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Relations and Functions Test - 9

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Relations and Functions Test - 9
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Weekly Quiz Competition
  • Question 1
    1 / -0

    A relation R in a set A is called universal relation, if

    Solution

    The relation R = A x A is called Universal relation.

  • Question 2
    1 / -0

    A relation R in a set A is called reflexive,

    Solution

    A relation R on a non empty set A is said to be reflexive if x Rx for all x ∈ R , Therefore , R is reflexive.

  • Question 3
    1 / -0

    A relation R on a non-empty set A is said to be an equivalence relation if

    Solution

    A relation R on a non empty set A is said to be reflexive if xRx for all x ∈ A . .
    A relation R on a non empty set A is said to be symmetric if xRy ⇔ yRx, for all x , y ∈A .
    A relation R on a non empty set A is said to be transitive if xRy and yRz ⇒ x Rz, for all x,y,z ∈ A.
    An equivalence relation satisfies all these three properties.

  • Question 4
    1 / -0

    Let A be the set of all students of a boys school. The relation R on A is given by R = {(a, b) : a is sister of b} is

    Solution

    Because, the relation is defined over the set A be the set of all students of a boys school.

  • Question 5
    1 / -0

    Let L be the set of all lines in a plane and R be the relation on L defined as R = {(L1, L2): L1 is perpendicular to L2}. Then R is

    Solution

    The relation R is symmetric only , because if L1 is perpendicular to L2 ,then L2 is also perpendicular to L1,but no other cases that is reflexive and transitive is not possible.

  • Question 6
    1 / -0

    The relation R on the set Z of integers given by R = {(a, b): 2 divides a – b} ,∀ a, b ∈ Z is

    Solution
    1. Since a – a = 0 , and 0 is divisible by 2 , therefore, aRa i.e. R is reflexive.
    2. If aRb , then a – b is divisible by 2. ⇒ - ( a- b ) is divisible by 2. ⇒(b – a ) is divisible by 2. ⇒bRa i.e. R is symmetric. .
    3. .If aRb and bRc , then a – b is divisible by 2 and b – c is divisible by 2 ⇒ a – b = 2q and b – c = 2q’ where q and q’ are integers. ⇒( a – b ) + ( b – c ) = 2 ( q + q’) ⇒ a – c =2( q + q’), but (q +q’) is an integer. ⇒(a –c ) is divisible by 2. ⇒aRc i.e. R is transitive. .
  • Question 7
    1 / -0

    Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Then R is

    Solution

    Consider any a ,b , c ∈A .

    1. Since both a and a must be either even or odd, so (a , a) ∈R ⇒R is reflexive.
    2. Let (a ,b) ∈ R ⇒ both a and b must be either even or odd, ⇒ both b and a must be either even or odd, ⇒ ( b ,a) ∈R .Thus , (a ,b) ∈R ⇒ ( b ,a) ∈R ⇒R is symmetric.
    3. Let (a ,b) ∈R and (b ,c) ∈R ⇒ both a and b must be either even or odd, also ,both b and c must be either even or odd, ⇒ all elements a, b and c must be either even or odd, ⇒ ( a ,c) ∈ R . Thus , (a ,b) ∈ R ⇒ ( b ,c) ∈ R ⇒ (a ,c) ∈ R ⇒R is transitive.
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