RD Sharma Class 12 Solutions Chapter 1 Relations

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Page No 1.10:

Question 1:

Let A be the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive:

(i) R = {(x, y) : x and y work at the same place}
(ii) R = {(x, y) : x and y live in the same locality}
(iii) R = {(x, y) : x is wife of y}
(iv) R = {(x, y) : x is father of and y}

Answer/Explanation

Answer:

(i) Reflexivity:

Letx be an arbitrary element of R . Then, x \inR \Rightarrowx and x work at the same place is true since they are the same . \Rightarrow(x , x)\inR So , R isa reflexive relation .

Symmetry:

Let(x , y)\inR \Rightarrowx and y work at the same place \Rightarrowy and x work at the same place\Rightarrow(y , x)\inR So , R isa symmetric relation .

Transitivity:

Let(x , y)\inR and(y , z)\inR . Then , x and y work at the same place . y and z also work at the same place . \Rightarrowx , y and z all work at the same place . \Rightarrowx and z work at the same place . \Rightarrow (x , z) \inR So , R is a transitive relation .

(ii) Reflexivity:

Letx be an arbitrary element of R . Then, x \inR \Rightarrowx and x live in the same locality is true since they are the same . So , R isa reflexive relation .

Symmetry:

Let(x , y)\inR \Rightarrowx and y live in the same locality\Rightarrowy and x live in the same locality\Rightarrow(y , x)\inR So , R isa symmetric relation .

Transitivity:

Let(x , y)\inR and(y , z)\inR . Then , x and y live in the same locality and y and z live in the same locality \Rightarrowx , y and z all live in the same locality\Rightarrowx and z live in the same locality \Rightarrow(x , z)\inR So , R isa transitive relation .

(iii)
Reflexivity:

Let x be an element of R . Then , x is wife of x cannot be true . ⇒(x ,x)\notinR So , R is nota reflexive relation .

Symmetry:

Let(x , y)\inR \Rightarrowx is wife of y \Rightarrowx is female and y is male\Rightarrowy cannot be wife of x as y is husband of x\Rightarrow(y , x)\notinR So , R is nota symmetric relation .

Transitivity:

Let(x , y)\inR , but(y , z)\notinR Since x is wife of y , but y cannot be the wife of z , y is husband of x . \Rightarrowx is not the wife of z \Rightarrow (x , z) \inR So , R is a transitive relation .

(iv)
Reflexivity:

Let x be an arbitrary element of R . Then , x is father of x cannot be true since no one can be father of himself . So , R is nota reflexive relation .

Symmetry:

Let(x , y)\inR \Rightarrowx is father of y\Rightarrowy is son /daughter of x\Rightarrow(y , x)\notinR So , R is nota symmetric relation .

Transitivity:

Let(x , y)\inR and(y , z)\inR . Then , x is father of y and y is father of z \Rightarrowx is grandfather of z\Rightarrow(x , z)\notinR So , R is nota transitive relation .

Page No 1.10:

Question 2:

Three relations R1, R2 and R3 are defined on a set A = {a, b, c} as follows:
R1 = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}
R2 = {(a, a)}
R3 = {(b, c)}
R4 = {(a, b), (b, c), (c, a)}.

Find whether or not each of the relations R1, R2, R3, R4 on A is (i) reflexive (ii) symmetric and (iii) transitive.

Answer/Explanation

Answer:

(i) R1
Reflexive:
Clearly, (a, a), (b, b) and (c, c)

\in

R1
So, R1 is reflexive.

Symmetric:
We see that the ordered pairs obtained by interchanging the components of R1 are also in R1.
So, R1 is symmetric.

Transitive:
Here,

(a , b) \in R_{1}, (b , c) \in R_{1} and also (a , c) \in R_{1}

So, R1 is transitive.

(ii) R2
Reflexive: Clearly

(a ,a) \in R_{2}

. So, R2 is reflexive.
Symmetric: Clearly

(a ,a) \inR \Rightarrow (a ,a) \inR

. So, R2 is symmetric.
Transitive: R2 is clearly a transitive relation, since there is only one element in it.

(iii) R3
Reflexive:
Here,

(b , b) \notin R_{3} neither (c , c) \notin R_{3}

So, R3 is not reflexive.

Symmetric:
Here,

(b , c) \in R_{3}, but (c ,b) \notin R_{3} So ,R_{3}is not symmetric .

Transitive:
Here, R3 has only two elements. Hence, R3 is transitive.

(iv) R4
Reflexive:
Here,

(a , a) \notin R_{4}, (b , b) \notin R_{4} (c , c) \notin R_{4} So ,R_{4}is not reflexive .

Symmetric:
Here,

(a , b) \in R_{4}, but (b ,a) \notin R_{4} . So , R_{4} is not symmetric .

Transitive:
Here,

(a , b) \in R_{4}, (b , c) \in R_{4}, but (a , c) \notin R_{4} So ,R_{4}is not transitive .

Page No 1.10:

Question 3:

Test whether the following relations R1, R2, and R3 are (i) reflexive (ii) symmetric and (iii) transitive:
(i) R1 on Q0 defined by (a, b) ∈ R1 ⇔ a = 1/b.
(ii) R2 on Z defined by (a, b) ∈ R2 ⇔ |a – b| ≤ 5
(iii) R3 on R defined by (a, b) ∈ R3 ⇔ a2 – 4ab + 3b2 = 0.

Answer/Explanation

Answer:

(i) Reflexivity:
Let a be an arbitrary element of R1. Then,

a ∈R_{1}\Rightarrowa \neq\frac{1}{a}for all a \inQ_{0}So ,R_{1}is not reflexive .

Symmetry:
Let (a, b)

\in R_{1}

. Then,

(a , b) \in R_{1} \Rightarrowa = \frac{1}{b} \Rightarrowb = \frac{1}{a} \Rightarrow (b , a) \in R_{1} So ,R_{1}is symmetric .

Transitivity:
Here,

(a , b) \in R_{1} and (b , c) \in R_{2} \Rightarrowa = \frac{1}{b} and b = \frac{1}{c} \Rightarrowa = \frac{1}{\frac{1}{c}} =c\Rightarrowa \neq\frac{1}{c}\Rightarrow(a , c)\notinR_{1}So ,R_{1}is not transitive .

(ii)
Reflexivity:
Let a be an arbitrary element of R2. Then,

a \in R_{2} \Rightarrow |a -a| = 0 \leq 5 So ,R_{1}is reflexive .

Symmetry:

Let(a , b)\inR_{2}\Rightarrow|a -b|\leq5\Rightarrow|b -a|\leq5[Since ,|a -b|=|b -a|]\Rightarrow(b , a)\inR_{2}So ,R_{2}is symmetric .

Transitivity:

Let(1, 3)\inR_{2}and(3, 7)\inR_{2}\Rightarrow|1 - 3|\leq5and|3 - 7|\leq5But|1 - 7|≰5\Rightarrow(1, 7)\notinR_{2}So ,R_{2}is not transitive .

(iii)
Reflexivity: Let a be an arbitrary element of R3. Then,

a \in R_{3} \Rightarrow a^{2} - 4 a ×a +3a^{2}=0So ,R_{3}is reflexive .

Symmetry:

Let(a , b)\inR_{3}\Rightarrowa^{2}-4ab +3b^{2}=0Butb^{2}-4ba +3a^{2}\neq0for all a , b \inR So ,R_{3}is not symmetric .

Transitivity:

(1, 2) \in R_{3} and (2, 3) \in R_{3} \Rightarrow 1 - 8 + 6 = 0 and 4 - 24 + 27 = 0 But1-12+9\neq0So ,R_{3}is not transitive .

Page No 1.10:

Question 4:

Let A = {1, 2, 3}, and let R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}, R2 = {(2, 2), (3, 1), (1, 3)}, R3 = {(1, 3), (3, 3)}. Find whether or not each of the relations R1, R2, R3 on A is (i) reflexive (ii) symmetric (iii) transitive.

Answer/Explanation

Answer:

(1)

R1
Reflexivity:
Here,

(1, 1), (2, 2), (3, 3) \inR So , R_{1} is reflexive .

Symmetry:

Here ,(2, 1)\inR_{1}, but(1, 2)\notinR_{1}So ,R_{1}is not symmetric .

Transitivity:

Here ,(2, 1)\inR_{1}and(1, 3)\inR_{1}, but(2, 3)\notinR_{1}So ,R_{1}is not transitive .

(2)

R2
Reflexivity:

Clearly ,(1, 1)and(3, 3)\notinR_{2}So ,R_{2}is not reflexive .

Symmetry:

Here ,(1, 3)\inR_{2}and(3, 1)\inR_{2}So ,R_{2}is symmetric .

Transitivity:

Here ,(1, 3)\inR_{2}and(3, 1)\inR_{2}But(3, 3)\notinR_{2}So ,R_{2}is not transitive .

(3)

R3
Reflexivity:

Clearly ,(1, 1)\notinR_{3}So ,R_{3}is not reflexive .

Symmetry:

Here ,(1, 3)\inR_{3}, but(3, 1)\notinR_{3}So ,R_{3}is not symmetric .

Transitivity:

Here ,(1, 3)\inR_{3}and(3, 3)\inR_{3}Also ,(1, 3)\inR_{3}So ,R_{3}is transitive .

Page No 1.11:

Question 5:

The following relations are defined on the set of real numbers.
(i) aRb if a – b > 0
(ii) aRb if 1 + ab > 0
(iii) aRb if |a| ≤ b

Find whether these relations are reflexive, symmetric or transitive.

Answer/Explanation

Answer:

(i)
Reflexivity: Let a be an arbitrary element of R. Then,

a \inR But a -a = 0 ≯ 0 So , this relation is not reflexive .

Symmetry:

Let(a , b)\inR \Rightarrowa -b > 0 \Rightarrow - (b -a ) > 0 \Rightarrowb -a < 0 So , the given relation is not symmetric .

Transitivity:

Let(a , b)\inR and(b , c)\inR . Then , a -b >0and b -c >0Adding the two , we get a -b +b -c >0\Rightarrowa -c >0\Rightarrow(a , c)\inR . So , the given relation is transitive .

(ii)
Reflexivity: Let a be an arbitrary element of R. Then,

a \inR\Rightarrow1+a \timesa >0i . e .1+a^{2}>0[Since , square of any number is positive]So , the given relation is reflexive .

Symmetry:

Let(a , b)\inR \Rightarrow 1 +ab > 0 \Rightarrow 1 +ba > 0 \Rightarrow (b , a) \inR So , the given relation is symmetric .

Transitivity:

Let(a , b)\inR and(b , c)\inR \Rightarrow 1 +ab > 0 and 1 +bc > 0 But1+a c ≯0\Rightarrow(a , c)\notinR So , the given relation is not transitive .

