Rd Sharma XII Vol 1 2020 for Class 12 Science Maths Chapter 1 - Relation

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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:

(i) Reflexivity:
$Let x be an arbitrary element of R . Then, x \in R \Rightarrow x and x work at the same place is true since they are the same . \Rightarrow (x, x) \in R So, R is a reflexive relation .$

Symmetry:
$Let (x, y) \in R \Rightarrow x and y work at the same place \Rightarrow y and x work at the same place \Rightarrow (y, x) \in R So, R is a symmetric relation .$

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

(ii) Reflexivity:
$Let x be an arbitrary element of R . Then, x \in R \Rightarrow x and x live in the same locality is true since they are the same . So, R is a reflexive relation .$

Symmetry:
$Let (x, y) \in R \Rightarrow x and y live in the same locality \Rightarrow y and x live in the same locality \Rightarrow (y, x) \in R So, R is a symmetric relation .$

Transitivity:
$Let (x, y) \in R and (y, z) \in R . Then, x and y live in the same locality and y and z live in the same locality \Rightarrow x, y and z all live in the same locality \Rightarrow x and z live in the same locality \Rightarrow (x, z) \in R So, R is a transitive relation .$

(iii)
Reflexivity:
$Let x be an element of R . Then, x is wife of x cannot be true . \Rightarrow (x, x) \notin R So, R is not a reflexive relation .$

Symmetry:
$Let (x, y) \in R \Rightarrow x is wife of y \Rightarrow x is female and y is male \Rightarrow y cannot be wife of x as y is husband of x \Rightarrow (y, x) \notin R So, R is not a symmetric relation .$

Transitivity:
$Let (x, y) \in R, but (y, z) \notin R Since x is wife of y, but y cannot be the wife of z, y is husband of x . \Rightarrow x is not the wife of z \Rightarrow (x, z) \in R So, R is a transitive relation .$

(iv)
Reflexivity:
$Let x be an arbitrary element of R . Then, x is father of x cannot be true since no one can be father of himself . So, R is not a reflexive relation .$

Symmetry:
$Let (x, y) \in R \Rightarrow x is father of y \Rightarrow y is son / daughter of x \Rightarrow (y, x) \notin R So, R is not a symmetric relation .$

Transitivity:
$Let (x, y) \in R and (y, z) \in R . Then, x is father of y and y is father of z \Rightarrow x is grandfather of z \Rightarrow (x, z) \notin R So, R is not a transitive relation .$

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:

(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 $(a, a) \in R \Rightarrow (a, a) \in R$ . So, R₂ is symmetric.
Transitive: R₂ is clearly a transitive relation, since there is only one element in it.

(iii) R₃
Reflexive:
Here,
$(b, b) \notin R_{3} n e i t h e r (c, c) \notin R_{3}$
So, R₃is 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,
$(a, b) \in R_{4}, b u t (b, a) \notin R_{4} . So, R_{4} is not symmetric .$

Transitive:
Here,
$(a, b) \in R_{4}, (b, c) \in R_{4}, b u t (a, c) \notin R_{4} So, R_{4} is not transitive .$

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) R₃on R defined by (a, b) ∈ R₃ ⇔ a²– 4ab + 3b²= 0.

Answer:

(i) Reflexivity:
Let a be an arbitrary element of R₁. Then,
$a \in R_{1} \Rightarrow a \neq \frac{1}{a} for all a \in Q_{0} So, R_{1} is not reflexive .$

Symmetry:
Let (a, b) $\in R_{1}$ . Then,

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

Transitivity:
Here,
$(a, b) \in R_{1} and (b, c) \in R_{2} \Rightarrow a = \frac{1}{b} and b = \frac{1}{c} \Rightarrow a = \frac{1}{\frac{1}{c}} = c \Rightarrow a \neq \frac{1}{c} \Rightarrow (a, c) \notin R_{1} So, R_{1} is not transitive .$

(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:
$Let (a, b) \in R_{2} \Rightarrow |a - b| \leq 5 \Rightarrow |b - a| \leq 5 [Since, |a - b| = |b - a|] \Rightarrow (b, a) \in R_{2} So, R_{2} is symmetric .$

