1. ## commutative, associative

p*q = p+q-pq

1. determine whether * is commutative
2. determine whether * is associative.

Ok im not sure i understand the question. Do i just work out the answers from the right side of the equation or both sides?

2. Originally Posted by djmccabie
p*q = p+q-pq

1. determine whether * is commutative
2. determine whether * is associative.

Ok im not sure i understand the question. Do i just work out the answers from the right side of the equation or both sides?
It is not an equation, it is a definition of the binary operation "*" which is not the usual product.

CB

3. ahh, so i work out commutative and associative for p+q-pq ??

and did i post this in the wrong place?

4. Originally Posted by djmccabie
ahh, so i work out commutative and associative for p+q-pq ??

and did i post this in the wrong place?
1. Yes.

2. Probably, but I will move it, on second thoughts it could be OK where it is.

CB

5. ok so i would write -

1.

p+q-pq does not equal p-q+pq

therefore not commutative?

and then similarly to associative?

(p+q)-pq does not equal p-(q+pq)

6. Originally Posted by djmccabie
ok so i would write -

1.

p+q-pq does not equal p-q+pq

therefore not commutative?
p*q = p+q-pq

q*p = q+p-qp

but normal addition and multiplication are commutative so:

q*p = q+p-qp = p+q-pq = p*q

so "*" is commutative.

CB

7. Originally Posted by djmccabie

and then similarly to associative?

(p+q)-pq does not equal p-(q+pq)

To determine if "*" is associative you need to determine if:

p*(q*r) = (p*q)*r

CB

8. thanks a lot i understand it now, did i do the associativity right?

9. Originally Posted by djmccabie
thanks a lot i understand it now, did i do the associativity right?
No see the earlier post that crossed yours in the aether

CB