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Relations Test 9

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Relations Test 9
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  • Question 1
    1 / -0
    Let $$R$$ be a reflexive relation on a finite set $$A$$ having $$n$$ elements, and let there be $$m$$ ordered pairs in $$R,$$ then:
    Solution
    By the definition for a relation with $$n$$ elements to be  reflexive relation  it must have at least $$n$$ ordered pairs.
    it has $$m$$ ordered pairs therefore $$\mathrm{m}\geq \mathrm{n}$$.
  • Question 2
    1 / -0
    Total number of equivalence relations defined in the set $$S =\{a,b,c\}$$ is
    Solution
    The smallest equivalence relation is the identity relation $$R_1=\{(a,a),(b,b),(c,c)\}$$
    Then two ordered pairs of two distinct elements can be added to give three more equivalence relations.
    $$R_2=\{(a,a),(b,b),( c,c),(a,b),(b,a)\}$$
    Similarly $$R_3$$and $$R_4$$ can be made by taking $$(b,c),(c,b)$$ and $$(a,c),(c,a)$$ respectively.
    Finally the largest equivalence relation i.e., the universal relation
    $$R_5=\{(a,a),(b, b),(c, c),(a, b),(b, a),(a, c),(c,a ),(b, c),(c, b) \}$$
    Hence, total $$5$$ equivalence relations can be created.
  • Question 3
    1 / -0
    If relation $$R$$ is defined by $$\mathrm{R}=\{(\mathrm{x},\ \mathrm{y}):2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6\}$$, then the domain of $$\mathrm{R}$$ is
    Solution
     Domain $$=$$ value of $$\mathrm{x}$$ in $$f=2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6$$  then
    $$2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6$$ = $$\displaystyle \frac{\mathrm{x}^{2}}{3}+\frac{\mathrm{y}^{2}}{2}\leq 1$$ 
    Since it is an equation of ellipse then value of $$x$$ $$ \Rightarrow  \mathrm{x}\in[-\sqrt{3},\sqrt{3}]$$
  • Question 4
    1 / -0
    The domain and range of relation $$R=\{(x,y) | x, y \in N$$, $$x+2y=5\} $$ is?
    Solution
    The possible values of $$x$$ which satisfies the given relation is the domain of the relation
    The possible values of $$y$$ which satisfies the given relation is the range of the relation
    Given that $$x,y$$ are natural numbers and the relation is $$x+2y=5$$
    For $$x=1$$ , the value of $$y$$ is $$2$$
    For $$x=3$$ , the value of $$y$$ is $$1$$
    If $$x\geq 5$$, then $$y$$ is negative which does not belong to naturals.
    If $$x=2,4$$ then $$y$$ becomes rational number which is not in naturals.
    Thus, only $$1$$ and $$3$$ gives $$y$$ as natural numbers $$2$$ and $$1$$.
    Therefore, the domain is $$\{1,3\}$$ and the range is $$\{2,1\}$$
  • Question 5
    1 / -0
    Let $$A$$ be a non-empty set such that $$A \times A$$ has $$9 $$ elements among which are found $$(-1, 0)$$ and $$(0, 1)$$, then
    Solution
    Given $$A\times A$$ has $$9$$ elements $$\Rightarrow n(A)=3$$
    Elements of $$A\times A  $$ are  $$(-1,0)$$  &  $$(0,1)$$.
    $$\therefore$$ $$A$$ needs to contain $$-1, 0$$ and $$1$$.
    $$\therefore$$ Since $$n(A)=3 \Rightarrow  A=\left \{ -1,0,1 \right \}$$
  • Question 6
    1 / -0
    Given the relation $$R =\{(1,2) (2,3)\}$$ on the set $$A=\{1,2,3\}$$, the minimum number of ordered pairs which when added to $$R$$ to make it an equivalence relation is
    Solution

    $$R$$ is reflexive if it contains $$(1, 1), (2, 2), (3, 3)$$.