Transitivity:
$Let (1, 3) \in R_{2} and (3, 7) \in R_{2} \Rightarrow |1 - 3| \leq 5 and |3 - 7| \leq 5 But |1 - 7| ≰ 5 \Rightarrow (1, 7) \notin R_{2} So, R_{2} is not transitive .$

(iii)
Reflexivity: Let a be an arbitrary element of R₃. Then,
_{$a \in R_{3} \Rightarrow a^{2} - 4 a \times a + 3 a^{2} = 0 So, R_{3} is reflexive .$}

Symmetry:
$Let (a, b) \in R_{3} \Rightarrow a^{2} - 4 a b + 3 b^{2} = 0 But b^{2} - 4 b a + 3 a^{2} \neq 0 for all a, b \in R So, R_{3} is not symmetric .$

Transitivity:
$(1, 2) \in R_{3} and (2, 3) \in R_{3} \Rightarrow 1 - 8 + 6 = 0 and 4 - 24 + 27 = 0 But 1 - 12 + 9 \neq 0 So, R_{3} is not transitive .$

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:

$(1)$ R₁
Reflexivity:
Here,
$(1, 1), (2, 2), (3, 3) \in R So, R_{1} is reflexive .$

Symmetry:
$Here, (2, 1) \in R_{1}, but (1, 2) \notin R_{1} So, R_{1} is not symmetric .$

Transitivity:
$Here, (2, 1) \in R_{1} and (1, 3) \in R_{1}, but (2, 3) \notin R_{1} So, R_{1} is not transitive .$

$(2)$ R₂
Reflexivity:

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

Symmetry:
$Here, (1, 3) \in R_{2} and (3, 1) \in R_{2} So, R_{2} is symmetric .$

Transitivity:
$Here, (1, 3) \in R_{2} and (3, 1) \in R_{2} But (3, 3) \notin R_{2} So, R_{2} is not transitive .$

$(3)$ R₃
Reflexivity:
$Clearly, (1, 1) \notin R_{3} So, R_{3} is not reflexive .$

Symmetry:
$Here, (1, 3) \in R_{3}, but (3, 1) \notin R_{3} So, R_{3} is not symmetric .$

Transitivity:
$Here, (1, 3) \in R_{3} and (3, 3) \in R_{3} Also, (1, 3) \in R_{3} So, R_{3} is transitive .$

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:

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

$a \in R But a - a = 0 ≯ 0 So, this relation is not reflexive .$

Symmetry:

$Let (a, b) \in R \Rightarrow a - b > 0 \Rightarrow - (b - a) > 0 \Rightarrow b - a < 0 So, the given relation is not symmetric .$

Transitivity:

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

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

$a \in R \Rightarrow 1 + a \times a > 0 i . e . 1 + a^{2} > 0 [Since, square of any number is positive] So, the given relation is reflexive .$

Symmetry:

$Let (a, b) \in R \Rightarrow 1 + a b > 0 \Rightarrow 1 + b a > 0 \Rightarrow (b, a) \in R So, the given relation is symmetric .$

Transitivity:

$Let (a, b) \in R and (b, c) \in R \Rightarrow 1 + a b > 0 and 1 + b c > 0 But 1 + a c ≯ 0 \Rightarrow (a, c) \notin R So, the given relation is not transitive .$

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

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

Symmetry:

$Let (a, b) \in R \Rightarrow |a| \leq b \Rightarrow |b| ≰ a for all a, b \in R \Rightarrow (b, a) \notin R So, R is not symmetric .$

Transitivity:

$Let (a, b) \in R and (b, c) \in R \Rightarrow |a| \leq b and |b| \leq c Multiplying the corresponding sides, we get |a| |b| \leq b c \Rightarrow |a| \leq c \Rightarrow (a, c) \in R Thus, R is transitive .$

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:

Reflexivity:

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

Symmetry:
$Let (a, b) \in R \Rightarrow b = a + 1 \Rightarrow - a = - b + 1 \Rightarrow a = b - 1 Thus, (b, a) \notin R So, R is not symmetric on A .$

