Relations and Functions Test - 19

Since the binary operation * is defined on set of natural numbers i.e. N And a * b = b * a is true for all natural numbers, because H.C.F. of a and b is same as b and a. Similarly , (a * b) * c = a * (b * c). Therefore * is both commutative as well as associative. 

Let ∗ be a binary operation on the set Q of rational numbers defined by a * b = a² + b², then ∗ is commutative, because for any two rational numbers a * b = a² + b² = b² + a² = b * a .

Let ∗ be a binary operation on the set Q of rational numbers defined by a ∗ b = a + ab, then ∗ is not commutative, because for any two rational numbers ab = ba , therefore, a ∗ b = a + ab not equal to  b + ba = b * a.

Let ∗ be a binary operation on the set Q of rational numbers defined by a * b = (a - b)², then ∗ is commutative because, for any two rational numbers

a * b = (a - b)²= (b - a)² = b * a.

Consider the non – empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is not symmetric, because aRb means a is brother of b, then, it is not necessary that b is also brother of a , it can be the sister of a. Therefore, bRa is not true. Therefore, the relation is not symmetric . Again, if aRb and bRc is true, then aRc is also true. Therefore, R is transitive only.

Let n be the number of elements in the set A , then the maximum number of equivalence relations defined on A are given by : 2n – 1. Here A={1,2,3} ⇒ n(A)=3, then the number of relations of A is given by 2 x 3 -1 = 5.

Given R = {(1,2)} defined on the set A={1,2,3},where (1,2) is the only ordered pair defined on R. Then,we have 

Reflexive property:  for (1,2) ∈ R ⇒ (1,1) not belongs to R  ;  ∀ 1,2 ∈ A ⇒ R is not reflexive
Symmetric property:  for (1,2) ∈ R ⇒ (2,1) not belongs to R , ∀ 1,2 ∈ A ⇒ R is not symmetric
Transitive property:  for (1,2) and (1,2) ∈ R ⇒ (1,2) ∈ R; ∀ 1,2 ∈ A ⇒ R is  transitive

Reflexivity : aRa a ⩾ a , which is true. Symmetry :aRb a ⩾ b, but, it doesn’t mean that b ⩾ a ,therefore Transitivity : aRb a ⩾ b and bRc b ⩾ c.  Therefore, a ⩾ c, which means aRc.