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

>>9158085
Because it's defined that way.

>>9158085
PRIME NUMBERS

>>9158088
also 15/5=3 and 15/3=5

>>9158092
so a/b=c and a/c=b

Say you have 3 bags, each of which has 5 apples in them. How many apples do you have?

Now say you have 5 bags, each of which has 3 apples in them. How many apples do you have?

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Imagine a [math]A \times B[/math] rectangle. It's area is [math]A \cdot B[/math] like they taught you at school. Now, if you rotate the rectangle 90 degrees, you get a [math]B \times A[/math] rectangle, but its area is still the same.

>>9158085
Subtract 1 from each five. You took away three total and are left with 4+4+4. Repeat.

>>9158112
Why do people always use apples as an example? Do math educators really fucking love apples that much.

>>9158552
Apples are a reason why we know about gravity in the first place :thinking:

>>9158085
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).

>>9158565
ISA bravo is that you son?

>>9158085
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

>>9158590
>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

>>9158085
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.

>>9158087
Yeah, you don't need to look at it any harder than this.

>>9158085
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.

>>9158145
Oiii that's nice

>>9158634
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.

>>9158639
>>9158661

<3

>>9158141
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

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>>9158085

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

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>>9158088
>t. Pucci

>>9158996
I'm sorta glad that Jojo is equal parts stupid and intellectual

>>9158946
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.
https://proofwiki.org/wiki/Natural_Number_Multiplication_is_Commutative

>>9158141
this might be the best and most simple explanation I've seen.

>>9158141
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

>>9158141
Are you a brainlet? That doesn't explain multiplication like this guy said >>9159600

Dumb chink spammer.

>>9159600
>>9159633
Pick up a book.
Rotate it.
Explain why the area would be any different.

>>9159642
You're assuming you can neglect the Lorentz contraction of the book when you rotate it.

>>9159645
why rotate book when you can rotate your coordinate system

>>9159647
I don't think that would make much of a difference because there is relative motion anyway.

>>9159649
I believe that was the point he was trying to make

>>9159533
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

>>9159647
read the fucking book already anon

>>9159654
You define -1*X = -X and then you can make any -n*X into n*-1*X

>>9159658
I don't get it
using the recursive definition he gave earlier
multiplication of negative numbers would yield infinity

>>9159647
why rotate your coordinate system when you can rotate your abstract view point of said coordinate system

>>9159600
because rotation is an isometry and thus measure preserving, brainlet

>>9159654
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.

>>9159662
Read the link you cuck.
The recursive definition is only for positive integers.

>>9159664
use his recursive definition to give me the result of
-1*-2

>>9159666
fuck the link
the whole point is to get to it yourself

>>9159666
and if the recursive definition only works for the positive integers
then prove to me that -a*-b = ab

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>>9159600
Vectors do not change magnitude upon rotation, you absolute brainlet.

>>9159670
why?

>>9159670
yes they do

if they were horizontal, their mangitude becomes uppier

>>9159669
>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

>>9159670
Prove it.

>>9159678
And more importantly, prove it without relying on the commutativity of multiplication.

>>9159677
ok nice
but what proves the distributive property for multiplication in Z?

>>9159680
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.

>>9159688
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)

>>9159693
>Proves rotation invariance by saying "Look when you rotate it it doesn't change!"
Great proof there.

>>9159698
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

>>9159700
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

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

>>9159705
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.

>>9159720
ok then prove to me that the distributive property of multiplication holds in Z

>>9159721
It holds in Q. Z is subset of Q. It holds in Z. Wow that was hard.

>>9159727
good luck in life
you're gonna need it

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>>9159729
>hurr durr i can't use the tools i have

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>>9159734
>to prove

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>>9159738
>a property that's given as introductory excercise

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>>9159741
>to freshmen undergrads

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>>9159746
>around the world
In other words, russian highschooler can do this. Why can't you?

>>9159748
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
>>9159677

I'm trying to get to it naturally

>>9159714
show the fault anon

https://youtu.be/elvOZm0d4H0?t=4m

>>9159750
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.

>>9159756
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

>>9159757
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.

>>9159533
You forgot to prove the distributive. And don't look it up.

>>9159762
fine with me
I'll be watching this thread later

>>9159767
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.

>>9159777
thanks but I just want someone to prove to me that -1*-1=1
it will prove everything else
you assumed it

>>9159698
You need distributive first to prove commutative.

>>9159782
-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

>>9159782
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.

>>9159784
I was using additive commutativity, not multiplicative commutativity. I don't need distributivity to use it.

>>9159782
e^(i*pi)*e^(ipi)=e^2*i*pi=e^0=1

>>9159795
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
?

>>9159782

-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.