(iii)
Reflexivity: Let a be an arbitrary element of R. Then,

a \inR \Rightarrow |a| ≮a [Since ,|a|=a] So , R is not reflexive .

Symmetry:

Let(a , b)\inR \Rightarrow |a| \leqb \Rightarrow |b| ≰a for all a , b \inR\Rightarrow(b , a)\notinR So , R is not symmetric .

Transitivity:

Let(a , b)\inR and(b , c)\inR \Rightarrow |a| \leqb and |b| \leqc Multiplying the corresponding sides , we get|a||b|\leqbc\Rightarrow|a|\leqc\Rightarrow(a , c)\inR Thus , R is transitive .

Page No 1.11:

Question 6:

Check whether the relation R defined on the set A = {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Answer/Explanation

Answer:

Reflexivity:

Let a be an arbitrary element of R. Then, a =a + 1 cannot be true for all a \inA . \Rightarrow (a , a) \notinR So , R is not reflexive on A .

Symmetry:

Let(a , b)\inR \Rightarrowb =a + 1 \Rightarrow -a = -b + 1 \Rightarrowa =b - 1 Thus ,(b , a)\notinR So , R is not symmetric on A .

Transitivity:

Let(1, 2)and(2, 3)\inR \Rightarrow 2 = 1 + 1 and 3 2 + 1 is true . But 3 \neq 1 + 1 \Rightarrow (1, 3) \notinR So , R is not transitive on A .

Page No 1.11:

Question 7:

Check whether the relation R on R defined by R = {(a, b) : a ≤ b3} is reflexive, symmetric or transitive.

Answer/Explanation

Answer:

Reflexivity:

Since\frac{1}{2}>{(\frac{1}{2})}^{3},(\frac{1}{2}, \frac{1}{2})\notinR So , R is not reflexive .

Symmetry:

Since(\frac{1}{2}, 2)\inR ,\frac{1}{2}<2^{3}But2>{(\frac{1}{2})}^{3}\Rightarrow(2, \frac{1}{2})\inR So , R is not symmetric .

Transitivity:

Since(7, 3)\inR and(3, 3^{\frac{1}{3}})\inR, 7 < 3^{3} and 3 = {(3^{\frac{1}{3}})}^{3} But7>{(3^{\frac{1}{3}})}^{3}\Rightarrow(7, 3^{\frac{1}{3}})\notinR So , R is not transitive .

Page No 1.11:

Question 8:

Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.

Answer/Explanation

Answer:

Let A be a set. Then,

Identity relation IA =I_{A}is reflexive , since(a , a)\inA \foralla

The converse of it need not be necessarily true.
Consider the set A = {1, 2, 3}

Here,
Relation R = {(1, 1), (2, 2) , (3, 3), (2, 1), (1, 3)} is reflexive on A.
However, R is not an identity relation.

Page No 1.11:

Question 9:

If A = {1, 2, 3, 4} define relations on A which have properties of being
(i) reflexive, transitive but not symmetric
(ii) symmetric but neither reflexive nor transitive
(iii) reflexive, symmetric and transitive.

Answer/Explanation

Answer:

(i) The relation on A having properties of being reflexive, transitive, but not symmetric is
R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}

Relation R satisfies reflexivity and transitivity . ⇒(1, 1),(2, 2),(3, 3)\inR and(1, 1),(2, 1)\inR \Rightarrow(1, 1)\inR However ,(2, 1)\inR , but(1, 2)\notinR

(ii) The relation on A having properties of being symmetric, but neither reflexive nor transitive is
R = {(1, 2), (2, 1)}
The relation R on A is neither reflexive nor transitive, but symmetric.

(iii) The relation on A having properties of being symmetric, reflexive and transitive is
R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1)}
The relation R is an equivalence relation on A.

Page No 1.11:

Question 10:

Let R be a relation defined on the set of natural numbers N as
R = {(x, y) : x, y ∈ N, 2x + y = 41}
Find the domain and range of R. Also, verify whether R is (i) reflexive, (ii) symmetric (iii) transitive.

Answer/Explanation

Answer:

Domain of R is the values of x and range of R is the values of y that together should satisfy 2x+y = 41.
So,
Domain of R = {1, 2, 3, 4, … , 20}
Range of R = {1, 3, 5, … , 37, 39}

Reflexivity: Let x be an arbitrary element of R. Then,

x ∈R \Rightarrow 2 x +x =41cannot be true . \Rightarrow(x , x)\notinR So , R is not reflexive .

Symmetry:

Let(x , y)\inR . Then ,2x +y =41\Rightarrow2y +x =41\Rightarrow(y , x)\notinR So , R is not symmetric .

Transitivity:

Let(x , y)and(y , z)\inR \Rightarrow 2 x +y =41and2y +z =41\Rightarrow2x +z =2x +41-2y41-y -2y =41-3y\Rightarrow(x , z)\notinR Thus , R is not transitive .

Page No 1.11:

Question 11:

Is it true that every relation which is symmetric and transitive is also reflexive? Give reasons.

Answer/Explanation

Answer:

No, it is not true.

Consider a set A = {1, 2, 3} and relation R on A such that R = {(1, 2), (2, 1), (2, 3), (1, 3)}
The relation R on A is symmetric and transitive. However, it is not reflexive.

(1, 1), (2, 2) and (3, 3) \notin R

Hence, R is not reflexive.

Page No 1.11:

Question 12:

An integer m is said to be related to another integer n if m is a multiple of n.Check if the relation is symmetric, reflexive and transitive.

Answer/Explanation

Answer:

$R =\{(m , n) : m , n \inZ , m =k\} n , where k ∈N Reflexivity : Let m be an arbitrary element of R . Then , m =k m is true for k =1\Rightarrow(m , m)\inR Thus , R is reflexive .Symmetry : Let(m , n)\inR\Rightarrowm =k n for some k ∈N \ton = \frac{1}{k} m \Rightarrow (n , m) \notinR Thus , R is not symmetric . Transitivity : Let(m , n)and(n , o)\inR \Rightarrowm =kn and n =lo for some k , l \inN\Rightarrowm = (kl ) o Here , kl \inR\Rightarrow(m , o)\inR Thus , R is transitive .$

Page No 1.11:

Question 13:

Show that the relation ‘≥’ on the set R of all real numbers is reflexive and transitive but not symmetric.

Answer/Explanation

Answer:

Let R be the set such that R = {(a, b) : a, b

\inR ; a \geqb

}

Reflexivity:

Let a be an arbitrary element of R . \Rightarrowa \inR\Rightarrowa =a \Rightarrowa \geqa is true for a =a\Rightarrow(a , a)\inRHence , R is reflexive .

Symmetry:

Let(a , b)\inR \Rightarrowa \geqb is same as b \leqa , but not b \geqa Thus , (b , a) \notinR Hence , R is not symmetric .

Transitivity:

Let(a , b)and(b , c)\inR \Rightarrowa \geqb and b \geqc\Rightarrowa \geqb \geqc\Rightarrowa \geqc\Rightarrow(a , c)\inR Hence , R is transitive .

Page No 1.11:

Question 14:

Give an example of a relation which is
(i) reflexive and symmetric but not transitive;
(ii) reflexive and transitive but not symmetric;
(iii) symmetric and transitive but not reflexive;
(iv) symmetric but neither reflexive nor transitive.
(v) transitive but neither reflexive nor symmetric.

Answer/Explanation

Answer:

Suppose A be the set such that A = {1, 2, 3}

(i) Let R be the relation on A such that
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3)}
Thus,
R is reflexive and symmetric, but not transitive.

(ii) Let R be the relation on A such that
R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3)}
Clearly, the relation R on A is reflexive and transitive, but not symmetric.

(iii) Let R be the relation on A such that
R = {(1, 2), (2, 1), (1, 3), (3, 1), (2, 3)}
We see that the relation R on A is symmetric and transitive, but not reflexive.

(iv) Let R be the relation on A such that
R = {(1, 2), (2, 1), (1, 3), (3, 1)}
The relation R on A is symmetric, but neither reflexive nor transitive.

(v) Let R be the relation on A such that
R = {(1, 2), (2, 3), (1, 3)}
The relation R on A is transitive, but neither symmetric nor reflexive.

Page No 1.11:

Question 15:

Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, add a minimum number of ordered pairs so that the enlarged relation is symmeteric, transitive and reflexive.

Answer/Explanation

Answer:

We have,

R = {(1, 2), (2, 3)}

R can be a transitive only when the elements (1, 3) is added

R can be a reflexive only when the elements (1, 1), (2, 2), (3, 3) are added

R can be a symmetric only when the elements (2, 1), (3, 1) and (3, 2) are added

So, the required enlarged relation, R’ = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} = A

\times

Page No 1.11:

Question 16:

Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. What minimum number of ordered pairs may be added to R so that it may become a transitive relation on A.

Answer/Explanation

Answer:

We have,

A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)}

To make R a transitive relation on A, (1, 3) must be added to it.

So, the minimum number of ordered pairs that may be added to R to make it a transitive relation is 1.

Page No 1.11:

Question 17:

Let A = {a, b, c} and the relation R be defined on A as follows: R = {(a, a), (b, c), (a, b)}. Then, write minimum number of ordered pairs to be added in R to make it reflexive and transitive. [NCERT EXEMPLAR]

Answer/Explanation

Answer:

We have,

A = {a, b, c} and R = {(a, a), (b, c), (a, b)}

R can be a reflexive relation only when elements (b, b) and (c, c) are added to it

R can be a transitive relation only when the element (a, c) is added to it

So, the minmum number of ordered pairs to be added in R is 3.