Transitivity:
$Let (1, 2) and (2, 3) \in R \Rightarrow 2 = 1 + 1 and 3 2 + 1 is true . But 3 \neq 1 + 1 \Rightarrow (1, 3) \notin R So, R is not transitive on A .$

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:

Reflexivity:

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

Symmetry:

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

Transitivity:
$Since (7, 3) \in R and (3, 3^{\frac{1}{3}}) \in R, 7 < 3^{3} and 3 = {(3^{\frac{1}{3}})}^{3} But 7 > {(3^{\frac{1}{3}})}^{3} \Rightarrow (7, 3^{\frac{1}{3}}) \notin R So, R is not transitive .$

Page No 1.11:

Question 8:

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

Answer:

Let A be a set. Then,

$Identity relation IA = I_{A} is reflexive, since (a, a) \in A \forall a$

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:

(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)}

$Relation R satisfies reflexivity and transitivity . \Rightarrow (1, 1), (2, 2), (3, 3) \in R and (1, 1), (2, 1) \in R \Rightarrow (1, 1) \in R However, (2, 1) \in R, but (1, 2) \notin R$

(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:

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,
$x \in R \Rightarrow 2 x + x = 41 cannot be true . \Rightarrow (x, x) \notin R So, R is not reflexive .$

Symmetry:
$Let (x, y) \in R . Then, 2 x + y = 41 \Rightarrow 2 y + x = 41 \Rightarrow (y, x) \notin R So, R is not symmetric .$

Transitivity:
$Let (x, y) and (y, z) \in R \Rightarrow 2 x + y = 41 and 2 y + z = 41 \Rightarrow 2 x + z = 2 x + 41 - 2 y 41 - y - 2 y = 41 - 3 y \Rightarrow (x, z) \notin R Thus, R is not transitive .$

Page No 1.11:

Question 11:

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

Answer:

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:

$R = \{(m, n) : m, n \in Z, m = k n, where k \in N\} Reflexivity : Let m be an arbitrary element of R . Then, m = k m is true for k = 1 \Rightarrow (m, m) \in R Thus, R is reflexive . Symmetry : Let (m, n) \in R \Rightarrow m = k n for some k \in N \to n = \frac{1}{k} m \Rightarrow (n, m) \notin R Thus, R is not symmetric . Transitivity : Let (m, n) and (n, o) \in R \Rightarrow m = k n and n = l o for some k, l \in N \Rightarrow m = (k l) o Here, k l \in R \Rightarrow (m, o) \in R Thus, R is transitive .$

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:

Let R be the set such that R = {(a, b) : a, b $\in R; a \geq b$ }

Reflexivity:
$Let a be an arbitrary element of R . \Rightarrow a \in R \Rightarrow a = a \Rightarrow a \geq a is true for a = a \Rightarrow (a, a) \in R Hence, R is reflexive .$

Symmetry:
$Let (a, b) \in R \Rightarrow a \geq b is same as b \leq a, but not b \geq a Thus, (b, a) \notin R Hence, R is not symmetric .$

Transitivity:
$Let (a, b) and (b, c) \in R \Rightarrow a \geq b and b \geq c \Rightarrow a \geq b \geq c \Rightarrow a \geq c \Rightarrow (a, c) \in R Hence, R is transitive .$

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:

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:

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 $\times$ A

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:

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:

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 $\in$ N

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

Answer:

(i) We have,

R = {(x, y) : x > y, x, y $\in$ N}

$As, x = x \forall x \in N \Rightarrow (x, x) \notin R So, R is not a reflexive relation Let (x, y) \in R \Rightarrow x > y but y < x \Rightarrow (y, x) \notin R So, R is not a symmeteric relation Let (x, y) \in R and (y, z) \in R \Rightarrow x > y and y > z \Rightarrow x > z \Rightarrow (x, z) \in R So, R is a transitive relation$