>>9159025
it's from one of those many stories from Neumann, I'm not sure if the story is true or not

https://math.stackexchange.com/questions/11267/what-are-some-interpretations-of-von-neumanns-quote
https://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%20von%20neumann%20felix%20smith&f=false

>>9159716
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.

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>>9159823
No.

>>9159782
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
>>9158634 seems to explain it best thanks

>>9159944
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

>>9160035
Nope.

>>9159663
why rotate your abstract view point when you can just kill yourself

>>9159944
this is exactly what I wanted!
thanks!
>>9160035
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?

>>9158085

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.

>>9158970

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

>>9159645

>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

>>9159678
>>9159680

The rotation matrices are unitary

>>9160114
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.

>>9160207
ohh I get it now
thanks!

>>9160207
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.

>>9160272

wtf are you talking about? N is the only one there without negatives and that's by definition, it's just a set not even a group or ring

>>9158085
Omni numeri ad posibilii final de metricquint quid even primi in calculer.

>>9160272
Well he was so adamant about everything arising "naturally" and you can only have "natural" after constructing Q, you get division on "N" and "Z" from that. The same goes for subtraction and N. If he wants a rigorous proof, he can't have these arbitrary assumptions, it can be proved without them. I think he just didn't go through anal because you start at peano axioms and construct R in every course in yuropoor (and i assume burgerland too), so he can't wrap his head around the way (-1)(-1)=1 is just a consequence of negation. Your proof is valid, but a bit hand-wavy because of that unneccessary assumption.
And the highschool progression is N to Z to Q to R, but N is Set Z is CRing, R is Field so they're not equal and don't have the same properties. Division isn't closed on N and Z so there's no easy way to define it. It is closed on Q, so if you want to divide something from N, you take their counterparts in Q and divide them, this is division on "N".

>>9159756
just ask your mom if you want an introductory course on real anal

lmao at all the cucks here. Just define mn to be the cardinality of the m \times n with m and n defined in the usual way: 0 = \phi and S(k) = k \cup \{ k \}.

>>9158552
Because retards get it faster that way

I actually don't get why addition is conmutative, at a fundamental level.

Why shouldn't the order of counting alter the result?

Any help?

>>9158085
You can define [math]5 \times 3[\math] as the number of elements in the set of pairs [math]\{a, b\}[\math] such that [math]1 \leq a \leq[\math] and [math]1 \leq b \leq 3[\math]. This is commutative because pairs are unique up to order.

>>9160580
Intuitively, addition is just repeated incrementation. It doesn't matter if you increment 2 five times or 5 two times, you will get 7.
More rigorously, let S(n) be the successor of n and assuming n+0=n
We want to prove n+m=m+n
We use induction, with base case n=0
m+0=0+m
m=m so base case holds
We want to show that S(n)+m=m+S(n)
S(n)+m=S(n+m) LHS
m+S(n)=S(m+n)=S(n+m) RHS
So it holds for all N.

>>9160580
ℕ ≡ { {}, {{}}, {{},{{}}}, {{},{{}},{{},{{}}}}, ... }

0 ≡ {}
1 ≡ {{}}
2 ≡ {{},{{}}}
3 ≡ {{},{{}},{{},{{}}}}
...

= ≡ ~( α , β ) , ~r ∧ ~s ∧ ~t

~r ≡ ∀αεΑ : ~( α , α )
~s ≡ ∀α,βεΑ : ~( α , β ) ⇔ ~( β , α )
~t ≡ ∀α,β,γεΑ : ~( α , β ) ∧ ~( β , γ ) ⇒ ~( α , γ )


+ ≡ φ( m , n ) : ℕ x ℕ → ℕ : Sn(m) ≡
S0(m) : ℕ → ℕ : m → m
S1(m) : ℕ → ℕ : m → ∪( m , {m} ) ⇔ S1(S0(m))
S2(m) : ℕ → ℕ : m → ∪( ∪( m , {m} , {∪( m , {m} )} ) ⇔ S1(S1(S0(m)))
...

⇒

S0(m) ⇔ Sm(0)
S1(m) ⇔ Sm(1)
S2(m) ⇔ Sm(2)
...

⇒

+( 0 , m ) ⊆ +( m , 0 ) ∧ +( 0 , m ) ⊇ +( m , 0 ) ⇔ +( 0 , m ) = +( m , 0 )
+( 1 , m ) ⊆ +( m , 1 ) ∧ +( 0 , m ) ⊇ +( m , 0 ) ⇔ +( 1 , m ) = +( m , 1 )
+( 2 , m ) ⊆ +( m , 2 ) ∧ +( 0 , m ) ⊇ +( m , 0 ) ⇔ +( 2 , m ) = +( m , 2 )
...
+( n , m ) ⊆ +( m , n ) ∧ +( n , m ) ⊇ +( m , n ) ⇔ +( n, m ) = +( m , n )

⇒

n + m = m + n

>>9161407
oops, i mean

+( 1 , m ) ⊆ +( m , 1 ) ∧ +( 1 , m ) ⊇ +( m , 1 ) ⇔ +( 1 , m ) = +( m , 1 )
+( 2 , m ) ⊆ +( m , 2 ) ∧ +( 2 , m ) ⊇ +( m , 2 ) ⇔ +( 2 , m ) = +( m , 2 )

tired af

>>9158085
primes are like weed, okay?

