1. Defining an operaor

Hai,

I have some toubles with this question.

Define an operator * on $R$ by

$x*y = 2xy -x -y$

a) is * commutative?
b) is * associative?

I can easily see that * is commutative, but how do i test for associativity?

The rule states that (x*y)*z = x*(y*z)

But what is z ?

2. Originally Posted by Jones
Hai,

I have some toubles with this question.

Define an operator * on $R$ by

$x*y = 2xy -x -y$

a) is * commutative?
b) is * associative?

I can easily see that * is commutative, but how do i test for associativity?

The rule states that (x*y)*z = x*(y*z)

But what is z ?
* is associative if $\forall x,y,z \in \mathbb{R} : (x*y)*z = x*(y*z)$

$(x*y)*z = (2xy - x - y)*z$
$x*(y*z) = x*(2yz - y - z)$

Can you finish?

Spoiler:
$(x*y)*z = (2xy - x - y)*z$ $= 2(2xy-x-y)z - (2xy - x - y) - z = 4xyz -2xz -2yz -2xy +x +y-z$
$x*(y*z) = x*(2yz - y - z)$ $= 2x(2yz - y - z) - x - (2yz - y - z) = 4xyz - 2xy - 2xz -x -2yz + y + z$
$4xyz - 2xy - 2xz -2yz -x + y + z \neq 4xyz -2xy -2xz -2yz +x +y-z$

3. Ah, thank you. I think i understand now. So the conclusion must be that the function is indeed associative because the two results are different?

4. Originally Posted by Jones
Ah, thank you. I think i understand now. So the conclusion must be that the function is indeed associative because the two results are different?
No - a function is associative if the two results are equal, not different.

The function is not associative (if Defunkt did his working right).

5. Originally Posted by Defunkt
* is associative if $\forall x,y,z \in \mathbb{R} : (x*y)*z = x*(y*z)$

$(x*y)*z = (2xy - x - y)*z$
$x*(y*z) = x*(2yz - y - z)$

Can you finish?

Spoiler:
$(x*y)*z = (2xy - x - y)*z$ $= 2(2xy-x-y)z - (2xy - x - y) - z = 4xyz -2xz -2yz -2xy +x +y-z$
$x*(y*z) = x*(2yz - y - z)$ $= 2x(2yz - y - z) - x - (2yz - y - z) = 4xyz - 2xy - 2xz -x -2yz + y + z$
$4xyz - 2xy - 2xz -2yz -x + y + z \neq 4xyz -2xy -2xz -2yz +x +y-z$
Im sorry, some of the steps i do not understand.
$(x*y)*z = (2xy - x - y)*z$
= 2 $(2xy-x-y)z - (2xy - x - y) - z$

Where did you get 2 from?