Question

Let S be a set, and * be a binary operation on S that satisfies two laws:
x*x = x for all x in S, and
(x*y)*z = (y*z)*x for all x, y, z in S.

Prove that * is commutative and associative.

Answer 1

I think i have an idea of how to prove it, I just want to see what other people have to say. Just for clarification purposes, We want to prove that:
x*y = y*x
and
(x*y)*z = x*(y*z)

Answer 2

Groupy stuff...

Answer 3

yeah >.< i thought i had an idea, but as I was typing it, i saw a hole in the logic and it fell apart =/

Answer 4

If it was a*a it would be GA.

Answer 5

x*x I mean...

Answer 6

x^2

Answer 7

if i end up with a statement like:

(x*y)*x = (y*x)*x

does that imply that x*y = y*x ?

Answer 8

Yes because of x*x

Answer 9

the xyx stuff IS associative, isn't it?

Answer 10

it has to be shown. just from that second law its not clear. My plan is to prove commutative with just those two laws, then prove associative with commutative + two laws.

Answer 11

Whatdy'u call it, bilinear wotsit..

Answer 12

if (x*y)*x = (y*x)*x proves x*y = y*x, then thats commutative down. Then associative shouldnt be too bad...i think lolol

Answer 13

it just seems a little iffy to me >.<

Answer 14

That must be right if x*x = x for all x.

Answer 15

How so? im a little confused.

Answer 16

Oh, I see, moment, anticommutative is still possible....

Answer 17

I guess i should state how i got that statement. I started with x*y, and did:

x*y = (x*x)*y

Then by law 2 we can change the order:

(x*x)*y = (x*y)*x

Using law 2 once more:

(x*y)*x = (y*x)*x

Answer 18

From here I want to say something like, "because binary operations are one-to-one, and the object x is on the right side of both statements, this implies that x*y = y*x, yada yada yada".

Answer 19

I think this is correct. "If * is a binary operation, and a*b = a*c, then b = c."

Answer 20

Transitivity.

Answer 21

i need to verify that though >.< ima search the web lol

Answer 22

I really much prefer to use the algebras than prove that the algebra exists, lol

Answer 23

me too >.< this is a for a problem solving class. We just pick random problems to solve. This one looked do-able so i picked it.

Answer 24

I'm just thinking because GA is noncommutative algebra but associative and I am trying to see what exactly it is in yours makes it commutative and it can only be the x*x = x

Answer 25

Whereas in GA it's x^2

Answer 26

That allows u to pull that substitution trick...

Answer 27

right right. I think i have enough info to write out the proof now. thanks :)

Answer 28

Pick something less, um... well u know, next ime, lol:-)

Answer 29

Alright, i came up with better reasoning. This looks so much better lol.

Start with x*y. Using property 1 we get:

x*y = (x*y)*(x*y)

Then by property 2 we have:

(x*y)*(x*y) = [(x*y)*x]*y

Using property 2 on the inside argument backwards we obtain:

[(x*y)*x]*y = [(x*x)*y]*y = [x*y]*y

Then one more property 2 gives:

[x*y]*y = (y*y)*x = y*x

Thus x*y = y*x, and * is commutative.

For Associative, we start with property 2:

(x*y)*z = (y*z)*x

Then using the commutative property we just proved, we obtain:

(y*z)*x = x*(y*z)

So (x*y)*z = x*(y*z), and * is associative.