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?
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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
ahh, so i work out commutative and associative for p+q-pq ?? and did i post this in the wrong place?
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
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) your help is much appreciated
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
Originally Posted by djmccabie and then similarly to associative? (p+q)-pq does not equal p-(q+pq) your help is much appreciated To determine if "*" is associative you need to determine if: p*(q*r) = (p*q)*r CB
thanks a lot i understand it now, did i do the associativity right?
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
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