Discrete math, defining an operator

[James889]

Hi,
I have some troubles with this question.

Define an operator * on $$R$$ by

$$x*y = 2xy -x -y +1$$

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 ?

 

[mathie.girl]

For associativity, you should look at both sides of the equation you listed ((x*y)*z = x*(y*z)) and use your rule to expand. For your associativity condition, z is any third element of R, as both x and y must be. So if you're asked to do y*z, then you fill in y and z in the place of x and y in your rule. Then check to see if you get the same thing on both sides.

 

[statdad]

$$(x \ast y) \ast z$$ refers to the expression you get by first computing $$x \ast y = 2xy -x -y - 1$$, then calculating $$(2xy - x -y -1) \ast z$$ - z is another real number.

 

[James889]

$$(x \ast y) \ast z$$ refers to the expression you get by first computing $$x \ast y = 2xy -x -y - 1$$, then calculating $$(2xy - x -y -1) \ast z$$ - z is another real number.

Hi,

$$(x \ast y) \ast z = (2xy -x -y +1)z \astz \longrightarrow (2xyz -xz -yz +z)$$

Now i have to test for the expression $$x \ast (y \ast z)$$

But how do i do that when the operator is defined for x and y ?

More precisely, i don't know what $$(y \ast z) is$$

 

[lurflurf]

The rule for * has been given, so the question is simply
compute
(x*y)*z-x*(y*z)
and see if it is zero
your operator is defined for all x and y in particular
y*z=2yz-y-z+1

 

[James889]

The rule for * has been given, so the question is simply
compute
(x*y)*z-x*(y*z)
and see if it is zero
your operator is defined for all x and y in particular
y*z=2yz-y-z+1

Thanks, i think i understand. Is this correct?:

$$(x \ast y) = 2xy -x -y +1$$

$$(x \ast y)\ast z = 2(2xy -x -y +1)z -2xy -x -y -z +1$$

$$= 4xyz -2xy -2xz +x -2yz -y -z +2$$

$$(y \ast z) = 2yz - y -z +1$$

$$x \ast(y \ast z) = 2x(2yz -y -z +1) -x -2yz -y -z +2 = 4xyz -2xy -2xz +x -2yz -y -z +2$$

They are equal

 

[HallsofIvy]

Yes, that is correct.

Referring to your earlier question, "But how do i do that when the operator is defined for x and y ?" or basically, "how do I combine three things with a binary operator?" The answer was "do them two at a time". To find (x*y)*z, first you find x*y then you treat that result as a single "object" with z. To find x*(y*z) you first find y*z and treat that as a single object with z. The whole point of the "associative law", (x*y)*z= x*(y*z), is that we can do it either way and so don't really need the parentheses: we could write x*y*z to mean either (x*y)*z or x*(y*z) and get the same answer. And, we can extend that to any number of objects, writing u*v*w*x*y*z, just like we do with sums or products of numbers!