Page No 1.11:

Question 18:

Each of the following defines a relation on N:

(i) x > y, x, y

\in

N
(ii) x + y = 10, x, y

\in

N
(iii) xy is square of an integer, x, y

\in

N
(iv) x + 4y = 10, x, y

\in

Determine which of the above relations are reflexive, symmetric and transitive. [NCERT EXEMPLAR]

Answer/Explanation

Answer:

(i) We have,

R = {(x, y) : x > y, x, y

\in

As , x =x ∀x ∈N\Rightarrow(x ,x)\notinR So , R is nota reflexive relationLet(x ,y)\inR\Rightarrowx >y but y <x \Rightarrow (y ,x) \notinR So , R is not a symmeteric relation Let(x ,y)\inR and(y ,z)\inR \Rightarrowx >y and y >z\Rightarrowx >z\Rightarrow(x ,z)\inR So , R isa transitive relation

(ii) We have,

R = {(x, y) : x + y = 10, x, y

\in

R =\{(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)\}As ,(1, 1)\notinR So , R is nota reflexive relationLet(x ,y)\inR\Rightarrowx +y =10\Rightarrowy +x =10\Rightarrow(y ,x)\inR So , R isa symmeteric relationAs ,(1, 9)\inR and(9, 1)\inR but(1, 1)\notinR So , R is nota transitive relation

(iii) We have,

R = {(x, y) : xy is square of an integer, x, y

\in

As , x ×x =x^{2}, which isa square of an integer x\Rightarrow(x ,x)\inR So , R isa reflexive relationLet(x ,y)\inR\Rightarrowx y is square of an integer \Rightarrowyx is also a square of an integer \Rightarrow (y ,x) \inR So , R is a symmeteric relation Let(x ,y)\inR and(y ,z)\inR \Rightarrowxy is square of an integer and yz is also a square of an interger \Rightarrowxz must be a square of an integer \Rightarrow (x ,z) \inR So , R is a transitive relation

(iv) We have,

R = {(x, y) : x + 4y = 10, x, y

\in

R =\{(2, 4), (6, 1)\}As ,(2, 2)\notinR So , R is nota reflexive relationAs ,(2, 4)\inR but(4, 2)\notinR So , R is nota symmeteric relationAs ,(2, 4)\inR but4is not related to any natural number So , R isa transitive relation

Page No 1.26:

Question 1:

Show that the relation R defined by R = {(a, b) : a – b is divisible by 3; a, b ∈ Z} is an equivalence relation.

Answer/Explanation

Answer:

We observe the following relations of relation R.

Reflexivity:

Let a be an arbitrary element of R . Then , a -a = 0 = 0 \times 3 \Rightarrowa -a is divisible by 3 \Rightarrow (a , a) \inR for all a \inZ So , R is reflexive on Z .

Symmetry:

Let(a , b)\inR \Rightarrowa -b is divisible by 3 \Rightarrowa -b 3 p for some p ∈Z \Rightarrowb -a = 3 (-p) Here , -p \inZ\Rightarrowb -a is divisible by3\Rightarrow(b , a)\inR for all a , b \inZ So,R is symmetric on Z .

Transitivity:

Let(a , b)and(b , c)\inR \Rightarrowa -b and b -c are divisible by 3 \Rightarrowa -b = 3 p for some p ∈Z and b -c =3q for some q ∈Z Adding the above two , we geta -b +b -c =3p +3q\Rightarrowa -c =3(p +q)Here , p +q ∈Z\Rightarrowa -c is divisible by3\Rightarrow(a , c)\inR for all a , c \inZ So , R is transitive on Z .

Hence, R is an equivalence relation on Z.

Page No 1.26:

Question 2:

Show that the relation R on the set Z of integers, given by
R = {(a, b) : 2 divides a – b}, is an equivalence relation.

Answer/Explanation

Answer:

We observe the following properties of relation R.

Reflexivity:

Let a be an arbitrary element of the set Z . Then , a ∈R \Rightarrowa -a = 0 = 0 \times 2 \Rightarrow 2 divides a -a\Rightarrow(a , a)\inR for all a \inZ So , R is reflexive on Z .

Symmetry:

Let(a , b)\inR \Rightarrow 2 divides a -b\Rightarrow\frac{a -b}{2}=p for some p \inZ\Rightarrow\frac{b -a}{2}= -p Here , -p \inZ\Rightarrow2divides b -a\Rightarrow(b , a)\inR for all a , b \inZ So , R is symmetric on Z .

Transitivity:

Let(a , b)and(b , c)\inR \Rightarrow 2 divides a -b and 2 divides b -c\Rightarrow\frac{a -b}{2}=p and\frac{b -c}{2}=q for some p , q \inZ Adding the above two , we get\frac{a -b}{2}+\frac{b -c}{2}=p +q\Rightarrow\frac{a -c}{2}=p +q Here , p +q \inZ\Rightarrow2divides a -c\Rightarrow(a , c)\inR for all a , c \inZ So , R is transitive on Z .

Hence, R is an equivalence relation on Z.

Page No 1.26:

Question 3:

Prove that the relation R on Z defined by
(a, b) ∈ R ⇔ a − b is divisible by 5
is an equivalence relation on Z.

Answer/Explanation

Answer:

We observe the following properties of relation R.

Reflexivity:

Let a be an arbitrary element of R . Then , ⇒a -a =0=0\times5\Rightarrowa -a is divisible by5\Rightarrow(a , a)\inR for all a \inZ So , R is reflexive on Z .

Symmetry:

Let(a , b)\inR \Rightarrowa -b is divisible by 5 \Rightarrowa -b = 5 p for some p ∈Z \Rightarrowb -a = 5 (-p) Here , -p \inZ [Since p \inZ ] \Rightarrowb -a is divisible by 5 \Rightarrow (b , a) \inR for all a , b \inZ So , R is symmetric on Z .

Transitivity:

Let(a , b)and(b , c)\inR \Rightarrowa -b is divisible by 5 \Rightarrowa -b = 5 p for some Z Also , b -c is divisible by5\Rightarrowb -c =5q for some Z Adding the above two , we get a -b +b -c =5p +5q\Rightarrowa -c =5(p +q ) \Rightarrowa -c is divisible by5Here , p +q ∈Z\Rightarrow(a , c)\inR for all a , c \inZ So , R is transitive on Z .

Hence, R is an equivalence relation on Z.

Page No 1.26:

Question 4:

Let n be a fixed positive integer. Define a relation R on Z as follows:
(a, b) ∈ R ⇔ a − b is divisible by n.
Show that R is an equivalence relation on Z.

Answer/Explanation

Answer:

We observe the following properties of R. Then,
Reflexivity:

Let a ∈N Here , a -a =0=0\times n \Rightarrowa -a is divisible by n\Rightarrow(a , a)\inR\Rightarrow(a , a)\inR for all a \inZ So , R is reflexive on Z .

Symmetry:

Let(a , b)\inR Here , a -bis divisible byn\Rightarrowa -b =np for some p \inZ\Rightarrowb -a =n(-p)\Rightarrowb -a is divisible by n [p \inZ \Rightarrow -p \inZ ] \Rightarrow(b , a)\inR So , R is symmetric on Z .

Transitivity:

Let(a , b)and(b , c)\inR Here , a -b is divisible by n and b -c is divisible by n . \Rightarrowa -b =n p for some p ∈Z and b -c =n q for some q ∈Z Adding the above two , we get a -b +b -c =n p +n q \Rightarrowa -c =n (p +q ) Here , p +q \inZ\Rightarrow(a , c)\inR for all a , c \inZ So , R is transitive on Z .

Hence, R is an equivalence relation on Z.

Page No 1.26:

Question 5:

Let Z be the set of integers. Show that the relation
R = {(a, b) : a, b ∈ Z and a + b is even}
is an equivalence relation on Z.

Answer/Explanation

Answer:

We observe the following properties of R.

Reflexivity:

Let a be an arbitrary element of Z . Then , a ∈R Clearly , a +a =2a is even for all a ∈Z . ⇒(a , a)\inR for all a \inZ So , R is reflexive on Z .

Symmetry:

Let(a , b)\inR \Rightarrowa +b is even\Rightarrowb +a is even\Rightarrow(b , a)\inR for all a , b \inZ So , R is symmetric on Z .

Transitivity:

Let(a , b)and(b , c)\inR \Rightarrowa +b and b +c are even Now , let a +b = 2 x for some x ∈Z and b +c =2y for some y ∈Z Adding the above two , we geta +2b +c =2x +2y\Rightarrowa +c =2(x +y -b ) , which is even for all x , y , b \inZ Thus ,(a , c)\inR So , R is transitive on Z .

Hence, R is an equivalence relation on Z.

Page No 1.26:

Question 6:

m is said to be related to n if m and n are integers and m − n is divisible by 13. Does this define an equivalence relation?

Answer/Explanation

Answer:

We observe the following properties of relation R.

Let R = {(m , n): m , n \inZ : m -nisdivisibleby13} Relexivity : Let m be an arbitrary element of Z . Then , m \inR\Rightarrowm -m =0=0\times13\Rightarrowm -m is divisible by13\Rightarrow(m , m)is reflexive on Z .Symmetry : Let(m , n)\inR . Then , m -n is divisible by13\Rightarrowm -n =13p Here , p ∈Z\Rightarrown -m =13(-p)Here , -p \inZ\Rightarrown -m is divisible by13\Rightarrow(n , m)\inR for all m , n \inZ So , R is symmetric on Z .Transitivity : Let(m , n)and(n , o)\inR\Rightarrowm -n and n -o are divisible by13\Rightarrowm -n =13p and n -o =13q for some p , q ∈ZAdding the above two , we getm -n +n -o =13p +13q\Rightarrowm -o =13(p +q)Here , p +q ∈Z\Rightarrowm -o is divisible by13\Rightarrow(m , o)\inR for all m , o \inZ So , R is transitive on Z .

Hence, R is an equivalence relation on Z.

Page No 1.26:

Question 7:

Let R be a relation on the set A of ordered pair of integers defined by (x, y) R (u, v) if xv = yu. Show that R is an equivalence relation.

Answer/Explanation

Answer:

We observe the following properties of R.

Reflexivity : Let(a , b)be an arbitrary element of the setA . Then ,(a , b)\inA\Rightarrowa b =b a ⇒(a , b)R(a , b)Thus , R is reflexive on A .Symmetry : Let(x , y)and(u , v)\inA such that(x , y)R(u , v). Then , xv =yu\Rightarrowvx =uy\Rightarrowu y =v x \Rightarrow (u , v) R (x , y) So , R is symmetric on A .Transitivity : Let(x , y),(u , v)and(p , q)\inR such that(x , y)R(u , v)and(u , v)R(p , q). \Rightarrowxv =yu and uq =vpMultiplying the corresponding sides , we get xv × uq =yu × vp\Rightarrowx q =y p \Rightarrow (x , y) R (p , q) So , R is transitive on A .

Hence, R is an equivalence relation on A.

Page No 1.26:

Question 8:

Show that the relation R on the set A = {x ∈ Z ; 0 ≤ x ≤ 12}, given by R = {(a, b) : a = b}, is an equivalence relation. Find the set of all elements related to 1.

Answer/Explanation

Answer:

We observe the following properties of R.

Reflexivity: Let a be an arbitrary element of A. Then,

a ∈R \Rightarrowa =a [Since , every element is equal to itself] \Rightarrow (a , a) \inR for all a \inA So , R is reflexive on A . Symmetry : Let(a , b)\inR \Rightarrowa b\Rightarrowb =a\Rightarrow(b , a)\inR for all a , b \inA So , R is symmetric on A .Transitivity : Let(a , b)and(b , c)\inR\Rightarrowa =b and b =c\Rightarrowa =b c\Rightarrowa =c\Rightarrow(a , c)\inR So , R is transitive on A .

