Q2 of 27 Page 78

Define * on Z by a * b = a – b + ab. Show that * is a binary operation on Z which is neither commutative nor associative.

* is an operation as a*b = a-b + ab where a, b ∈ Z. Let and b = 2 two integers.


So, * is a binary operation from .


For commutative,



Since a*b ≠ b*a, hence * is not commutative operation.


Again for associative,


a*(b*c) = a*(b-c+ bc)


= a- (b- c+ bc) +a (b- c+ bc)


= a-b+ c- bc+ ab- ac+ abc


(a*b) *c = (a-b+ ab) *c


= a- b+ ab-c+ (a- b+ ab) c


= a-b-c+ ab+ ac- bc+ abc


As a*(b*c) ≠ (a*b) *c, hence * not an associative operation.


More from this chapter

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17

A binary operation * on the set (0, 1, 2, 3, 4, 5) is defined as


Show that 0 is the identity for this operation and each element a has an inverse (6 - a)


 1

Define * on N by m * n = 1 cm (m, n). Show that * is a binary operation which is commutative as well as associative.

 3

Define * on Z by a * b = a + b - ab. Show that * is a binary operation on Z which is commutative as well as associative.

 4

Consider a binary operation on Q – {1}, defined by a * b = a + b - ab.

(i) Find the identity element in Q – {1}.


(ii) Show that each a ∈ Q - {1} has its inverse.