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Relations Test 20

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Relations Test 20
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  • Question 1
    1 / -0
    Write the properties that the relation "is greter that" satisfies in the set of all positive integers
    Solution
    If in a relation, an element $$ a $$ is related to $$ b $$, and $$ b $$ is related to a third element $$ c $$,  the relation $$ R $$ is said to be transitive if $$ a $$ is also related to $$ c $$

    Clearly, when $$ a > b $$ and $$ b > c $$ then $$ a $$  $$ > c $$
    Hence, the relation is transitive.
  • Question 2
    1 / -0
    If $$\displaystyle R=R^{-1}$$ then the relation R is ________
    Solution
    When  a relation is equal to its inverse, then the relation is symmetric.
  • Question 3
    1 / -0
    The domain of the relation R = $$\displaystyle \left \{ \left ( x,y \right ):x,y\epsilon N \ and\ x+y\leq 3 \right \}$$ is____
    Solution
    The only natrual numbers, less than equal to $$ 3 $$ are $$ 1, 2, 3 $$

    From these, only the sum of $$ 1 ,2 $$ gives $$ 3 $$

    So the domain of he relation is {$$ 1,2$$}
  • Question 4
    1 / -0
    If $$\displaystyle n\left ( A\times B \right )=36$$ then n(A) can possibly be____
    Solution
    $$n(A\times B)=n(A)\times n(B)$$

    Hence $$n(A)$$ must be a factor of $$36$$. only possible answer is $$B:9$$
  • Question 5
    1 / -0
    The relation 'is a sister of' in the set of human beings is____
    Solution
    A relation is said to be transitive when an element $$ a $$ is related to $$ b $$ and element $$ b $$ is related to $$ c $$, then $$ c $$ is also related to $$ a $$

    We know that if $$ a $$ is the sister of $$ b $$ and $$ b $$ is sister of $$ c $$, then $$ c $$ is also a sister of $$ a $$

    Thus, the relation " is a sister of " is a transitive relation.
  • Question 6
    1 / -0
    The relation 'is a factor of' on the set of natural numbers is not___________
    Solution
    A relation is said to be symmetric if an element a is related to b then b is also related to a.

    Suppose factor of a number say $$ 10 = 1,2,5,10 $$
    We can notice that factors of $$ 10 $$  are all these numbers, but the reverse is not true for each factor. That is $$ 10 $$ is not a factor of $$ 1 $$ or $$ 2 $$ or $$ 5 $$.

    So, this relation is not symmetric.
  • Question 7
    1 / -0
    If $$\displaystyle n\left ( P\times Q \right )=0$$  then n(P) can possibly be
    Solution
    $$ n(P) $$ can be of any value as we are not sure of $$ n(Q) $$
    Hence, $$ n(P) $$ can take any of the given values.
  • Question 8
    1 / -0
    If a R b, b R c and a R c, and a,b,c,$$\displaystyle \epsilon $$ A, then the relation R on the set A is said to be a/an_____ relation
    Solution
    If in a relation, an element $$ a $$ is related to $$ b $$, and $$ b $$ is related to a third element $$ c $$,  the relation $$ R $$ is said to be transitive if $$ a $$ is also related to $$ c $$

  • Question 9
    1 / -0
    Let $$R$$ be the relation in the set $$\left\{1, 2, 3, 4\right\}$$ given by $$R=\left\{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\right\}$$. Choose the correct answer.
    Solution
    $$R=\left\{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\right\}$$
    It is seen that $$(a, a)\in R$$, for every $$a\in \left\{1, 2, 3, 4\right\}$$.
    $$\therefore $$ $$R$$ is reflexive.
    It is seen that $$(1, 2)\in R$$, but $$(2, 1)\notin R$$.
    $$\therefore$$ $$R$$ is not symmetric.
    Also, it is observed that $$(a, b), (b, c)\in R\Rightarrow (a, c)\in R$$ for all $$a, b, c \in \left\{1, 2, 3, 4\right\}$$
    $$\therefore$$ $$R$$ is transitive.
    Hence,$$R$$ is reflexive and transitive but not symmetric.
    The correct answer is B.
  • Question 10
    1 / -0
    If $$A=\left \{1, 2,3\right \}$$ and $$B=\left \{3,8\right \}$$, then $$(A\cup B)\times (A\cap B)$$ is equal to
    Solution
    Given, $$A=\left \{1, 2,3\right \}$$ and $$B=\left \{3,8\right \}$$,
    Therefore, $$A\cup B=\left \{1,2,3\right \}\cup \left \{3,8\right \}=\left \{1,2,3,8\right \}$$
    and $$A\cap B=\left \{1,2,3\right \}\cap \left \{3,8\right \}=\left \{3\right \}$$
    $$\therefore (A\cup B)\times (A\cap B)=\left \{1,2,3,8\right \}\times \left \{3\right \}$$
    $$=\left \{(1,3), (2,3), (3,3), (8,3)\right \}$$
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