(ii) We have,

R = {(x, y) : x + y = 10, x, y $\in$ N}

$R = \{(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)\} As, (1, 1) \notin R So, R is not a reflexive relation Let (x, y) \in R \Rightarrow x + y = 10 \Rightarrow y + x = 10 \Rightarrow (y, x) \in R So, R is a symmeteric relation As, (1, 9) \in R and (9, 1) \in R but (1, 1) \notin R So, R is not a transitive relation$

(iii) We have,

R = {(x, y) : xy is square of an integer, x, y $\in$ N}

$As, x \times x = x^{2}, which is a square of an integer x \Rightarrow (x, x) \in R So, R is a reflexive relation Let (x, y) \in R \Rightarrow x y is square of an integer \Rightarrow y x is also a square of an integer \Rightarrow (y, x) \in R So, R is a symmeteric relation Let (x, y) \in R and (y, z) \in R \Rightarrow x y is square of an integer and y z is also a square of an interger \Rightarrow x z must be a square of an integer \Rightarrow (x, z) \in R So, R is a transitive relation$

(iv) We have,

R = {(x, y) : x + 4y = 10, x, y $\in$ N}

$R = \{(2, 4), (6, 1)\} As, (2, 2) \notin R So, R is not a reflexive relation As, (2, 4) \in R but (4, 2) \notin R So, R is not a symmeteric relation As, (2, 4) \in R but 4 is not related to any natural number So, R is a transitive relation$

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:

We observe the following relations of relation R.

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

Symmetry:
$Let (a, b) \in R \Rightarrow a - b is divisible by 3 \Rightarrow a - b 3 p for some p \in Z \Rightarrow b - a = 3 (- p) Here, - p \in Z \Rightarrow b - a is divisible by 3 \Rightarrow (b, a) \in R for all a, b \in Z So, R is symmetric on Z .$

Transitivity:
$Let (a, b) and (b, c) \in R \Rightarrow a - b and b - c are divisible by 3 \Rightarrow a - b = 3 p for some p \in Z and b - c = 3 q for some q \in Z Adding the above two, we get a - b + b - c = 3 p + 3 q \Rightarrow a - c = 3 (p + q) Here, p + q \in Z \Rightarrow a - c is divisible by 3 \Rightarrow (a, c) \in R for all a, c \in Z So, R is transitive on Z .$

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:

We observe the following properties of relation R.

Reflexivity:
$Let a be an arbitrary element of the set Z . Then, a \in R \Rightarrow a - a = 0 = 0 \times 2 \Rightarrow 2 divides a - a \Rightarrow (a, a) \in R for all a \in Z So, R is reflexive on Z .$

Symmetry:
$Let (a, b) \in R \Rightarrow 2 divides a - b \Rightarrow \frac{a - b}{2} = p for some p \in Z \Rightarrow \frac{b - a}{2} = - p Here, - p \in Z \Rightarrow 2 divides b - a \Rightarrow (b, a) \in R for all a, b \in Z So, R is symmetric on Z .$

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

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:

We observe the following properties of relation R.

Reflexivity:
$Let a be an arbitrary element of R . Then, \Rightarrow a - a = 0 = 0 \times 5 \Rightarrow a - a is divisible by 5 \Rightarrow (a, a) \in R for all a \in Z So, R is reflexive on Z .$

Symmetry:
$Let (a, b) \in R \Rightarrow a - b is divisible by 5 \Rightarrow a - b = 5 p for some p \in Z \Rightarrow b - a = 5 (- p) Here, - p \in Z [Since p \in Z] \Rightarrow b - a is divisible by 5 \Rightarrow (b, a) \in R for all a, b \in Z So, R is symmetric on Z .$

Transitivity:
$Let (a, b) and (b, c) \in R \Rightarrow a - b is divisible by 5 \Rightarrow a - b = 5 p for some Z Also, b - c is divisible by 5 \Rightarrow b - c = 5 q for some Z Adding the above two, we get a - b + b - c = 5 p + 5 q \Rightarrow a - c = 5 (p + q) \Rightarrow a - c is divisible by 5 Here, p + q \in Z \Rightarrow (a, c) \in R for all a, c \in Z So, R is transitive on Z .$

