/sci/ - Science & Math

[Anonymous No.8753502]

Prove that a*b = b*a

[Anonymous No.8753545]

>>8753502
Go back to Instagram "Please"

[Anonymous No.8753561]

>>8753502
Assume a = a and b = b, divide both equations by a and you are left with b = b.

[Anonymous No.8753595]

>>8753561
You can't divide by a because we could have a=0.

[Anonymous No.8753672]

>>8753502
it's part of the definition.

[Anonymous No.8753682]

>>8753595
then just take the limit as a approaches 0, you dumb gay cuck

[Anonymous No.8753685]

>>8753502
Prove that I and I make two I's

This is the answer to all questions of this type so never come back, nigger

[Anonymous No.8753690]

>>8753502
it's a basic axiom of multiplication in a field (commutativity)

[Anonymous No.8753691]

>>8753502
just check that their commutator is 0, problem solved

[Anonymous No.8753728]

>>8753682
rude

[Anonymous No.8753730]

>>8753502
define the algebraic structure and I'll tell you.

[Anonymous No.8753739]

>>8753595
If a = 0 then both ab and ba are 0.
Otherwise divide by a

[Anonymous No.8753873]

>>8753730
First good answer in this thread.

[Anonymous No.8753883]

I don't need to.

[Anonymous No.8753888]

>>8753502
Let [math]G[/math] be a group under operation [math]*[/math] and let [math]a,b \in G[/math]. Then [math] a*b = b*a \Leftrightarrow G[/math] is abelian.

[Anonymous No.8753920]

>>8753502
(a*b)/a=(b*a)/a is equal to b=b
Conversly (a*b)/b=(b*a)/b is equal to a=a

[Anonymous No.8753941]

I think if you divide by a without assuming commutativity you would come up with b = (a^-1)*b*a or b = a*b*(a^-1)

[Anonymous No.8753945]

1 apple * 1 banana = 1 apple and 1 banana. 1 banana * 1 apple = 1 apple and 1 banana.

[Anonymous No.8753967]

>>8753502
That chick looks very practiced tbqh

[Anonymous No.8753970]

>>8753941
Yes.

[Anonymous No.8754134]

>>8753502
This is why communism will never work.

[Anonymous No.8754238]

>>8753502
>Pic depicting a young infant being sucked off, child porn?
MOOOOOOOOODDDDDDDDDDDSSSSSSSSSS!!!!!!!!!!!!!

[Anonymous No.8754354]

I

[Anonymous No.8754361]

>>8753502
Chemistry undergrad here so I don't know about mathmatical rigor. I would assume you can prove in which case the commutative law actually applies to multiplication (assuming that a and b don't have to be scalars but could just as well both be matrices, for example, and whatever rings or abstract algebra shit there are).

So could anyone please simply say how to go about this in those cases or just call it trivial?
(Haven't posted here before this one.)

[Anonymous No.8754366]

>>8753502
it's not always true

[Anonymous No.8754448]

>>8753730
brainlet here
is this because in some fields(?) of numbers(objects?) commutativity doesn't hold? like in the field of quaternions?

[Anonymous No.8754456]

>>8753945
>1 apple * 1 banana = 1 apple and 1 banana.
It does? What about 2 apples? Or half an apple.

[Anonymous No.8754465]

Ignoring all edits, this picture is straight creepy. Who would think it was a good idea to post it online?

[Anonymous No.8754591]

>>8754448
>>8753502


Consider the set containing the single element

[math]S=\{e\}[/math]

Then we can define a binary operation [math]\ast[/math] that takes 2 elements and returns a single element:

[math]\ast:S\mapstoS:e*e = e[/math]

We've defined what's called a group [math]\(S,\ast\)[/math], with 4 necessary properties:

Closure: [math]a\astb=c\quad\foralla,b,c\inS

Associativity: [math]a\astb=\bast\a[/math]

Identity: [math]e\asta=a=a\aste\quad\existse\inS[/math]

Inverse: [math]\foralla\existsa^{-1}\quada\ast\a^{-1}=e=a^{-1}\asta[/math]


Notice we don't have commutativity isn't here. The identity and inverse elements always commute by definition, but it is not necessary that a*b = b*a for 2 arbitrary elements in a group.

The group we just made though, [math]\(S,\ast\)[/math], is commutative. Naturally since, the only element is both the identity and it's own inverse. [math]e\aste=e[/math] is literally the only thing we can do.

A commutative group is called Abelian.

A group without commutativity is Non-abelian.

The smallest possible nonabelian group will have 6 elements, and be isomorphic to [math]D_3[/math] (dihedral group 3), or [math]S_3[/math] (symmetric group 3). Isomorphic meaning the algebra under some operation [math]\ast[/math] works the same, we really just renamed the elements.

[Anonymous No.8755025]

>>8753502

[Anonymous No.8755031]

>>8755025
Prove the 2d plane is rotation-invariant

[Anonymous No.8755049]

>>8753502

5*2 = 10
2*5 = 10

[Anonymous No.8755053]

Instream of answer: https://en.wikipedia.org/wiki/Commutative_ring

[Anonymous No.8755300]

>>8753595
catch(DivideByZeroException e)
{
//duh
}

[Anonymous No.8755332]

>>8753502
nice black man * nice black woman = nice black human^2

nice black woman * nice black man = nice black dik ;) ^2 (poopy dik XDDDDD)

prove me wrong /sci/!

[Anonymous No.8755335]

>>8754591
>>8755332

[Anonymous No.8755513]

>>8753502
You can prove this from the usual definition of multiplication on the peano axioms

[Anonymous No.8755523]

>>8753502
[math] a \cdot b \neq b \cdot a \\
a \neq \frac{b \cdot a}{b} \\
a \neq a [/math]

[Anonymous No.8755628]

>>8753888

[Anonymous No.8755633]

>>8753967
Indeed. Look at her hand. She instinctively holds the hand like it's a cock. Top fucking kek.

[Anonymous No.8755738]

You can't actually prove that a*b=b*a because you didn't define what a and b are.
For example: If a and b are matrices, the multiplication is in fact not commutative.
>>8753888
you just assumed G is an abelian group when you don't know that

[Anonymous No.8755903]

>>8753502
> a and b € {R}
> a != b
If ( a or b )= 0 》 a * b and b * a = 0;
a * b = b * a 》(a/a ) * (b/a) = (b/a) * (a/a) 》1 * (b/a) = (b/a) * 1 》(b/a) - (b/a) = 1 - 1 》0 = 0. C.v.d.

[Anonymous No.8755915]

>>8755903
>C.v.d.
spotted the italian
in the civilized world, we write q.e.d. (quod erat demonstrandum)