that n+0=0, prove that n+m=m+n

Are you a brainlet? That doesn't explain multiplication like this guy said Dumb chink spammer.

Pick up a book.
Rotate it.
Explain why the area would be any different.

You're assuming you can neglect the Lorentz contraction of the book when you rotate it.

why rotate book when you can rotate your coordinate system

I don't think that would make much of a difference because there is relative motion anyway.

I believe that was the point he was trying to make

nice
BUT
what about negative numbers?
using this method the answer is infinite
maybe you should have used the abs value

give me a second to think about it

read the fucking book already user

You define -1*X = -X and then you can make any -n*X into n*-1*X

I don't get it
using the recursive definition he gave earlier
multiplication of negative numbers would yield infinity

why rotate your coordinate system when you can rotate your abstract view point of said coordinate system

because rotation is an isometry and thus measure preserving, brainlet

You extend multiplication in N to Z, then to Q, then to R, then to C. That recursive definition shows that multiplication is commutative in N. Then you show that that extends to Z, Q and R. It's fucking first year abstract math shit, brainlets.

Read the link you cuck.
The recursive definition is only for positive integers.

use his recursive definition to give me the result of
-1*-2

fuck the link
the whole point is to get to it yourself

and if the recursive definition only works for the positive integers
then prove to me that -a*-b = ab

Vectors do not change magnitude upon rotation, you absolute brainlet.

why?

yes they do

if they were horizontal, their mangitude becomes uppier

>prove to me that -a*-b = ab
factoring out the a and b you have
0 = (-1)*0 = (-1)*(1 + (-1)) = (-1)*1 + (-1)*(-1) = -1+((-1)*(-1))
thus -1*-1 is the inverse of -1, since inverses are unique -1*-1 = 1

Prove it.

And more importantly, prove it without relying on the commutativity of multiplication.

ok nice
but what proves the distributive property for multiplication in Z?

Take any vector. The magnitude of this vector can be taken as the radius of a sphere centered at the origin, such that the tip of the vector touches the shell of the sphere. Rotation this sphere about the origin will not change the radius of this sphere. Because the radius is unchanged, so must be the magnitude of the vector we took in the first step. Ergo: magnitude does not change under rotation. QED.

Euler did this shit in 1776 by the way. Try not to fall behind schedule, brainlet.

a(b+c) = ab+ac over the integers because addition is commutative and associative (expand a(b+c) as (b+c)+... then use assoc and comm to get b+...+b+c+...+c which is the RHS).

I think he needs to show that he can factor out the a and the b out of -a*-b though (probably by proving the special case of commutativity for -1*a)

>Proves rotation invariance by saying "Look when you rotate it it doesn't change!"
Great proof there.

yeah
I mean we use it all the time
but it's surprisingly hard to prove

if we could only find some def for multiplication that is valid for all Z it would help a lot.
I think it might need a sign function

to be fair, any mathematician who accepts the faulty notion of "uncountable infinity" ought not to have a problem with his explanation, since it's on the same level of handwaveyness

>Veeky Forums can't even construct the reals out of the naturals and prove it satisfies the field axioms
lmaoing @ your life, brainlets

Construct Q from N. Prove the properties still hold. Done. It's chapter 2 in Tao, why do you think it's hard to prove? It is one of the first things you do in uni.

ok then prove to me that the distributive property of multiplication holds in Z

It holds in Q. Z is subset of Q. It holds in Z. Wow that was hard.

good luck in life
you're gonna need it

>hurr durr i can't use the tools i have

>to prove

>a property that's given as introductory excercise

>to freshmen undergrads

>around the world
In other words, russian highschooler can do this. Why can't you?

it pretty much assumes that -1*-1=1
it does not derive it naturally.

to say that is equivalent to saying that the distributive property of multiplication holds in Z.
like this guy showed
I'm trying to get to it naturally

show the fault user

youtu.be/elvOZm0d4H0?t=4m

Please do me favor and read the book. You're spouting nonsense, it is probably the most rigorous book on introductory real anal, there's no hand-waving (there is, but it is rigorously proved later). I skipped the proof of negation on integers, but i presume you can use your brain and see how the proof of -1(-1)=1 is consequence of composing negation and multiplication.

assume I can't
prove it to me

if your proof is "read the book" than
1) you are too stupid to remember the proof even though the book s in front of you
2) you enjoy writing nonsense while pretending to be in any way mentally superior to others

it could be either one or both

Or none. I'm not going to type it all out on phone, if the thread is alive when i get home, i will, nicely formatted in latex. Starting from peano axioms, i will show you how to prove distribution on multiplication for Z.
I don't like doing other peoples homework, and this thread coincides with the start of semester in many countries so i'm suspicious, but i'll give you the benefit of the doubt.

You forgot to prove the distributive. And don't look it up.

fine with me
I'll be watching this thread later

Not him but I'll give you the quick and dirty version of integer commutativity :
Let x and y be two integers. x = a-b and y = c-d where a,b,c,d are natural numbers. From there, expand out xy=(a-b)(c-d)=ac-ad-bc+bd and yx=(c-d)(a-b)=... (note that distributivity for integers was proven above in this thread). Then use the fact that we know natural numbers are commutative under addition and multiplication to conclude the proof that xy=yx for any two integers.

thanks but I just want someone to prove to me that -1*-1=1
it will prove everything else
you assumed it

You need distributive first to prove commutative.