Hence, R is an equivalence relation on A.

The set of all elements related to 1 is {1}.

Page No 1.27:

Question 9:

Let L be the set of all lines in XY-plane and R be the relation in L defined as R = {L1, L2) : L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Answer/Explanation

Answer:

We observe the following properties of R.

Reflexivity : LetL_{1}be an arbitrary element of the setL . Then ,L_{1}\inL\RightarrowL_{1}is parallel toL_{1}[Every line is parallel to itself]\Rightarrow(L_{1}, L_{1})\inR for allL_{1}\inL So , R is reflexive on L .Symmetry : Let(L_{1}, L_{2})\inR\RightarrowL_{1}is parallel toL_{2}\RightarrowL_{2}is parallel toL_{1}\Rightarrow(L_{2}, L_{1})\inR for allL_{1}andL_{2}\inL So , R is symmetric on L .Transitivity : Let(L_{1}, L_{2})and(L_{2}, L_{3})\inR\RightarrowL_{1}is parallel toL_{2}andL_{2}is parallel toL_{3}\RightarrowL_{1},L_{2}andL_{3}are all parallel to each other \Rightarrow L_{1} is parallel to L_{3} \Rightarrow (L_{1}, L_{3}) \inR So , R is transitive on L .

Hence, R is an equivalence relation on L.

Set of all the lines related to y = 2x+4
= L’ = {(x, y) : y = 2x+c, where c

\in

Page No 1.27:

Question 10:

Show that the relation R, defined on the set A of all polygons as
R = {(P1, P2) : P1 and P2 have same number of sides},
is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Answer/Explanation

Answer:

We observe the following properties on R.

Reflexivity : LetP_{1}be an arbitrary element of A . Then , polygonP_{1}andP_{1}have the same number of sides , since they are one and the same . \Rightarrow(P_{1}, P_{1})\inR for allP_{1}\inA So , R is reflexive on A .Symmetry : Let(P_{1}, P_{2})\inR\RightarrowP_{1}andP_{2}have the same number of sides . \RightarrowP_{2}andP_{1}have the same number of sides . \Rightarrow(P_{2}, P_{1})\inR for allP_{1},P_{2}\inA So , R is symmetric on A .Transitivity : Let(P_{1}, P_{2}),(P_{2}, P_{3})\inR\RightarrowP_{1}andP_{2}have the same number of sides andP_{2}andP_{3}have the same number of sides . \RightarrowP_{1},P_{2}andP_{3}have the same number of sides . \RightarrowP_{1}andP_{3}have the same number of sides . \Rightarrow(P_{1}, P_{3})\inR for allP_{1},P_{3}A So , R is transitive on A .

Hence, R is an equivalence relation on the set A.

Also, the set of all the triangles

\in

A is related to the right angle triangle T with the sides 3, 4, 5.

Page No 1.27:

Question 11:

Let O be the origin. We define a relation between two points P and Q in a plane if OP = OQ. Show that the relation, so defined is an equivalence relation.

Answer/Explanation

Answer:

Let A be the set of all points in a plane such that

A = {P : P isa point in the plane } Let R be the relation such that R =\{(P , Q) : P , Q \inA and O\}P =OQ , where O is the origin

We observe the following properties of R.

Reflexivity: Let P be an arbitrary element of R.
The distance of a point P will remain the same from the origin.
So, OP = OP

\Rightarrow (P , P) \inR So , R is reflexive on A . Symmetry : Let(P , Q)\inR \RightarrowOP =OQ\RightarrowOQ =OP\Rightarrow(Q , P)\inR So , R is symmetric on A .Transitivity : Let(P , Q),(Q , R)\inR\RightarrowOP =OQ and OQ=OR\RightarrowOP =OQ =OR\RightarrowOP =OR\Rightarrow(P , R)\inR So , R is transitive on A .

Hence, R is an equivalence relation on A.

Page No 1.27:

Question 12:

Let R be the relation defined on the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Show that R is an equivalence relation. Further, show that all the elements of the subset {1, 3, 5, 7} are related to each other and all the elements of the subset {2, 4, 6} are related to each other, but no element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}.

Answer/Explanation

Answer:

We observe the following properties of R.

Reflexivity:

Let a be an arbitrary element of R . Then, a ∈R \Rightarrow (a , a) \inR for all a \inA So , R is reflexive on A .Symmetry: Let(a , b)\inR \RightarrowBoth a and b are either even or odd . \RightarrowBoth b anda are either even or odd . \Rightarrow(b , a)\inR for all a , b \inA So , R is symmetric on A .Transitivity : Let (a , b) and (b , c) \inR\RightarrowBoth a and b are either even or odd and both b and c are either even or odd .\Rightarrowa , b and c are either even or odd .\Rightarrowa and c both are either even or odd . \Rightarrow(a , c)\inR for all a , c \inA So , R is transitive on A .

Thus, R is an equivalence relation on A.

We observe that all the elements of the subset {1, 3, 5, 7} are odd. Thus, they are related to each other.
This is because the relation R on A is an equivalence relation.

Similarly, the elements of the subset {2, 4, 6} are even. Thus, they are related to each other because every element is even.

Hence proved.

Page No 1.27:

Question 13:

Let S be a relation on the set R of all real numbers defined by
S = {(a, b) ∈ R × R : a2 + b2 = 1}
Prove that S is not an equivalence relation on R.

Answer/Explanation

Answer:

We observe the following properties of S.

Reflexivity : Let a be an arbitrary element of R . Then , a ∈R \Rightarrow a^{2} + a^{2} \neq 1 \foralla \inR\Rightarrow(a , a)\notinS So , S is not reflexive on R .Symmetry : Let (a , b) \inR\Rightarrowa^{2}+b^{2}=1\Rightarrowb^{2}+a^{2}=1\Rightarrow(b , a)\inS for all a , b \inR So , S is symmetric on R .Transitivity : Let (a , b) and (b , c) \inS\Rightarrowa^{2}+b^{2}=1andb^{2}+c^{2}=1Adding the above two , we geta^{2}+c^{2}=2-2b^{2}\neq1for all a , b , c \inR So , S is not transitive on R .

Hence, S is not an equivalence relation on R.

Page No 1.27:

Question 14:

Let Z be the set of all integers and Z0 be the set of all non-zero integers. Let a relation R on Z × Z0 be defined as
(a, b) R (c, d) ⇔ ad = bc for all (a, b), (c, d) ∈ Z × Z0,
Prove that R is an equivalence relation on Z × Z0.

Answer/Explanation

Answer:

We observe the following properties of R.

Reflexivity:

Let(a , b)be an arbitrary element ofZ ×Z_{0}. Then ,(a , b)\inZ ×Z_{0}\Rightarrowa , b \inZ ,Z_{0}\Rightarrowab =ba\Rightarrow(a , b)\inR for all(a , b)\inZ \timesZ_{0}So , R is reflexive on Z ×Z_{0}.

Symmetry:

Let(a , b),(c , d)\inZ \timesZ_{0}such that(a , b)R(c , d). Then ,(a , b)R(c , d)\Rightarrowa d =b c \Rightarrowcb =da\Rightarrow(c , d)R(a , b)Thus , (a , b) R (c , d) \Rightarrow (c , d) R (a , b) for all (a , b), (c , d) \inZ \times Z_{0} So , R is symmetric on Z ×Z_{0}.

Transitivity:

Let(a , b),(c , d),(e , f)\inN \timesN_{0}such that(a , b)R(c , d)and(c , d)R(e , f). Then ,\begin{array}{r} (a , b) R (c , d) \Rightarrowa \end{array}\} d =b c (c , d) R (e , f) \Rightarrowcf =de \Rightarrow (a d) (c f) = (b c) (d e) \Rightarrowaf =be\Rightarrow(a , b)R(e , f)Thus , (a , b) R (c , d) and (c , d) R (e , f) \Rightarrow (a , b) R (e , f) \Rightarrow (a , b) R (e , f) for all values (a , b), (c , d), (e , f) \inN \times N_{0} So , R is transitive on N ×N_{0}.

Page No 1.27:

Question 15:

If R and S are relations on a set A, then prove that
(i) R and S are symmetric ⇒ R ∩ S and R ∪ S are symmetric
(ii) R is reflexive and S is any relation ⇒ R ∪ S is reflexive.

Answer/Explanation

Answer:

(i) R and S are symmetric relations on the set A.

\RightarrowR \subsetA \timesA and S \subsetA \timesA\RightarrowR \capS \subsetA \timesA Thus , R \capS isa relation on A . Let a , b ∈A such that(a , b)\inR \capS . Then ,(a , b)\inR \capS \Rightarrow (a , b) \inR and (a , b) \inS\Rightarrow(b , a)\inR and(b , a)\inS[Since R and S are symmetric]\Rightarrow(b , a)\inR \capS Thus ,(a , b)\inR \capS\Rightarrow(b , a)\inR \capS for all a , b \inA So , R \capS is symmetric onA .

Also,

Let a , b ∈A such that(a , b)\inR \cupS \Rightarrow (a , b) \inR or (a , b) \inS\Rightarrow(b , a)\inR or(b , a)\inS[Since R and S are symmetric]\Rightarrow(b , a)\inR \cupS So , R \cupS is symmetric on A .

(ii) R is reflexive and S is any relation.

Suppose a ∈A . Then ,(a , a)\inR[Since R is reflexive]\Rightarrow(a , a)\inR \cupS \RightarrowR \cupS is reflexive on A .

Page No 1.27:

Question 16:

If R and S are transitive relations on a set A, then prove that R ∪ S may not be a transitive relation on A.

Answer/Explanation

Answer:

Let A = {a, b, c} and R and S be two relations on A, given by

R = {(a, a), (a, b), (b, a), (b, b)} and
S = {(b, b), (b, c), (c, b), (c, c)}

Here, the relations R and S are transitive on A.

(a , b) \inR \cupS and (b , c) \inR \cupS But (a , c) \notinR \cupS

Hence, R

\cup

S is not a transitive relation on A.

Page No 1.27:

Question 17:

Let C be the set of all complex numbers and C0 be the set of all no-zero complex numbers. Let a relation R on C0 be defined as

z_{1}

z_{2} \Leftrightarrow \frac{z_{1} - z_{2}}{z_{1} + z_{2}}

is real for all

z_{1}, z_{2} \in

C0.

Show that R is an equivalence relation.

Answer/Explanation

Answer:

(i) Test for reflexivity:

Since,

\frac{z_{1} - z_{1}}{z_{1} + z_{1}} = 0

, which is a real number.

So,

(z_{1}, z_{1}) \inR

Hence, R is relexive relation.

