For a group that has closure, associativity, a neutral number and an opposite number in relation to action *

For a group that has closure, associativity, a neutral number and an opposite number in relation to action *

It's known that if a*c = b*a then c=b
How do you prove that there'scommutativity?

It feels right intuitively but I can't really formulate it

Okay I think I got it

a*c=b*a
c=(b*a)*a-1 (opposite of a)
c = b*(a* a-1 ) because of associativity
c=b

can someone assure this is right?

let x be the inverse of a and e be the neutral element

a*c = b*a implies c=b iff
a*c*x = b*a*x implies c = b iff
a*c*x = b implies c = b

if a*c*x = b implies c = b then
a*c*x = c
but c is equal to e*c which is equal to a*x*c* so
a*c*x = c = a*x*c so
a*c*x = a*x*c

cancel a from the left to get c*x = x*c

Now, assuming that in the original equation a, b and c are arbitrary members of the group then this ends the proof as c is an arbitrary member of the group and x is a member that depends on the arbitrary member a, so x itself is also arbitrary, so this last equations is basically an equation that asserts commutativity.

>can someone assure this is right?

It isn't.

>c=(b*a)*a-1 (opposite of a)

This is weird. In the left you had a*c but then you canceled a which means that you multiplied a-1 on the left of c, which means that it should be also multiplied to the left of b to get

a-1*a*c = a-1*b*a
c = a-1*b*a

and there associativity wouldn't help you.

tldr, you cannot do what you just did because you can't multiply to the left on one side and to the right in another side because that is basically commutativity so you would be assuming what you seek to prove.

How do you reach this step?

>but c is equal to e*c which is equal to a*x*c*

Where is a*x*c ever mentioned?

Do you know group theory? This is a very elementary concept.

Okay, we know that x is the inverse of a and by definition of the inverse this means that a*x = e

just like 1 + (-1) = 0 in the group of integers

So then I have e*c but e is simply equal to a*x so I replace with and get e*c = a*x*c

nothing magical happened here, a times x is still e but in a more convenient form.

>For a group that has closure, associativity, a neutral number and an opposite number in relation to action *
Why do you need to say all that? It's obvious from the definition of a group

I just realized you said x was a's inverse, sorry. I was treating it as an additional member.
thanks for helping

What is a 'group' of numbers that doesn't have those qualities called in english.

I think this is error of translation from my language.

Let g,h any elements in the group then

g^(-1) * g * h = h = h * g * g^(-1)

Now if you use the assumption with
a = g^(-1)
b = h * g
c = g * h
then you get
g * h = h * g

q.e.d

learn to use [math]\LaTeX[/math] faggots, anyway, here's a relatively short answer :
[math]a*c=b*c\implies c=b[/math] is equivalent to [math]c\neq b\implies a*c\neq b*c[/math]
So let's say that [math]a^{-1}*x\neq x*a^{-1}[/math]
This implies [math]a*(a^{-1}*x)\neq (x*a^{-1})*a[/math] which is equivalent to [math]x\neq x[/math]
Thus, for all [math]a^{-1}[/math] and [math]x[/math] in your group G, you've commutativity, there you go friend.

You probably meant 'set,' then.

Group is a set of numbers along with an operator that has certain properties.
You basically listed the properties that define a group.

thanks senpai, already wrote this calculation though

I got a question on a related topic to OP:

If set A = all even whole numbers

and action * is a*b = a+b-ab

How do I mathematically explain there's closure?
It's obvious there is but idk how to write it.

To show it stays even I just compared it to the group of all whole numbers so 2(a+b-ab) is a+b-ab in that group and it divides by 2.

>a*(a^{-1}*x)\neq (x*a^{-1})*a

You are multiplying a on the left and on the right and still preserving the equality relation...

Did you just assume my gender?

I mean, did you just assume commutativity to prove commutativity

Nah, dropped. Back to precalc boiii.

You can't multiply a on the left on one side and a on the right on the other, that can only preserve relations if commutativity was already true.

>How do I mathematically explain there's closure?

Not OP but...

let c and k be arbitrary whole numbers, then fix
a=2c and b=2k

then

a*b = (2c)*(2k) = 2c + 2k - 2c2k
= 2c + 2k - 4ck

but then you can simply factor two and get
= 2(c + k - 2ck)

so this result is also even and thus it is still in the set A.

Because a and b are arbitrary then that means that the "product" of any a and b is still inside the set A, therefore there is closure.

damn son;
[math]x\neq y[/math] fucking implies [math]a*x\neq y*a[/math]
got it now ? that's the fucking contraposition of OP's statement, know basic logic you twat

I'm not talking about that.

I am talking about [math]a*(a^{-1}*x)\neq (x*a^{-1})*a[/math]

You claim this follows from

[math] a^{-1}*x\neq x*a^{-1} [/math]

but because you are multiplying a on different sides, you are assuming commutativity.

Back to precaaaalc boiiiiiiiiiiiiiiiii

it proves they're even, but how does it prove they're "whole"?

multiplication and addition of whole numbers still yields a whole number.

This follows from the fact that (Z,+,*) is a ring.

yeah, that's what I wrote but I figured that it might not be enough to just have a verbal explanation for that

how do I prove what qualities of a group the same action a*b = a+b-ab applies to the set of all rational numbers besides 1?

I have no idea how to prove it mathematically again but it obviously has closure.

okay, just concentrate for a moment and watch where does it come from :
OP assumed that in his group, [math]a*c=b*a\implies c=b[/math], the contraposition of that is [math]c\neq b\implies a*c\neq b*a[/math], right ?
now, assume that [math]a^{-1}*x\neq x*a^{-1}[/math], just assume that, got it ?
now, what you do is you take your contraposition and replace [math]c[/math] by [math]a^{-1}*x[/math] and [math]b[/math] by [math]x*a^{-1}[/math], okay ?
and now that you've replaced the members of the contraposition, two lines tell you that this leads to a contradiction, so the original assumption is false, which means that [math]a^{-1}*x=x*a^{-1}[/math]
now tell me what's wrong with that, i dare you

ok I think got it ?

a+b-ab = 1
a(a+b-ab) = a
a^2 = a
this can only be true for the number 1 which isn't in the group so this can't be

so who was wrong?

doesn't my math makes sense ?

>a(a+b-ab) = a
>a^2 = a

No

a^2 + ab - (a^2)b = a

oh shit right

so what do I do to prove it?

who one are you

>so what do I do to prove it?

I don't know, could you rewrite
>how do I prove what qualities of a group the same action a*b = a+b-ab applies to the set of all rational numbers besides 1?

I don't understand a single word.

I think I did it

any result of rational number addition subtraction and multiplication will be a rational number, that's ok to say right?

so after that to show that it's not 1 which isn't in the group

a +b -ab = 1
a = 1-b+ab
1-b+ab+b-ab = a
a = 1

this cannot be because 1 isn't in the group

qed

>1-b+ab+b-ab = a
>a = 1

what?

You added b-ab to one side but not the other here. Is this right?

if a +b -ab = 1 then a = 1-b +ab

no?

a*b=c
b*a=d

if a-b=e
a*(a-e)=c
b*(b+e) = d
a*a = a(e + b)

if c = d
a*a - a*e = b*b + b*e
a*a - b*b = b*e + a*e
a(e+b) = b*e + a*e + b*b
a*b = b*e + b*b
a = e + b

Thus correct

What mistakes have I made here?