Check the commutativity and associativity of each of the following binary operations: (ix) ‘*’ on Q defined by a * b = (a – b)2 for all a, b ∈ Q (x) ‘*’ on Q defined by a * b = a b + 1 for all a, b ∈ Q
Check the commutativity and associativity of each of the following binary operations: (ix) ‘*’ on Q defined by a * b = (a – b)2 for all a, b ∈ Q (x) ‘*’ on Q defined by a * b = a b + 1 for all a, b ∈ Q

(ix)  to check: commutativity of *

    \[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }Q,\text{ }then  \\ a\text{ }*\text{ }b\text{ }=\text{ }{{\left( a\text{ }-\text{ }b \right)}^{2}}  \\ =\text{ }{{\left( b\text{ }-\text{ }a \right)}^{2}}  \\ =\text{ }b\text{ }*\text{ }a  \\ a\text{ }*\text{ }b\text{ }=\text{ }b\text{ }*\text{ }a,\text{ }for\text{ }all\text{ }a,\text{ }b\text{ }\in \text{ }Q  \\ \end{array}\]

Thus, * is commutative on Q

to prove: associativity of * on Q

Let a, b, c ∈ Q, then

    \[\begin{array}{*{35}{l}} a\text{ }*\text{ }\left( b\text{ }*\text{ }c \right)\text{ }=\text{ }a\text{ }*\text{ }{{\left( b\text{ }-\text{ }c \right)}^{2}}  \\ =\text{ }a\text{ }*\text{ }\left( {{b}^{2}}~+\text{ }{{c}^{2}}~-\text{ }2\text{ }b\text{ }c \right)  \\ =\text{ }{{\left( a\text{ }-\text{ }{{b}^{2}}~-\text{ }{{c}^{2}}~+\text{ }2bc \right)}^{2}}  \\ \left( a\text{ }*\text{ }b \right)\text{ }*\text{ }c\text{ }=\text{ }{{\left( a\text{ }-\text{ }b \right)}^{2}}~*\text{ }c  \\ =\text{ }\left( {{a}^{2}}~+\text{ }{{b}^{2}}~-\text{ }2ab \right)\text{ }*\text{ }c  \\ =\text{ }{{\left( {{a}^{2}}~+\text{ }{{b}^{2}}~-\text{ }2ab\text{ }-\text{ }c \right)}^{2}}  \\ ~a\text{ }*\text{ }\left( b\text{ }*\text{ }c \right)\text{ }\ne \text{ }\left( a\text{ }*\text{ }b \right)\text{ }*\text{ }c  \\ \end{array}\]

Thus, * is not associative on Q.

(x) to check : commutativity of *

    \[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }Q,\text{ }then  \\ a\text{ }*\text{ }b\text{ }=\text{ }ab\text{ }+\text{ }1  \\ =\text{ }ba\text{ }+\text{ }1  \\ =\text{ }b\text{ }*\text{ }a  \\ a\text{ }*\text{ }b\text{ }=\text{ }b\text{ }*\text{ }a,\text{ }for\text{ }all\text{ }a,\text{ }b\text{ }\in \text{ }Q  \\ \end{array}\]

Thus, * is commutative on Q

Now we have to prove associativity of * on Q

    \[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b,\text{ }c\text{ }\in \text{ }Q,\text{ }then  \\ a\text{ }*\text{ }\left( b\text{ }*\text{ }c \right)\text{ }=\text{ }a\text{ }*\text{ }\left( bc\text{ }+\text{ }1 \right)  \\ =\text{ }a\text{ }\left( b\text{ }c\text{ }+\text{ }1 \right)\text{ }+\text{ }1  \\ =\text{ }a\text{ }b\text{ }c\text{ }+\text{ }a\text{ }+\text{ }1  \\ \left( a\text{ }*\text{ }b \right)\text{ }*\text{ }c\text{ }=\text{ }\left( ab\text{ }+\text{ }1 \right)\text{ }*\text{ }c  \\ =\text{ }\left( ab\text{ }+\text{ }1 \right)\text{ }c\text{ }+\text{ }1  \\ =\text{ }a\text{ }b\text{ }c\text{ }+\text{ }c\text{ }+\text{ }1  \\ ~a\text{ }*\text{ }\left( b\text{ }*\text{ }c \right)\text{ }\ne \text{ }\left( a\text{ }*\text{ }b \right)\text{ }*\text{ }c  \\ \end{array}\]

Thus, * is not associative on Q.