(ii) Test for symmetric:

Let

(z_{1}, z_{2}) \inR

Then,

\frac{z_{1} - z_{2}}{z_{1} + z_{2}} =x

, where x is real

\Rightarrow - (\frac{z_{1} - z_{2}}{z_{1} + z_{2}}) = -x\Rightarrow(\frac{z_{2} - z_{1}}{z_{2} + z_{1}})= -x , is alsoa real number

So,

(z_{2}, z_{1}) \inR

Hence, R is symmetric relation.

(iii) Test for transivity:

Let

(z_{1}, z_{2}) \inR and (z_{2}, z_{3}) \in R

Then,

\frac{z_{1} - z_{2}}{z_{1} + z_{2}} =x , where x is a real number . ⇒z_{1}-z_{2}=x z_{1} +xz_{2}\Rightarrowz_{1}-xz_{1}=z_{2}+xz_{2}\Rightarrowz_{1}(1 -x)=z_{2}(1 +x)\Rightarrow\frac{z_{1}}{z_{2}}=\frac{(1 +x)}{(1 -x)}. . .(1)

Also,

\frac{z_{2} - z_{3}}{z_{2} + z_{3}} =y , where y is a real number . ⇒z_{2}-z_{3}=y z_{2} +Yz_{3}\Rightarrowz_{2}-yz_{2}=z_{3}+yz_{3}\Rightarrowz_{2}(1 -y)=z_{3}(1 +y)\Rightarrow\frac{z_{2}}{z_{3}}=\frac{(1 +y)}{(1 -y)}. . .(2)

Dividing (1) and (2), we get

\frac{z_{1}}{z_{3}} = (\frac{1 +x}{1 -x}) \times (\frac{1 -y}{1 +y}) =z , wherez isa real number . \Rightarrow \frac{z_{1} - z_{3}}{z_{1} + z_{3}} = \frac{z –1}{z +1}, which is real\Rightarrow(z_{1}, z_{3})\inR

Hence, R is transitive relation.

From (i), (ii), and (iii),

R is an equivalenve relation.

Page No 1.29:

Question 1:

Let R be a relation on the set N given by
R = {(a, b) : a = b − 2, b > 6}. Then,
(a) (2, 4) ∈ R
(b) (3, 8) ∈ R
(c) (6, 8) ∈ R
(d) (8, 7) ∈ R

Answer/Explanation

Answer:

(6, 8) \inR Then , a =b –2\Rightarrow6=8-2and b =8>6Hence ,(6, 8)\inR

Page No 1.29:

Question 2:

If a relation R is defined on the set Z of integers as follows:
(a, b) ∈ R ⇔ a2 + b2 = 25. Then, domain (R) is
(a) {3, 4, 5}
(b) {0, 3, 4, 5}
(c) {0, ± 3, ± 4, ± 5}
(d) none of these

Answer/Explanation

Answer:

R = \{(a , b) : a^{2} + b^{2} = 25, a , b \inZ\} \Rightarrowa \in \{- 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5\} and b \in \{- 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5\}

So , domain (R ) =\{0, \pm 3, \pm 4, \pm 5\}

Page No 1.30:

Question 3:

R is a relation on the set Z of integers and it is given by
(x, y) ∈ R ⇔ | x − y | ≤ 1. Then, R is
(a) reflexive and transitive
(b) reflexive and symmetric
(c) symmetric and transitive
(d) an equivalence relation

Answer/Explanation

Answer:

(b) reflexive and symmetric

Reflexivity : Let x ∈R . Then , x -x =0<1\Rightarrow|x -x|\leq1\Rightarrow(x , x)\inR for all x \inZ So , R is reflexive on Z .Symmetry : Let(x , y)\inR . Then ,|x -y|\leq0\Rightarrow|- (y -x )|\leq1\Rightarrow|y -x|\leq1[Since|x -y|=|y -x|]\Rightarrow(y , x)\inR for all x , y \inZ So , Ris symmetri c on Z . Transitivity : Let(x , y)\inR and(y , z)\inR . Then ,|x -y|\leq1and|y -z|\leq1\RightarrowIt is not always true that|x -y|\leq1. \Rightarrow(x , z)\notinR So , R is not transitive on Z .

Page No 1.30:

Question 4:

The relation R defined on the set A = {1, 2, 3, 4, 5} by
R = {(a, b) : | a2 − b2 | < 16} is given by
(a) {(1, 1), (2, 1), (3, 1), (4, 1), (2, 3)}
(b) {(2, 2), (3, 2), (4, 2), (2, 4)}
(c) {(3, 3), (4, 3), (5, 4), (3, 4)}
(d) none of these

Answer/Explanation

Answer:

(d) none of these

R is given by {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (1, 3), (3, 1), (1, 4), (4, 1) ,(2, 4), (4, 2)}, which is not mentioned in (a), (b) or (c).

Page No 1.30:

Question 5:

Let R be the relation over the set of all straight lines in a plane such that l1 R l2 ⇔ l 1⊥ l2. Then, R is
(a) symmetric
(b) reflexive
(c) transitive
(d) an equivalence relation

Answer/Explanation

Answer:

(a) symmetric

A = Set of all straight lines in the plane

R =\{(l_{1,} l_{2}) : l_{1}, l_{2} \inA : l_{1} ⊥ l_{2}\}Reflexivity :l_{1}is not ⊥l_{1}\Rightarrow(l_{1}, l_{1})\notinR So , R is not reflexive on A .Symmetry : Let(l_{1}, l_{2})\inR\Rightarrowl_{1}⊥l_{2}\Rightarrowl_{2}⊥l_{1}\Rightarrow(l_{2}, l_{1})\inR So , R is symmetric on A .Transitivity : Let(l_{1}, l_{2})\inR ,(l_{2}, l_{3})\inR\Rightarrowl_{1}⊥l_{2}andl_{2}⊥l_{3}Butl_{1}is not ⊥l_{3}\Rightarrow(l_{1}, l_{3})\notinR So , R is not transitive on A .

Page No 1.30:

Question 6:

If A = {a, b, c}, then the relation R = {(b, c)} on A is
(a) reflexive only
(b) symmetric only
(c) transitive only
(d) reflexive and transitive only

Answer/Explanation

Answer:

The relation R = {(b,c)} is neither reflexive nor symmetric because every element of A is not related to itself. Also, the ordered pair of R obtained by interchanging its elements is not contained in R.

We observe that R is transitive on A because there is only one pair.

Page No 1.30:

Question 7:

Let A = {2, 3, 4, 5, …, 17, 18}. Let ‘≃’ be the equivalence relation on A × A, cartesian product of A with itself, defined by (a, b) ≃ (c, d) if ad = bc. Then, the number of ordered pairs of the equivalence class of (3, 2) is
(a) 4
(b) 5
(c) 6
(d) 7

Answer/Explanation

Answer:

The ordered pairs of the equivalence class of (3, 2) are {(3, 2), (6, 4), (9, 6), (12, 8), (15, 10), (18, 12)}.
We observe that these are 6 pairs.

Page No 1.30:

Question 8:

Let A = {1, 2, 3}. Then, the number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
(a) 1
(b) 2
(c) 3
(d) 4

Answer/Explanation

Answer:

(a) 1

The required relation is R.
R = {(1, 2), (1, 3), (1, 1), (2, 2), (3, 3), (2, 1), (3, 1)}

Hence, there is only 1 such relation that is reflexive and symmetric, but not transitive.

Page No 1.30:

Question 9:

The relation ‘R’ in N × N such that
(a, b) R (c, d) ⇔ a + d = b + c is
(a) reflexive but not symmetric
(b) reflexive and transitive but not symmetric
(c) an equivalence relation
(d) none of the these

Answer/Explanation

Answer:

We observe the following properties of relation R.

Reflexivity : Let (a , b ) ∈N ×N \Rightarrowa , b \inN\Rightarrowa +b =b +a\Rightarrow(a , b)\inR So , R is reflexive on N \timesN .Symmetry : Let(a , b),(c , d)\inN \times N such that(a , b)R(c , d)\Rightarrowa +d =b +c \Rightarrowd +a =c +b\Rightarrow(d , c),(b , a)\inR So , R is symmetric onN \times N .Transitivity : Let(a , b),(c , d),(e , f)\inN \timesN such that(a , b)R(c , d)and(c , d)R(e , f)\Rightarrowa +d =b +c and c +f =d +e\Rightarrowa +d +c +f =b +c +d +e \Rightarrowa +f =b +e\Rightarrow(a , b)R(e , f)So , R is transitive on N \timesN .

Hence, R is an equivalence relation on N.

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Question 10:

If A = {1, 2, 3}, B = {1, 4, 6, 9} and R is a relation from A to B defined by ‘x is greater than y’. The range of R is
(a) {1, 4, 6, 9}
(b) {4, 6, 9}
(c) {1}
(d) none of these

Answer/Explanation

Answer:

Here,

R =\{(x , y) : x \inA and y \inB : x > y\} \RightarrowR = \{(2, 1), (3, 1)\}

Thus,
Range of R = {1}

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Question 11:

A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : x R y ⇔ x is relatively prime to y. Then, domain of R is
(a) {2, 3, 5}
(b) {3, 5}
(c) {2, 3, 4}
(d) {2, 3, 4, 5}

Answer/Explanation

Answer:

(d) {2, 3, 4, 5}

The relation R is defined as

R = \{(x , y) : x \in \{2, 3, 4, 5\}, y \in \{3, 6, 7, 10\} : x is relatively prime to y\} \RightarrowR = \{(2, 3), (2, 7), (3, 7), (3, 10), (4, 7), (5, 3), (5, 7)\}

Hence, the domain of R includes all the values of x, i.e. {2, 3, 4, 5}.

Page No 1.30:

Question 12:

A relation ϕ from C to R is defined by x ϕ y ⇔ | x | = y. Which one is correct?
(a) (2 + 3 i) ϕ 13
(b) 3 ϕ (−3)
(c) (1 + i) ϕ 2
(d) i ϕ 1

Answer/Explanation

Answer:

(d) i ϕ 1

∵ |2 + 3 i| = \sqrt{13} \neq 13 |3| \neq - 3 |1 +i| = \sqrt{2} \neq 2 and|i|=1So ,(i ,1)\inϕ

Page No 1.30:

Question 13:

Let R be a relation on N defined by x + 2y = 8. The domain of R is
(a) {2, 4, 8}
(b) {2, 4, 6, 8}
(c) {2, 4, 6}
(d) {1, 2, 3, 4}

Answer/Explanation

Answer:

The relation R is defined as

R = \{(x , y) : x , y \inN and x + 2 y =8\} \RightarrowR = \{(x , y) : x , y \inN and y = \frac{(8 -x)}{2}\}

Domain of R is all values of x

\in

N satisfying the relation R. Also, there are only three values of x that result in y, which is a natural number. These are {2, 6, 4}.

