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Home » Check the commutativity and associativity of each of the following binary operations: (xi) ‘*’ on N defined by a * b = ab for all a, b ∈ N (xii) ‘*’ on Z defined by a * b = a – b for all a, b ∈ Z
Check the commutativity and associativity of each of the following binary operations: (xi) ‘*’ on N defined by a * b = ab for all a, b ∈ N (xii) ‘*’ on Z defined by a * b = a – b for all a, b ∈ Z
Binary Operations | CBSE | Class 12 | Exercise 3.2 | Maths | RD Sharma
Check the commutativity and associativity of each of the following binary operations: (xi) ‘*’ on N defined by a * b = ab for all a, b ∈ N (xii) ‘*’ on Z defined by a * b = a – b for all a, b ∈ Z

(xi)  to check : commutativity of *

    \[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }N,\text{ }then \\ a\text{ }*\text{ }b\text{ }=\text{ }{{a}^{b}} \\ b\text{ }*\text{ }a\text{ }=\text{ }{{b}^{a}} \\ ~a\text{ }*\text{ }b\text{ }\ne \text{ }b\text{ }*\text{ }a \\ \end{array}\]

Thus, * is not commutative on N.

to check : associativity of *

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

Thus, * is not associative on N

(xii)  to check :commutativity of *

    \[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }Z,\text{ }then \\ a\text{ }*\text{ }b\text{ }=\text{ }a\text{ }-\text{ }b \\ b\text{ }*\text{ }a\text{ }=\text{ }b\text{ }-\text{ }a \\ a\text{ }*\text{ }b\text{ }\ne \text{ }b\text{ }*\text{ }a \\ \end{array}\]

Thus, * is not commutative on Z.

to check : associativity of *

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

 

Thus, * is not associative on Z

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