>>9159533 Omni retardii calculori americani
>>9159677
>>9159698
>>9159750
>>9159752
>>9159782
>>9159794
>>9159795
>>9159802
>>9159823
>>9159944
>>9161407
>>9161418
Exego americanii troll-militi.

>>9160677 Non troll-militi
Omni rigtere abex non de

>We want to show that S(n)+m=m+S(n)
>S(n)+m=S(n+m) LHS
>m+S(n)=S(m+n)=S(n+m) RHS
>So it holds for all N.

Cogere calculere in abstracto.

Legionii homini et 4chan vadere legionii. Legionii conquere Amerika in present excursio aggressii, pretere! Ad victori, in glori regress.
Ego revelere non fals informari ergo non /x/.
Memento cancerii quanti de Amerika est in futur eradicarii.
>Cancerii quanti de Amerika est Republicani.

>>9158552
basically everyone has eaten/seen an apple
we can easily visualize them as singular, distinct objects

>>9158617
That's literally the definition of multiplication you brainlet, for any question of "why" you can ask of the answer another question of "why"

>>9161616
>learning a second language so no one will engage your retarded posts
nice

Because you touch yourself at night

>>9161616
Sumer de Latein

Singular Plural (utilere i sequent)

I est prim primi
II est secund i
III est terc i
IV est quart i
V est quint i
VI est sext i
VII est septim i
VIII est octav i
IX est nue i
X metric i

40 est quarti
10 est metrici
150 est metrici-quinti
386 est terci-octavi-sext
2796 est secundi-septimi-nuei- sext
Concludere infinitii sumeri.

Cogere calculere in latein.

>>9161649
Ego forgetere in past: utilere ego lateini sumeri.

>>9160279
>>9161616
>>9161649
>>9161673
What books did you learn latin from? Your composition is terrible, any italian highschooler would put out a more coherent sentence than you. Assuming you're learning latin for 3 weeks, lurk for two years before trying to write in latin. I shiver just thinking about how terrible your pronunciation must be.

>>9158725
Oh hey that makes sense
>perposes
dropped the ball there

>>9158085
if you break it down further to 1's it's obvious.
you have 5+5+5 you call it 3 fives, but you could also say it's a hidden 5 threes. 1+1+1 is 3 and you have 5 superimposed 1+1+1=3's there.
5+5+5 is 5 threes.

>>9158085

5+5+5 = 5(1+1+1) = 5 * 3 = 3 + 3+ 3 + 3 +3

>>9161742
More like using some shitty computer algorithm. Structure is all over the place, not even self-taught could have such bad composition.

>>9160088
Why kill yourself when you could just not be alive.

>>9161407
I didn't get a thing about those strange symbols. I bet it's rigurous and fundamental, I just can't read it.

>>9160677
Why is addition repeated increment? Why it must have to do anything with counting? Why it must assume that only one number, and a diferent one to boot, comes after another?

What if addition is defined, like, traveling on a network such as some sums lead to the same point, and others lead to very different points at the same time?

>>9162161
Because the number line is 1 dimensional. You can only go forwards and backwards so from any given point on the line, another point is uniquely determined by the first point and a length.

You'd have serious problems in mathematics if 1+ 1 = 2 and/or 3 for example.

break it all up into ones and count the ones:

5 + 5 + 5 = (1+1+1+1+1)+(1+1+1+1+1)+(1+1+1+1+1), 15 1s total
3+3+3+3+3 = (1+1+1)+(1+1+1)+(1+1+1)+(1+1+1)+(1+1+1), 15 1s total

>>9158085
Because its just [math] (1+1+1+1+1)+(1+1+1+1+1)+(1+1+1+1+1) = 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 = (1+1+1)+(1+1+1)+(1+1+1)+(1+1+1)+(1+1+1) [/math] Simples.

>>9159673
the education system has failed you son

>some people in this thread actually need vector spaces to prove commutativity of the integers

>>9162161
We work with peano axioms. In short, they give you these tools: successor, induction. That's it. Everything you can build in this axiomatic system reduces to these two tools. It turns out, you can construct a *lot* with just these tools.
If your definition of addition serves you better than the usual one and is valid, by all means use it. Of course you could prove it with set theory or category theory, but peano is easiest for the purposes of analysis.

>>9162161
Constructs a set ℕ containing the empty set and the union of the empty set with the set containing the empty set and so on, and defines names of the elements of this set.