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Question 14:

R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x − 3. Then, R−1 is
(a) {(8, 11), (10, 13)}
(b) {(11, 8), (13, 10)}
(c) {(10, 13), (8, 11)}
(d) none of these

Answer/Explanation

Answer:

(a) {(8, 11), (10, 13)}

The relation R is defined by

R = \{(x , y) : x \in \{11, 12, 13\}, y \in \{8, 10, 12\} : y = x - 3\} \RightarrowR = \{(11, 8), (13, 10)\} So ,R^{- 1}=\{(8, 11), (10, 13)\}

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Question 15:

Let R = {(a, a), (b, b), (c, c), (a, b)} be a relation on set A = a, b, c. Then, R is
(a) identify relation
(b) reflexive
(c) symmetric
(d) antisymmetric

Answer/Explanation

Answer:

(b) reflexive

Reflexivity : Since(a , a)\inR \forall a \inA , R is reflexive on A .Symmetry : Since(a , b)\inR but(b , a)\notinR , R is not symmetric on A .\RightarrowR is not antisymmetric on A . Also , R is not an identity relation on A .

Page No 1.30:

Question 16:

Let A = {1, 2, 3} and B = {(1, 2), (2, 3), (1, 3)} be a relation on A. Then, R is
(a) neither reflexive nor transitive
(b) neither symmetric nor transitive
(c) transitive
(d) none of these

Answer/Explanation

Answer:

Reflexivity : Since (1,1) \notinB , B is not reflexive on A .Symmetry : Since(1, 2)\inB but(2, 1)\notinB , B is not symmetric on A .Transitivity : Since(1, 2)\inB ,(2, 3)\inB and(1, 3)\inB , B is transitive on A .

Page No 1.30:

Question 17:

If R is the largest equivalence relation on a set A and S is any relation on A, then
(a) R ⊂ S
(b) S ⊂ R
(c) R = S
(d) none of these

Answer/Explanation

Answer:

(b) S ⊂ R

Since R is the largest equivalence relation on set A,

R ⊆ A × A

Since S is any relation on A,

S ⊂ A × A

So, S ⊂ R

Page No 1.30:

Question 18:

If R is a relation on the set A = {1, 2, 3, 4, 5, 6, 7, 8, 9} given by x R y ⇔ y = 3 x, then R =
(a) {(3, 1), (6, 2), (8, 2), (9, 3)}
(b) {(3, 1), (6, 2), (9, 3)}
(c) {(3, 1), (2, 6), (3, 9)}
(d) none of these

Answer/Explanation

Answer:

(d) none of these

The relation R is defined as

R = \{(x , y) : x , y \inA : y = 3 x\} \RightarrowR = \{(1, 3), (2, 6), (3, 9)\}

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Question 19:

If R is a relation on the set A = {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3)}, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) all the three options

Answer/Explanation

Answer:

(d) all the three options

R =\{(a , b) : a =b and a , b \inA\} Reflexivity : Let a ∈A . Then , a =a \Rightarrow (a , a) \inR for all a \inA So , R is reflexive on A . Symmetry : Let a , b ∈A such that(a , b)\inR . Then ,(a , b)\inR \Rightarrowa =b\Rightarrowb =a\Rightarrow(b , a)\inR for all a \inA So , R is symmetric on A .Transitivity : Let a , b , c \inA such that(a , b)\inR and(b , c)\inR . Then ,(a , b)\inR \Rightarrowa =b and(b , c)\inR \Rightarrowb =c\Rightarrowa =c\Rightarrow(a , c)\inR for all a \inA So , R is transitive on A .

Hence, R is an equivalence relation on A.

Page No 1.31:

Question 20:

If A = {a, b, c, d}, then a relation R = {(a, b), (b, a), (a, a)} on A is
(a) symmetric and transitive only
(b) reflexive and transitive only
(c) symmetric only
(d) transitive only

Answer/Explanation

Answer:

(a) symmetric and transitive only

Reflexivity : Since(b , b)\notinR , R is not reflexive on A .Symmetry : Since(a , b)\inR and(b , a)\inR , R is symmetric on A .Transitivity : Since(a , b)\inR ,(b , a)\inR and(a , a)\inR , R is transitive on A .

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Question 21:

If A = {1, 2, 3}, then a relation R = {(2, 3)} on A is
(a) symmetric and transitive only
(b) symmetric only
(c) transitive only
(d) none of these

Answer/Explanation

Answer:

The relation R is not reflexive because every element of A is not related to itself. Also, R is not symmetric since on interchanging the elements, the ordered pair in R is not contained in it.

R is transitive by default because there is only one element in it.

Page No 1.31:

Question 22:

Let R be the relation on the set A = {1, 2, 3, 4} given by
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then,
(a) R is reflexive and symmetric but not transitive
(b) R is reflexive and transitive but not symmetric
(c) R is symmetric and transitive but not reflexive
(d) R is an equivalence relation

Answer/Explanation

Answer:

(b) R is reflexive and transitive but not symmetric.

Reflexivity : Clearly , (a , a ) ∈R ∀ a ∈A So , R is reflexive on A .Symmetry : Since(1, 2)\inR , but(2, 1)\notinR , R is not symmetric on A .Transitivity : Since ,(1, 3),(3, 2)\inR and(1, 2)\inR , R is transitive on A .

Page No 1.31:

Question 23:

Let A = {1, 2, 3}. Then, the number of equivalence relations containing (1, 2) is
(a) 1
(b) 2
(c) 3
(d) 4

Answer/Explanation

Answer:

(b) 2

There are 2 equivalence relations containing {1, 2}.
R = {(1, 2)}
S = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)}

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Question 24:

The relation R = {(1, 1), (2, 2), (3, 3)} on the set {1, 2, 3} is
(a) symmetric only
(b) reflexive only
(c) an equivalence relation
(d) transitive only

Answer/Explanation

Answer:

R =\{(a , b) : a =b and a , b \inA\} Reflexivity : Let a ∈A Here , a =a \Rightarrow (a , a) \inR for all a \inA So , R is reflexive on A . Symmetry : Let a , b ∈A such that(a , b)\inR . Then ,(a , b)\inR \Rightarrowa =b\Rightarrowb =a\Rightarrow(b , a)\inR for all a \inA So , R is symmetric on A .Transitive : Let a , b , c \inA such that(a , b)\inR and(b , c)\inR . Then ,(a , b)\inR \Rightarrowa =b and(b , c)\inR \Rightarrowb =c\Rightarrowa =c\Rightarrow(a , c)\inR for all a \inA So , R is transitive on A .

Hence, R is an equivalence relation on A.

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Question 25:

S is a relation over the set R of all real numbers and it is given by
(a, b) ∈ S ⇔ ab ≥ 0. Then, S is
(a) symmetric and transitive only
(b) reflexive and symmetric only
(c) antisymmetric relation
(d) an equivalence relation

Answer/Explanation

Answer:

(d) an equivalence relation

Reflexivity: Let a

\in

R
Then,

a a =a^{2}>0\Rightarrow(a , a)\inR \forall a \inR

So, S is reflexive on R.

Symmetry: Let (a, b)

\in

S
Then,

(a , b) \inS\Rightarrowab \geq0\Rightarrowba \geq0\Rightarrow(b , a)\inS \forall a , b \inR

So, S is symmetric on R.

Transitivity:

If(a , b),(b , c)\inS\Rightarrowa b ≥0and b c ≥0\Rightarrowa b ×b c ≥0\Rightarrowa c ≥0[∵ b^{2} \geq 0]\Rightarrow(a , c)\inS for all a , b , c \inset R

Hence, S is an equivalence relation on R.

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Question 26:

In the set Z of all integers, which of the following relation R is not an equivalence relation?
(a) x R y : if x ≤ y
(b) x R y : if x = y
(c) x R y : if x − y is an even integer
(d) x R y : if x ≡ y (mod 3)

Answer/Explanation

Answer:

(a) x R y : if x ≤ y

Clearly, R is not symmetric because x < y does not imply y < x.

Hence, (a) is not an equivalence relation.

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Question 27:

Mark the correct alternative in the following question:

Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then, R is

(a) reflexive but not symmetric (b) reflexive but not transitive
(c) symmetric and transitive (d) neither symmetric nor transitive

Answer/Explanation

Answer:

We have,

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

As ,(a ,a)\inR \foralla \inA So , R is reflexive relation Also ,(1, 2)\inR but(2, 1)\notinR So , R is not symmetric relation And ,(1, 2)\inR ,(2, 3)\inR and(1, 3)\inR So , R is transitive relation

Hence, the correct alternative is option (a).

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Question 28:

Mark the correct alternative in the following question:

The relation S defined on the set R of all real number by the rule aSb iff a

\geq

b is

(a) an equivalence relation
(b) reflexive, transitive but not symmetric
(c) symmetric, transitive but not reflexive
(d) neither transitive nor reflexive but symmetric

Answer/Explanation

Answer:

We have,

S = {(a, b) : a

\geq

b; a, b

\in

As , a =a ∀a ∈R\Rightarrow(a ,a)\inS So , S is reflexive relation Let(a ,b)\inS \Rightarrowa \geqb But b \leqa\Rightarrow(b ,a)\notinS So , S is not symmetric relationLet(a ,b)\inS and(b ,c)\inS\Rightarrowa \geqb and b \geqc\Rightarrowa \geqc\Rightarrow(a ,c)\inS So , S is transitive relation

Hence, the correct alternative is option (b).

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Question 29:

Mark the correct alternative in the following question:

The maximum number of equivalence relations on the set A = {1, 2, 3} is

(a) 1 (b) 2 (c) 3 (d) 5

Answer/Explanation

Answer:

$Consider the relationR_{1}=\{(1, 1)\}It is clearly reflexive , symmetric and transitiveSimilarly ,R_{2}=\{(2, 2)\}andR_{3}=\{(3, 3)\}are reflexive , symmetric and transitiveAlso ,R_{4}=\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}It is reflexive as(a ,a)\inR_{4}for all a \in\{1, 2, 3\}It is symmetric as(a ,b)\inR_{4}\Rightarrow(b ,a)\inR_{4}for all a \in\{1, 2, 3\}Also , it is transitive as(1, 2)\inR_{4},(2, 1)\inR_{4}\Rightarrow (1,1) \inR_{4}The relation defined byR_{5}=\{(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)\}is reflexive , symmetric and transitive as well .Thus , the maximum number of equivalence relation on set A = {1,2,3} is5.$

Hence, the correct alternative is option (d).

