Relations Test 42

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Relations Test 42

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Weekly Quiz Competition

Question 1
1 / -0

Consider the set $$A=(1, 2, 3)$$ and the R be the smallest equivalence relation on R then R is

A
$${(1, 1), (2, 2), (3,3)}$$

B
$${(1,1), (2 ,2), (3, 3), (1, 2), (2, 1)}$$

C
$${(1, 1,(2 ,2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)}$$

D
None of these

Solution
Question 2
1 / -0

The relation $$R=\left\{ \left( 1,1 \right) ,\left( 2,2 \right) ,\left( 3,3 \right) \right\} $$ on the set $$\left\{ 1,2,3 \right\} $$ is

A
symmetric only

B
reflexive only

C
transitive only

D
an equivalence relation

Solution
Question 3
1 / -0

If $ A $ is a finite set containing n distinct elements, then the number of relations on A is equal to

A
$$ 2^{n} $$

B
$$ 2^{n^{2}} $$

C
$$ 2^{2} $$

D
none of these

Solution
Question 4
1 / -0

If R is an equivalence relation on a set A, then $$R^1$$ is

A
reflexive only

B
symmetric but not transitive

C
equivalence

D
transitive
Question 5
1 / -0

Let $$N$$ denotes the set of all natural numbers and $$R$$ be the relation on $$N \times N$$ defined by $$( a , b ) R ( c , d )$$ iff $$a d ( b + c ) = b c ( a + d ) ,$$ then $$R$$ is

A
symmetric only

B
reflexive only

C
transitive only

D
an equivalence relation

Solution
Question 6
1 / -0

Let $$R:N \to N$$ be defined by $$R = \{ (a,b):a,b \in N$$ and $$a = {b^2}\}$$ then, which of the following is true?

A
$$(a,a) \in R,\forall a \in N$$

B
$$(a,b) \in R, \Rightarrow (b,a) \in R$$

C
$$(a,b) \in R,(b,c) \in R \Rightarrow (a,c) \in R$$

D
None of the above

Solution
Question 7
1 / -0

Let $$A$$ be set of first ten natural numbers and $$R$$ be a relation on $$A$$ defined by (x , y) $$\in$$ $$R$$ $$\Rightarrow$$ x + 2y = 10 , then domain of $$R$$ is

A
$$\left \{ 1 , 2 , 3 ,............, 10 \right \}$$

B
$$\left \{ 2 , 4 , 6 , 8 \right \}$$

C
$$\left \{ 1 , 2 , 3 , 4 \right \}$$

D
$$\left \{ 2 , 4 , 6 , 8 , 10 \right \}$$

Solution

$$\begin{aligned} A &=\{1,2,3, \cdots \cdots, 10\} \\ R &=\{(x, y): x+2 y=10, x, y \in A\} \\ & \therefore y=\dfrac{10-x}{2} \\ R &=\{(2,4),(4,3),(6,2),(8,1)\} \\ \therefore & \text { Domain of } R \text { is }\{2,4,6,8\} \end{aligned}$$
Correct option is (B).
Question 8
1 / -0

If $$h=\left\{ ((x,y),(x-y, x+y))/ x,y \epsilon N \right\} $$ is a relation on NxN, then domain of h

A
$$\left\{ (x,y):\quad x

B
$$\left\{ (x,y):\quad x\le y,x,y\epsilon N \right\} $$

C
$$\left\{ (x,y):\quad x>y,x,y\epsilon N \right\} $$

D
$$\left\{ (x,y):\quad x \ge y,x,y\epsilon N \right\} $$
Question 9
1 / -0

If $$A=\left\{ x:{ x }^{ 2 }-3x+2=0 \right\} $$, and R is a universal relation on A, then R is

A
$${(1, 1), (2, 2)}$$

B
$${(1, 1)}$$

C
$$\{ \phi \} $$

D
None of these.

Solution
Question 10
1 / -0

Let $$R_1, R_2$$ are relation defined on $$Z$$ such that $$aR_1b \Longleftrightarrow (a - b)$$ is divisible by $$3$$ and $$a \,R_2 b \Longleftrightarrow (a - b)$$ is divisible by $$4$$. Then which of the two relation $$(R_1 \cup R_2), (R_1 \cap R_2)$$ is an equivalence relation?

A
$$(R_1 \cup R_2)$$ Only

B
$$(R_1 \cap R_2)$$ Only

C
Both $$(R_1 \cup R_2), (R_1 \cap R_2)$$

D
Neither $$(R_1 \cup R_2)$$ nor $$(R_1 \cap R_2)$$

Solution

Givn relations defined on z are,
$$R_1=aR_1b\leftrightarrow (a-b)$$ is divisible by 3.
and $$R_2=aR_2b\leftrightarrow (a-b)$$ is divisible by 4.

Consider,
$$aR_1b\leftrightarrow a-b$$ is divisible by 3.

Reflexive:
$$aR_1a\leftrightarrow a-a$$ is divisible by 3
True
$$\therefore$$ the given relation $$R_1$$ is a reflexive relation.

Symmetric:
$$aR_1b\leftrightarrow a-b$$ is divisible by 3.
$$\Rightarrow a-b=3k$$; where $$k$$ is an integer.
then
$$bR_1a\leftrightarrow b-a$$ is divisible by 3.
because $$b-a=-(a-b)=-3k$$ which is divisible by 3.
$$\therefore$$ the given relation $$R$$ is a symmetric relation.

Transitive:
$$aR_1b\leftrightarrow a-b$$ is divisible by 3.
$$\Rightarrow a-b=3k; k$$ is an integer.

$$b R_1c\leftrightarrow b-c$$ is divisible by 3.
$$\Rightarrow b-c=3p;p$$ is integer.

Then $$aR_1c\leftrightarrow a-c$$ is divisible by 3.
Because $$(a-b)+(b-c)=a-c$$
$$\Rightarrow 3k+3p=3(k+p)$$ which is divisible by 3.
$$\therefore$$ The given relation $$R_1$$ is transitive relation.
$$\therefore R_1$$ satisfies the reflexive, symmetric and transitive relation properties.
$$\therefore R_1$$ is an equivalence relation.

Similarly $$R_2$$ is also an equivalence relation.

And as we know that,
"If $$R_1$$ and $$R_2$$ are equivalence relations then $$R_1\cap R_2$$ is also an equivalence relation."

$$\therefore R_1\cap R_2$$ is an equivalence relation.