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Relations Test 6

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Relations Test 6
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  • Question 1
    1 / -0
    Let A={ 1, 2, 3, 4} and R= {( 2, 2), (3, 3), (4, 4), (1, 2)} be a relation on A. Then R is
    Solution
    $$\neq$$ R as every element is not related to itself.
    $$\neq$$ S as (1, 2) $$\in$$ R but (2, 1) $$\notin$$ R
    T (1,2), $$\in$$ R, (2, 2) $$\in$$ R
    $$\Rightarrow $$ (1, 2)o (2, 2) = (1, 2) $$\in$$ R
  • Question 2
    1 / -0
    Given the relation R= {(1,2), (2,3) } on the set {1, 2, 3}, the minimum number of ordered pairs which when added to R make it an equivalence relation is 
    Solution
    For R add ( 1,1) ( 2, 2) (3, 3) - 3
    For S add (2, 1) (3, 2)-  2
    For T add (1, 3) ( 3, 1) - 2 
    Thus the minimum number of ordered pairs to be added is 7.
  • Question 3
    1 / -0
    If $$A=\left\{2,4\right\}$$ and $$B=\left\{3,4,5\right\},$$ then
    $$\left( A\cap B \right) \times \left( A\cup B \right)$$ is
    Solution
    From given we get,
    $$A\cap B=\{ 4\} ,A\cup B=\{ 2,3,4,5\}$$

    $$\therefore \left( A\cap B \right) \times\left( A\cup B \right)$$$$=\left\{ (4,2),(4,3),(4,4),(4,5) \right\} $$

    Ans. $$(d)$$.
  • Question 4
    1 / -0
    Let $$A=\left\{1,2,3,4,5\right\}, B=\left\{2,3,6,7\right\}.$$ Then the number of elements in $$\left( A\times B \right) \cap \left( B\times A \right)$$ is
    Solution
    Ans. $$(c)$$.
    Common elements will be
    $$\left\{ { (2,2),(3,3),(2,3),(3,2) } \right\} $$
  • Question 5
    1 / -0
    Let R be a relation from a set A to a set B then 
    Solution
    Let R be a relation from a set A to a set B then 

    $$R\subseteq A\times B$$
  • Question 6
    1 / -0
    The union of two equivalence relations is: 
    Solution
    The union of two equivalence relation is not necessarily an equivalence relation.
    Union of reflexive relation is reflexive,
    Also, the union of symmetric relation is symmetric.
    But the union of a transitive relation is not necessarily transitive.
    Hence, the union of two equivalence relation is not equivalence.
  • Question 7
    1 / -0
    If $$n(A) = 2, n(B) = m$$ and the number of relation from $$A$$ to $$B$$ is $$64$$, then the value of $$m$$ is
    Solution
    The number of relations between sets can be calculated using $$ 2^{mn}$$ where m and n represent 
    the number of members in each set.
    Given that $$ n(A) = 2 $$ and $$ n(B) = m $$
    So, 
    $$ 2^{2m} = 64 $$
    $$ 2^{2m} = 2^{6} $$
    On comparing powers of 2 we get,
    $$ 2m = 6 $$
    $$ m = 3 $$

  • Question 8
    1 / -0
    If $$A$$ and $$B$$ are two sets containing four and two elements, respectively. Then the number of subsets of the set $$A\times B$$ each having at least three elements is
    Solution
    $$A=\left\{a,b,c,d\right\}$$
    $$B=\left\{x,y\right\}$$
    $$A\times B=2\times 4=8$$ Elements.
    Total number of subsets of $$A\times B$$ having $$3$$ or more elements $$=2^8=256$$
    $$\Rightarrow 256-(1$$ null set $$+$$ $$8$$ singleton set $$+$$ $$^8e_2$$having $$2$$ elements$$)$$
    $$=256-1-8-28$$
    $$=219$$
  • Question 9
    1 / -0
    Let $$R$$ be a relation such that $$R = \{(1,4), (3, 7), (4, 5), (4, 6), (7, 6) \}$$ then $$(R^{-1} oR)^{-1} =$$
    Solution

  • Question 10
    1 / -0
    If $$A$$ and $$B$$ are two sets, then $$A \times B  = B \times A$$ if
    Solution

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