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Question 30:

Mark the correct alternative in the following question:

Let R be a relation on the set N of natural numbers defined by nRm iff n divides m. Then, R is

(a) Reflexive and symmetric (b) Transitive and symmetric
(c) Equivalence (d) Reflexive, transitive but not symmetric [NCERT EXEMPLAR]

Answer/Explanation

Answer:

We have,

R = {(m, n) : n divides m; m, n

\in

As , m divides m \Rightarrow (m ,m) \inR \forallm \in N So , R is reflexive Since ,(2, 1)\inRi .e .1divides2but2cannot divide1i .e .(2, 1)\notinR So , R is not symmetricLet(m ,n)\inRand(n ,p)\inR . Then , n divides mand p divides n\Rightarrowp divides m\Rightarrow(m ,p)\inR So , R is transitive

Hence, the correct alternative is option (d).

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Question 31:

Mark the correct alternative in the following question:

Let L denote the set of all straight lines in a plane. Let a relation R be defined by lRm iff l is perpendicular to m for all l, m

\in

L. Then, R is

(a) reflexive (b) symmetric (c) transitive (d) none of these
[NCERT EXEMPLAR]

Answer/Explanation

Answer:

$We have , R =\{(l ,m) :l is perpendicular to m ; l ,m \inL\} As , l is not perpencular to l \Rightarrow (l ,l) \notinR So , R is not reflexive relationLet (l ,m) \inR\Rightarrowl is perpendicular to m\Rightarrowm is also perpendicular to l\Rightarrow(m ,l)\inR So , R is symmetric relationLet(l ,m)\inR and(m ,n)\inR\Rightarrowl is perpendicular to m and m is perpendicular to n\Rightarrowl is parallel to n(Lines perpendicular to same line are parallel)\Rightarrow(m ,l)\notinR So , R is not transitive relation$

Hence, the correct alternative is option (b).

Page No 1.32:

Question 32:

Mark the correct alternative in the following question:

Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b for all a, b

\in

T. Then, R is

a) reflexive but not symmetric (b) transitive but not symmetric
c) equivalence (d) none of these

Answer/Explanation

Answer:

$We have , R =\{(a ,b) :a is congruent to b ; a ,b \inT\} As , a ≅a \Rightarrow (a ,a) \inR So , R is reflexive relationLet (a ,b) \in R . Then , a ≅b \Rightarrowb ≅a\Rightarrow(b ,a)\inR So , R is symmetric relationLet(a ,b)\inR and(b ,c)\inR . Then , a ≅b and b ≅c\Rightarrowa ≅c\Rightarrow(a ,c)\inR So , R is transitive relation∴ R is an equivalence relation$

Hence, the correct alternative is option (c).

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Question 33:

Mark the correct alternative in the following question:

Consider a non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then, R is

(a) symmetric but not transitive (b) transitive but not symmetric
(c) neither symmetric nor transitive (d) both symmetric and transitive

Answer/Explanation

Answer:

$We have , R =\{(a ,b) :a is brother of b\} Let(a ,b)\inR . Then , a is brother of b but b is not necessary brother of a(As , b can be sister of a)\Rightarrow(b ,a)\notinR So , R is not symmetric Also , Let(a ,b)\inR and(b ,c)\inR \Rightarrowa is brother of b and b is brother of c\Rightarrowa is brother of c\Rightarrow(a ,c)\inR So , R is transitive$

Hence, the correct alternative is option (b).

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Question 34:

Mark the correct alternative in the following question:

For real numbers x and y, define xRy iff

x -y +\sqrt{2}

is an irrational number. Then the relation R is

(a) reflexive (b) symmetric (c) transitive (d) none of these

Answer/Explanation

Answer:

$We have , R =\{(x ,y) 😡 -y + \sqrt{2} is an irrational number ; x ,y \in R\}As , x -x +\sqrt{2}=\sqrt{2}, which is an irrational number\Rightarrow(x ,x)\inR So , R is reflexive relation Since ,(\sqrt{2}, 2)\inR i .e .\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2, which is an irrational number but2-\sqrt{2}+\sqrt{2}=2, which isa rational number\Rightarrow(2, \sqrt{2})\notinR So , R is not symmetric relationAlso ,(\sqrt{2}, 2)\inR and(2, 2 \sqrt{2})\inRi .e .\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2, which is an irrational number and2-2\sqrt{2}+\sqrt{2}=2-\sqrt{2}, which is also an irrational number But\sqrt{2}-2\sqrt{2}+\sqrt{2}=0, which isa rational number\Rightarrow(\sqrt{2}, 2 \sqrt{2})\notinR So , R is not transitive relation$

Hence, the correct alternative is option (a).

Page No 1.32:

Question 35:

If a relation R on the set (1, 2, 3) be defined by R = {(1, 2)}, then R is
(a) reflexive (b) transitive (c) symmetric (d) none of these

Answer/Explanation

Answer:

Given: A relation R on the set {1, 2, 3} be defined by R = {(1, 2)}.

R = {(1, 2)}

Since, (1, 1) ∉ R
Therefore, It is not reflexive.

Since, (1, 2) ∈ R but (2, 1) ∉ R
Therefore, It is not symmetric.

But there is no counter example to disapprove transitive condition.
Therefore, it is transitive.

Hence, the correct option is (b).

Page No 1.32:

Question 1:

If R =

\{(x , y) : x^{2} + y^{2} \leq 4, x , y \inZ\}

is a relation in Z, then the domain of R is ______________________.

Answer/Explanation

Answer:

Given: R =

\{(x , y) : x^{2} + y^{2} \leq 4, x , y \inZ\}

R = {(−2, 0), (2, 0), (0, 2), (0, −2), (−1, 1), (−1, −1), (1, −1), (1, 1), (0, 1), (1, 0), (−1, 0), (0, −1), (0, 0)}

Therefore, Domain of R = {−2, −1, 0, 1, 2}

Hence, if R =

\{(x , y) : x^{2} + y^{2} \leq 4, x , y \inZ\}

is a relation in Z, then the domain of R is {−2, −1, 0, 1, 2}.

Page No 1.32:

Question 2:

Let R be a relation in N defined by R ={(x, y): x + 2y = 8}, then the range of R is ___________________.

Answer/Explanation

Answer:

Given: R = {(x, y): x + 2y = 8} where x, y ∈ N

R = {(6, 1), (4, 2), (2, 3)}

Therefore, Range of R = {1, 2, 3}

Hence, the range of R is {1, 2, 3}.

Page No 1.32:

Question 3:

The number of relations on a finite set having 5 elements is __________________.

Answer/Explanation

Answer:

Let R be a relation on A, where A contains 5 elements.

R is a subset of A × A.

Number of elements in A × A = 5 × 5 = 25

Number of relations = Number of subsets of A × A = 225

Hence, the number of relations on a finite set having 5 elements is 225.

Page No 1.32:

Question 4:

Let A = {1, 2, 3, 4} and R be the relation on A defined by {(a, b): a, b ∈ A, a×b is an even number}, then the range of R is __________________.

Answer/Explanation

Answer:

Given: R = {(a, b): a, b ∈ A, a × b is an even number}, where A = {1, 2, 3, 4}.

R = {(1, 2), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}

Therefore, Range of R = {1, 2, 3, 4}

Hence, the range of R is {1, 2, 3, 4}.

Page No 1.32:

Question 5:

Let A = {1, 2, 3, 4, 5} The domain of the relation on A defined by R ={(x,y): y = 2x-1},is__________________.

Answer/Explanation

Answer:

Given: R = {(x, y): y = 2x − 1}, where A = {1, 2, 3, 4, 5} and x, y ∈ A.

R = {(1, 1), (2, 3), (3, 5)}

Therefore, Domain of R = {1, 2, 3}.

Hence, the domain of the relation on A defined by R = {(x, y): y = 2x − 1}, is {1, 2, 3}.

Page No 1.32:

Question 6:

If R s a relation defined on set A ={1, 2, 3} by the rule (a,b)

\inR \Leftrightarrow |a^{2} - b^{2}| \leq 5,

then R-1 =____________________.

Answer/Explanation

Answer:

Given: R = {(a, b):

|a^{2} - b^{2}| \leq 5

}, where A = {1, 2, 3} and a, b ∈ A.

R = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3)}

Therefore, R−1 = {(1, 1), (2, 1), (1, 2), (2, 2), (3, 2), (2, 3), (3, 3)} = R

Hence, if R is a relation defined on set A = {1, 2, 3} by the rule (a,b)

\inR \Leftrightarrow |a^{2} - b^{2}| \leq 5,

then R-1 = R.

Page No 1.32:

Question 7:

If R is a relation from A = {11, 12, 13} to B = {8, 10 12} defined by y = x-3, then R-1 =_______________________.

Answer/Explanation

Answer:

Given: R = {(x, y): y = x − 3, x ∈ A and y ∈ B}, where A = {11, 12, 13} and B = {8, 10 12}.

R = {(11, 8), (13, 10)}

Therefore, R−1 = {(8, 11), (10, 13)}

Hence, R-1 = {(8, 11), (10, 13)}.

Page No 1.32:

Question 8:

The smallest equivalence relation on the set A = {a, b, c, d} is _________________________.

Answer/Explanation

Answer:

Given: A = {a, b, c, d}

Identity relation is the smallest equivalence relation.

Therefore, R = {(a, a), (b, b), (c, c)} is the smallest equivalence relation.

Hence, the smallest equivalence relation on the set A = {a, b, c, d} is {(a, a), (b, b), (c, c)}.

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Question 9:

The largest equivalence relation on the set A = {1, 2, 3} is ___________________.

Answer/Explanation

Answer:

Given: A = {1, 2, 3}

The largest equivalence relation contains all the possible ordered pairs.

Therefore, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)} is the largest equivalence relation.

Hence, the largest equivalence relation on the set A = {1, 2, 3} is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)}.

Page No 1.32:

Question 10:

Let R be the equivalence relation on the set Z of integers given by R = {(a, b): 3 divides a-b}. Then the equivalence class [0] is equal to ____________________.

Answer/Explanation

Answer:

Given: R is the equivalence relation on the set Z of integers given by R = {(a, b): 3 divides a − b}.

To find the equivalence class [0], we put b = 0 in the given relation and find all the possible values of a.

Thus,
R = {(a, 0): 3 divides a − 0}
⇒ a − 0 is a multiple of 3
⇒ a is a multiple of 3
⇒ a = 3n , where n ∈ Z
⇒ a = 0, ±3, ±6, ±9, ….

Therefore, equivalence class [0] = {0, ±3, ±6, ±9, ….}

Hence, the equivalence class [0] is equal to {0, ±3, ±6, ±9, ….}.

Page No 1.32:

Question 11:

Let R be a relation on the set Z of all integers defined as (x, y) ∈ R ⇔ x-y is divisible by 2. Then, the equivalence class [1] is _________________.