-1(-1)=1 we want to prove, by definition we know -(a-b)=(b-a)
-(a-b)*-(a-b)
(b-a)(b-a)
bb-ba-ab+aa
we know ab=ba
bb+aa
Our case is 1. Assign a=1, b=0 since 1=1-0
0*0+1*1
1
thus -1(-1)=1

Alright. Stealing from Rosenlicht's book.
First, -1*a = -a: (-1)*a + a = (-1+1)*a = 0 => -1*a and -a are both solutions of x + a = 0 => they are equal. In particular, -1*-1 = -(-1)

Second, -(-a) = a: they're both solutions of the equation x + (-a) = 0 => they are equal. This means -(-1) = 1, and so -1*-1 = 1.

I was using additive commutativity, not multiplicative commutativity. I don't need distributivity to use it.

e^(i*pi)*e^(ipi)=e^2*i*pi=e^0=1

sorry I forgot to mention
using the distributive property for multiplication in Z is equivalent to stating that -1*-1=1
all examples so far show that
is there a way to prove the distributive property for multiplication in Z without assuming -1*-1=1
?

-1/-1 = 1
-1/1 = -1
-1*1 = -1
(a/b)/c = (c/b)/a

-1/1 = (-1/1)/1 = (1/1)/-1 = 1/-1

-1 * -1 = -1/1 * 1/-1 = -1/-1 = 1

Easy.

it's from one of those many stories from Neumann, I'm not sure if the story is true or not

math.stackexchange.com/questions/11267/what-are-some-interpretations-of-von-neumanns-quote
books.google.com.br/books?id=AInDCgAAQBAJ&pg=PA1&lpg=PA1&dq=john von neumann felix smith&source=bl&ots=flxodQKa3X&sig=HTict3YDyKAugJH2jzGyotJI9_o&hl=pt-BR&sa=X&ved=0ahUKEwjI7val3ZrWAhVLkJAKHc--A_YQ6AEIaTAN#v=onepage&q=john von neumann felix smith&f=false

I think (and hope) everybody in here knows the the construction of the reals out of Dedekind cuts or Cauchy sequences of the rationals.
This thread is more about the intuition behind the multiplication over the naturals being commutative.

No.

Starting from:

-1/-1 = 1
-1/1 = -1
-1*1 = -1

is insufficient because it is true for any number n when you replace -1 with n, ie, it doesn't define -1:

n/n = 1
n/1 = n
n*1 = n

If you define -1 with a simple subtraction like 1-1 = 0 then you are done:

1-1 = 0
-1/-1 -1 = 0
[-1 + (-1*-1)] / -1 = 0
[-1 + ( -1 * -1 ) ] = 0
-1 * -1 = 1

Tada!

OP here, looks like I've created a bit of a shit storm!
What I meant was how come 15 partitions so well into 3 lots of 5 and 5 lots of 3 as in the OP post
seems to explain it best thanks

i suspect this isn't what he wants because division isn't well-behaved on Z and you have to introduce this weird would-be-division
if he knows how to construct Z from N, then it's just a matter of proving negation and applying its definition on multiplication

Nope.

why rotate your abstract view point when you can just kill yourself

this is exactly what I wanted!
thanks!
oh god please no
I was finally pleased to know that there is a simple proof of the fact that -1*-1=1
please don't tell me that there is something wrong again...
I actually don't know about that "weird would-be-division" could you clarify?

Because if you transpose a rectangle so that it lays the other way, it obviously still has the same area. That's literally it, and it has the happy effect of covering the real cases as well.

Let x and y be relatively prime

Then it's impossible that x*x*x... y times = y*y*y ... x times because none of those factors can cancel or divide, and because of the fundamental theorem of arithmetic this means the left and right sides form unique, non-equal numbers

>trying to apply special relativity in a rotating and therefore non-inertial reference frame
>not realizing any effects would be gone when the book stops anyway

quit trying to seem smart you absolute freshman

The rotation matrices are unitary

Division isn't well-behaved on Z (is -4/3 in Z?) so you have to artificialy set n/n=1 and n/1=n, a would-be-division. So it doesn't "naturally" arise. But if you're happy with that, it's all good.
The easiest rigorous proof is applying negation to multiplication. But what is rigorous is context-dependent here, because if you accept Z is commutative ring, you don't have to prove anything, it just holds automatically because for (Z,*) inverse element is -1, composing two inverse elements yields identity element, which is 1. But of course, to be able to see Z as commutative ring, you have to prove it is one first and we're back, so what exactly can YOU assume about Z? Depending on that, we choose the easiest proof.

ohh I get it now
thanks!

Ahahaha
It is not even limited to Z in the first place. There is no mention of it. Thinking of your high school progressive order from Z to N, N to Q, Q to R is Arbitrary because you can pretty much talk about Q and R without negatives. The division isn't even general division, all it's parts are defined and the divisor passes to the other side anyway.