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# Let $\ast$ be a binary operation on the set $Q$ of rational numbers as follows: $(iv)\;\; a \ast b = (a-b)^2$ Find which of the binary operations are commutative and which are associative.

Note: This is part 4 of a 6 part question, split as 6 separate questions here.

Toolbox:
• An operation $\ast$ on $A$ is commutative if $a\ast b = b \ast a\; \forall \; a, b \in A$
• An operation $\ast$ on $A$ is associative if $a\ast ( b \ast c) = (a \ast b) \ast c\; \forall \; a, b, c \in A$
Given a binary operation $\ast$ in Q defined by $a \ast b=(a-b)^2$:
$\textbf {Step 1: Checking if the operation is Commutative}$:
For an operation $\ast$ to be commutative $a\ast b = b \ast a$.
$a \ast b = (a-b)^2 = a^2+b^2 - 2ab$
$b \ast a = (b-a)^2 = b^2+a^2 - 2ba$
Since $a\ast b = b \ast a$, $\ast$ is commutative.
$\textbf {Step 2: Checking if the operation is Associative}$:
For an operation $\ast$ on $A$ is associative $a\ast ( b \ast c) = (a \ast b) \ast c\; \forall \; a, b, c \in A$
$(a \ast b) \ast c = (a-b)^2 \ast c = ((a-b)^2 - c)^2$
$a \ast (b \ast c = a \ast (b-c)^2 = (a -(b - c)^2)^2$
Since $a\ast ( b \ast c) \neq (a \ast b) \ast c \; \ast$ is not associative
edited Mar 19, 2013