Answer/Explanation

Answer:

Given: R is the equivalence relation on the set Z of integers defined as (x, y) ∈ R ⇔ x − y is divisible by 2.

To find the equivalence class [1], we put y = 1 in the given relation and find all the possible values of x.

Thus,
R = {(x, 1): x − 1 is divisible by 2}
⇒ x − 1 is divisible by 2
⇒ x = ±1, ±3, ±6, ±9, ….

Therefore, equivalence class [0] = {±1, ±3, ±6, ±9, ….}

Hence, the equivalence class [1] is {±1, ±3, ±6, ±9, ….}.

Page No 1.32:

Question 12:

The relation R = {(1, 2,), (1, 3)} on set A = [1, 2, 3] is _________________ only.

Answer/Explanation

Answer:

Given: A relation R on the set {1, 2, 3} be defined by R = R = {(1, 2,), (1, 3)}.

R = {(1, 2,), (1, 3)}

Since, (1, 1) ∉ R
Therefore, It is not reflexive.

Since, (1, 2) ∈ R but (2, 1) ∉ R
Therefore, It is not symmetric.

But there is no counter example to disapprove transitive condition.
Therefore, it is transitive.

Hence, The relation R = {(1, 2,), (1, 3)} on set A = {1, 2, 3} is transitive only.

Page No 1.33:

Question 1:

Write the domain of the relation R defined on the set Z of integers as follows:
(a, b) ∈ R ⇔ a2 + b2 = 25

Answer/Explanation

Answer:

Domain of R is the set of values satisfying the relation R.
As a should be an integer, we get the given values of a:

0, \pm 3, \pm 4, \pm 5 Thus , Domain of R =\{0, \pm 3, \pm 4, \pm 5\}

Page No 1.33:

Question 2:

If R = {(x, y) : x2 + y2 ≤ 4; x, y ∈ Z} is a relation on Z, write the domain of R.

Answer/Explanation

Answer:

Domain of R is the set of values of x satisfying the relation R.
As x must be an integer, we get the given values of x:

0, \pm 1, \pm 2 Thus , Domain of R =\{0, \pm 1, \pm 2\}

Page No 1.33:

Question 3:

Write the identity relation on set A = {a, b, c}.

Answer/Explanation

Answer:

Identity set of A is
I = {(a, a), (b, b), (c, c)}

Every element of this relation is related to itself.

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Question 4:

Write the smallest reflexive relation on set A = {1, 2, 3, 4}.

Answer/Explanation

Answer:

Here,
A = {1, 2, 3, 4}
Also, a relation is reflexive iff every element of the set is related to itself.

So, the smallest reflexive relation on the set A is
R = {(1, 1), (2, 2), (3, 3), (4, 4)}

Page No 1.33:

Question 5:

If R = {(x, y) : x + 2y = 8} is a relation on N by, then write the range of R.

Answer/Explanation

Answer:

R = {(x, y) : x + 2y = 8, x, y

\in

N}
Then, the values of y can be 1, 2, 3 only.
Also, y = 4 cannot result in x = 0 because x is a natural number.

Therefore, range of R is {1, 2, 3}.

Page No 1.33:

Question 6:

If R is a symmetric relation on a set A, then write a relation between R and R−1.

Answer/Explanation

Answer:

Here, R is symmetric on the set A.

Let(a , b)\inR \Rightarrow (b , a) \inR [Since R is symmetric] \Rightarrow (a , b) \in R^{- 1} [By definition of inverse relation] \RightarrowR \subset R^{- 1} Let(x , y)\inR^{- 1}\Rightarrow(y , x)\inR[By definition of inverse relation]\Rightarrow(x , y)\inR[Since R is symmetric]\RightarrowR^{- 1}\subsetR Thus , R =R^{- 1}

Page No 1.33:

Question 7:

Let R = {(x, y) : |x2 − y2| <1) be a relation on set A = {1, 2, 3, 4, 5}. Write R as a set of ordered pairs.

Answer/Explanation

Answer:

R is the set of ordered pairs satisfying the above relation. Also, no two different elements can satisfy the relation; only the same elements can satisfy the given relation.

So, R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}

Page No 1.33:

Question 8:

If A = {2, 3, 4}, B = {1, 3, 7} and R = {(x, y) : x ∈ A, y ∈ B and x < y} is a relation from A to B, then write R−1.

Answer/Explanation

Answer:

Since R = {(x, y) : x

\in

A, y

\in

A and x < y},
R = {(2, 3), (2, 7), (3, 7), (4, 7)}

So, R-1 = {(3, 2), (7, 2), (7, 3), (7, 4)}

Page No 1.33:

Question 9:

Let A = {3, 5, 7}, B = {2, 6, 10} and R be a relation from A to B defined by R = {(x, y) : x and y are relatively prime}. Then, write R and R−1.

Answer/Explanation

Answer:

R = {(x, y) : x and y are relatively prime}
Then,

R = {(3, 2), (5, 2), (7, 2), (3, 10), (7, 10), (5, 6), (7, 6)}

So, R-1 = {(2, 3), (2, 5), (2, 7), (10, 3), (10, 7), (6, 5), (6, 7)}

Page No 1.33:

Question 10:

Define a reflexive relation.

Answer/Explanation

Answer:

A relation R on A is said to be reflexive iff every element of A is related to itself.

i.e. R is reflexive

\Leftrightarrow (a , a) \inR for all a \inA

Page No 1.33:

Question 11:

Define a symmetric relation.

Answer/Explanation

Answer:

A relation R on a set A is said to be symmetric iff

(a , b) \inR\Rightarrow(b , a)\inR for all a , b \inAi .e . aRb \RightarrowbRa for all a , b \inA

Page No 1.33:

Question 12:

Define a transitive relation.

Answer/Explanation

Answer:

A relation R on a set A is said to be transitive iff

(a , b) \inR and (b , c) \inR\Rightarrow(a , c)\inR for all a , b , c \inRi .e . aRb and bRc\RightarrowaRc for all a , b , c \inR

Page No 1.33:

Question 13:

Define an equivalence relation.

Answer/Explanation

Answer:

A relation R on set A is said to be an equivalence relation iff
(i) it is reflexive,
(ii) it is symmetric and
(iii) it is transitive.

Relation R on set A satisfying all the above three properties is an equivalence relation.

Page No 1.33:

Question 14:

If A = {3, 5, 7} and B = {2, 4, 9} and R is a relation given by “is less than”, write R as a set ordered pairs.

Answer/Explanation

Answer:

$Since , R =\{(x , y) : x , y \inN and x <y\}, R = {(3, 4), (3 , 9) , (5 , 9) , (7, 9)}$

Page No 1.33:

Question 15:

A = {1, 2, 3, 4, 5, 6, 7, 8} and if R = {(x, y) : y is one half of x; x, y ∈ A} is a relation on A, then write R as a set of ordered pairs.

Answer/Explanation

Answer:

Since R = {(x, y) : y is one half of x; x, y

\in

So, R = {(2, 1), (4, 2), (6, 3), (8, 4)}

Page No 1.33:

Question 16:

Let A = {2, 3, 4, 5} and B = {1, 3, 4}. If R is the relation from A to B given by a R b if “a is a divisor of b”. Write R as a set of ordered pairs.

Answer/Explanation

Answer:

$Since R =\{(a , b) : a , b \inN : a is a divisor of b\}$

So, R = {(2, 4), (3, 3), (4, 4)}

Page No 1.33:

Question 17:

State the reason for the relation R on the set {1, 2, 3} given by R = {(1, 2), (2, 1)} to be transitive.

Answer/Explanation

Answer:

$Since(1, 2)\inR ,(2, 1)\inR but(1, 1)\notinR , R is not transitive on the set\{1, 2, 3\}.For R to be ina transitive relation , we must have(1, 1)\inR .$

Page No 1.33:

Question 18:

Let R = {(a, a3) : a is a prime number less than 5} be a relation. Find the range of R. [CBSE 2014]

Answer/Explanation

Answer:

We have,
R = {(a, a3) : a is a prime number less than 5}
Or,
R = {(2, 8), (3, 27)}

So, the range of R is {8, 27}.

Page No 1.33:

Question 19:

Let R be the equivalence relation on the set Z of the integers given by R = {(a, b) : 2 divides a

-

b}. Write the equivalence class [0].
[NCERT EXEMPLAR]

Answer/Explanation

Answer:

We have,
An equivalence relation, R = {(a, b) : 2 divides a

-

If b =0, then a -b =a -0=a As ,2dividesa –b And , the set of integers which are divided by2is\{0, \pm 2, \pm 4, \pm 6, . . .\}So , the equivalence class [0] =\{0, \pm 2, \pm 4, \pm 6, . . .\}

Page No 1.33:

Question 20:

For the set A = {1, 2, 3}, define a relation R on the set A as follows:
R = {(1, 1), (2, 2), (3, 3), (1, 3)}
Write the ordered pairs to be added to R to make the smallest equivalence relation.

Answer/Explanation

Answer:

We have,
R = {(1, 1), (2, 2), (3, 3), (1, 3)}

As, (a, a)

\in

R, for all values of a

\in

A
So, R is a reflexive relation

R can be a symmetric and transitive relation only when element (3, 1) is added

Hence, the ordered pairs to be added to R to make the smallest equivalence relation is (3, 1).

Page No 1.33:

Question 21:

Let A = {0, 1, 2, 3} and R be a relation on A defined as
R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}
Is R reflexive? symmetric? transitive?

Answer/Explanation

Answer:

We have,
R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}

As ,(a ,a)\inR \foralla \inA So , R isa reflexive relationAlso ,(a ,b)\inR and(b ,a)\inR So , R isa symmetric relation as wellAnd ,(0, 1)\inR but(1, 2)\notinR and(2, 3)\notinR So , R is nota transitive relation

Page No 1.33:

Question 22:

Let the relation R be defined on the set A = {1, 2, 3, 4, 5} by R = {(a, b) : |a2

-

b2| < 8}. Write R as a set of ordered pairs.

Answer/Explanation

Answer:

As, R = {(a, b) : |a2

-

b2| < 8}
So, R = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (3, 4), (4, 3), (4, 4), (5, 5)}

Page No 1.34:

Question 23:

Let the relation R be defined on N by aRb iff 2a + 3b = 30. Then write R as a set of ordered pairs.

Answer/Explanation

Answer:

As, R = {(a, b) : 2a + 3b = 30; a, b

\in

So, R = {(3, 8), (6, 6), (9, 4), (12, 2)}

Page No 1.34:

Question 24:

Write the smallest equivalence relation on the set A = {1, 2, 3}.

Answer/Explanation

Answer:

The smallest equivalence relation on the set A = {1, 2, 3} is R = {(1, 1), (2, 2), (3, 3)}