Defines = as a binary relation quantifier ~ that is reflexive, symmetric, and transitive.

Defines addition the collection of maps φ that map ℕ x ℕ → ℕ that takes either element of an element of the domain and maps it to the union of that element with the set containing that element and so on depending on what the other element of the element of the domain is. Except in the case of using 0 as the "map selector" which maps the other element to itself.

The targets φ( m , n ) and φ( n , m ) are subsets of each other, so :

∀m,nεℕ : =( φ( m , n ) , φ( n , m ) )

This reads that for all m and n in the set ℕ , φ( m , n ) and φ( n , m ) have the = relation, which was previously defined.

Then we define new notation that is easier to use, where m + n is defined as φ( m , n ) , n + m is defined as φ( n , m ) , and m + n = n + m is defined as =( m + n , n + m) .


This mapping is an important subtlety, because the sets { m , n } and { a , b } are not always subsets of each other, yet it is still possible that the targets of these sets after φ have the = relation.

Although this isn't necessary to prove commutativity of addition in ℕ , it is useful for the construction of equivalence classes, e.g. { 1 , 19 } , { 12 , 8 } , ...

>mathfags can't prove why multiplication is conmutative without recurring to "muh count fruits"
The absolute state of pure math students

>>9159533
Wrong

>>9159533
The mistake here is it assumes b(a-1) = (a-1)b

>>9158085
>I still don't get why multiplication is commutative?!

It is not trivial, OP. Let's define multiplication:

-------

5 x dollar = dollar + dollar + dollar + dollar + dollar

5 x 4 = 4 + 4 + 4 + 4 + 4 = 20

a x b = b + b + b... + b {a times}

a x (b+c) = b+c+...+b+c {a times} = b+...+b {a times} + c+c...+c {a times} = ab + ac
-------

5 x dollar + 2 x dollar = 7 dollar

5 x 4 + 17 x 4 = (5+17)4 = 23 x 4 = 812

a x b + c x b = (a+c)b

-------

(a+b)(c+d) = a(c+d) + b(c+d) = ac + ad + bc + bd

-------

ab = ba ?

c = a-b
a = b+c
b = a-c

ab - ba = a(a-c) - b(b+c) = aa - ac - bb - bc = aa - bb - (a+b)c = aa - bb - ((b+c) + (a-c))(a-b) = aa - bb - ((b+c)a - (a-c)b) = aa - bb - ((ba + ca) - (ab - cb)) = aa - bb - (ba + ca - ab + cb) = aa - bb - ba - ca + ab - cb = (a-b-c)a + (a-b-c)b = (c-c)a + (c-c)b = 0

ab - ba = 0
ab = ba

>>9164646
>The mistake here is it assumes b(a-1) = (a-1)b
no it doesnt, did you even read the post?
it assumes
>definition of multiplication in piano axioms: a*b = a + a*(b-1)
>multiplicative identity: a = 1*a
>one side of distributive property: (a + b)c = ac + bc
that all. you can derive the 2nd and 3rd property from the axioms if you want.

>>9158085
Multiplication is a shortcut for addition, and addition never lose value, always adds up, therefore the order does not affect the result since the value is never lost, it's always increased rgardless of the order.

File: multi.png (4KB, 453x182px) Image search: [Google]

4KB, 453x182px

>>9158634
This here and the pic I just made are the easiest explanation

>>9158102
Yes, because a = b*c. Divide a by either to get the other.

>>9158085

Let y be the set of natural numbers such that for all x, xy = yx.

Then y contains 0 and 1.

And if nx = xn for all n, then (n+1)x = nx + x = xn + x = x(n+1).

So for all x, y in N, xy = yx by induction QED

>>9158085
induction.
Also, because your put your stuff in a rectangular array. number of squares = number of columnss times number of rows = number of rows times number of columns.

Thinking of it as just a string of ones makes it pretty intuitive for me.

3 * 2 = (1 + 1 + 1) + (1 + 1 + 1) = (1 + 1) + (1 + 1) + (1 + 1)

>>9165530
Nice proof of a special case. But I want to know if the multiplication of the numbers 928373737372718118 and 287347372747372728 is also commutative? Can you show me the string of 1s argument here?

>>9161407
Is this a computer language?

>>9165669
That's set theory with first order logic

>>9165178
No, you fucking dumbarse, take a look at :

(1 + 1*(a-1))*b = b + (a-1)*b

and compare it with the definition

a + a*(b-1)

a comes before (b-1) while b comes after it, so you cant move on to =ba

if you are saying that a(b-1) = (b-1)a then you are already assuming what you should have proven. What a fucking brainlet

>>9158085
Set theory

File: images.jpg (4KB, 224x225px) Image search: [Google]

4KB, 224x225px

Is computer science relatet to math?

>>9166020
No. Math is related to computer science. Next question.