    $$\because (1, 2)\in R$$ and $$ (2,3)\in R$$
    $$\therefore R$$ is symmetric if $$(2, 1), (3, 2) \in R.$$
    As $$(1, 2), (2,3)\in R$$
    Then, $$R$$ is transitive is $$(1,3)\in R$$
    Also, $$(3,1)\in R$$ for symmetric property.
    Now, $$R=\{(1,1),(2,2),(3,3),(2,1),(3,2),(2,3),(1,2),(1,3),(3,1)\}$$
    Thus, $$R$$ becomes an equivalence relation by adding
    $$(1,1),(2,2),(3,3),(2,1),(3,2),(1,3),(3,1)$$
    Hence, the total number of ordered pairs is $$7$$.
  • Question 7
    1 / -0
    Which one of the following relations on Z is equivalence relation?
    Solution
    A relation is equivalence if it is symmetric, transitive and reflexive
    Consider, $$xR_1y\Leftrightarrow \left | x \right |=\left | y \right |$$
    For reflexive:
    $$xR_1x\Leftrightarrow \left | x \right |=\left | x \right |$$, clearly it satisfies $$R_{1}$$
    Thus, $$R_{1}$$  is reflexive.
    For symmetric:
    $$xR_{1}y\Leftrightarrow \left | x \right |=\left | y \right |$$ $$\implies yR_{1}x\Rightarrow \left | y \right |=\left | x \right |$$  as equality communicative 
    So, $$ R_{1}$$ is symmetric.
    For transitive:
    Let $$(x,y)$$ and $$(y,z)$$ satisfy $$R_{1}$$
    $$\implies xR_{1}y\Leftrightarrow \left | x \right |=\left | y \right |$$ and $$yR_{1}z\Leftrightarrow \left | y \right |=\left | z \right |$$
    $$\implies |x|=|y|$$ and $$|y |=|z|$$
    $$\implies |x |=|z|$$
    $$\therefore xR_{1}z\Leftrightarrow \left | x \right |=\left | z \right |$$
    $$\therefore \left ( x,z \right )$$ satisfies $$R_{1}$$
    Thus, $$R_1$$ is a transitive relation.
    Hence, $$  R_{1}$$ is equivalent.
    Also, relations $$R_{2},\ R_{3}$$ and $$R_{4}$$ are not symmeteric on $$Z,$$ hence not an equivalence relations.        
  • Question 8
    1 / -0
    Which of the following are not equivalence relations on $$I$$?
    Solution
    Option A, $$aRb$$ if $$a+b$$ is an even integer.

    Let $$a, b,c \in I$$
    Clearly, $$a+a =2a$$ is an even integer.
    Hence, $$aRa$$ for all $$a\in I$$

    Now, let $$aRb$$ 
    $$\Rightarrow a+b$$ is an even integer.
    or $$b+a$$ is an even integer.
    $$\Rightarrow bRa$$

    Next, let $$aRb, bRc$$
    $$\Rightarrow a+b $$ in an even integer and $$ b+c$$ is an even integer
    $$\Rightarrow  a$$ and $$b$$ both are even or both are odd . Also, $$b$$ and $$c$$ both are even or both are odd.
    Hence, if $$a$$ and $$b$$ are even .So $$c$$ is also even . Hence, $$a+c$$ is even.
    And if $$a$$ and $$b$$ are odd.So, $$c$$ is also odd . Again $$a+c$$ is even.
    Hence, $$aRc$$

    Hence, option A is equivalence relation.

    Option B, $$aRb$$ if $$a-b$$ is an even integer.

    Let $$a, b,c \in I$$
    Clearly, $$a-a =0$$ is an even integer.
    Hence, $$aRa$$ for all $$a\in I$$

    Now, let $$aRb$$ 
    $$\Rightarrow a-b$$ is an even integer.
    or $$-(a-b)$$ is also an even integer.
    $$\Rightarrow \ bRa$$

    Next, let $$aRb, bRc$$
    $$\Rightarrow a-b $$ in an even integer and $$ b-c$$ is an even integer
    $$a+c=a-b+b-c=$$ even+even 
    Hence, $$a+c$$ is even.
    Hence, $$aRc$$

    Hence, option $$B$$ is equivalence relation.

    Option C $$aRb$$ if $$a<b$$

    Let $$a, b,c \in I$$
    Clearly,symmetry does not hold here 
    Now, let $$aRb$$ 
    $$\Rightarrow a<b$$ 
    But it does not implies $$b<a$$

    Hence, option C is not an equivalence relation.

    Option D, $$aRb$$ if $$a=b$$ 

    Let $$a, b,c \in I$$
    Clearly, $$a=a$$ 
    Hence, $$aRa$$ for all $$a\in I$$

    Now, let $$aRb$$ 
    $$\Rightarrow a=b$$ 
    or $$b=a$$ is an even integer.
    $$\Rightarrow bRa$$

    Next, let $$aRb, bRc$$
    $$\Rightarrow a=b $$and $$ b=c$$ 
    $$\Rightarrow a=b=c$$
    Hence, $$aRc$$

    Hence, option D is equivalence relation.
  • Question 9
    1 / -0
    The number of relations from $$ A=\left \{ 1,2,3 \right \} $$ to $$B = \left \{ 4,6,8,10 \right \}$$ is
    Solution
    $$n\left ( B \right )=4$$
    $$n\left ( A \right )=3$$
    Therefore there exists $$2^{3 \times 4}$$ relations from $$A$$ to $$B$$
    i.e $$2^{12}$$ relations
  • Question 10
    1 / -0
    The range of the function $$f(x)=\dfrac{\sin(\pi [x])}{x^{2}+1}$$ (where $$[.]$$ denotes greatest integer function) is
    Solution
    $$f(x)=\dfrac {\sin(\pi[x])}{x^2+1}$$
    $$\because$$ $$[x]$$ always gives integer value and $$\sin n\pi =0$$
    $$\therefore f(x)=0$$
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