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:

We observe the following properties of R. Then,
Reflexivity:
$Let a \in N Here, a - a = 0 = 0 \times n \Rightarrow a - a is divisible by n \Rightarrow (a, a) \in R \Rightarrow (a, a) \in R for all a \in Z So, R is reflexive on Z .$

Symmetry:
$Let (a, b) \in R Here, a - b is divisible by n \Rightarrow a - b = n p for some p \in Z \Rightarrow b - a = n (- p) \Rightarrow b - a is divisible by n [p \in Z \Rightarrow - p \in Z] \Rightarrow (b, a) \in R So, R is symmetric on Z .$

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

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:

We observe the following properties of R.

Reflexivity:
$Let a be an arbitrary element of Z . Then, a \in R Clearly, a + a = 2 a is even for all a \in Z . \Rightarrow (a, a) \in R for all a \in Z So, R is reflexive on Z .$

Symmetry:
$Let (a, b) \in R \Rightarrow a + b is even \Rightarrow b + a is even \Rightarrow (b, a) \in R for all a, b \in Z So, R is symmetric on Z .$

Transitivity:
$Let (a, b) and (b, c) \in R \Rightarrow a + b and b + c are even Now, let a + b = 2 x for some x \in Z and b + c = 2 y for some y \in Z Adding the above two, we get a + 2 b + c = 2 x + 2 y \Rightarrow a + c = 2 (x + y - b), which is even for all x, y, b \in Z Thus, (a, c) \in R So, R is transitive on Z .$

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:

We observe the following properties of relation R.

$Let R = {(m, n) : m, n \in Z : m - n i s d i v i s i b l e b y 13} Relexivity : Let m be an arbitrary element of Z . Then, m \in R \Rightarrow m - m = 0 = 0 \times 13 \Rightarrow m - m is divisible by 13 \Rightarrow (m, m) is reflexive on Z . Symmetry : Let (m, n) \in R . Then, m - n is divisible by 13 \Rightarrow m - n = 13 p Here, p \in Z \Rightarrow n - m = 13 (- p) Here, - p \in Z \Rightarrow n - m is divisible by 13 \Rightarrow (n, m) \in R for all m, n \in Z So, R is symmetric on Z . Transitivity : Let (m, n) and (n, o) \in R \Rightarrow m - n and n - o are divisible by 13 \Rightarrow m - n = 13 p and n - o = 13 q for some p, q \in Z Adding the above two, we get m - n + n - o = 13 p + 13 q \Rightarrow m - o = 13 (p + q) Here, p + q \in Z \Rightarrow m - o is divisible by 13 \Rightarrow (m, o) \in R for all m, o \in Z So, R is transitive on Z .$

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:

We observe the following properties of R.

$Reflexivity : Let (a, b) be an arbitrary element of the set A . Then, (a, b) \in A \Rightarrow a b = b a \Rightarrow (a, b) R (a, b) Thus, R is reflexive on A . Symmetry : Let (x, y) and (u, v) \in A such that (x, y) R (u, v) . Then, x v = y u \Rightarrow v x = u y \Rightarrow u y = v x \Rightarrow (u, v) R (x, y) So, R is symmetric on A . Transitivity : Let (x, y), (u, v) and (p, q) \in R such that (x, y) R (u, v) and (u, v) R (p, q) . \Rightarrow x v = y u and u q = v p Multiplying the corresponding sides, we get x v \times u q = y u \times v p \Rightarrow x q = y p \Rightarrow (x, y) R (p, q) So, R is transitive on A .$

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:

We observe the following properties of R.

Reflexivity: Let a be an arbitrary element of A. Then,
$a \in R \Rightarrow a = a [Since, every element is equal to itself] \Rightarrow (a, a) \in R for all a \in A So, R is reflexive on A . Symmetry : Let (a, b) \in R \Rightarrow a b \Rightarrow b = a \Rightarrow (b, a) \in R for all a, b \in A So, R is symmetric on A . Transitivity : Let (a, b) and (b, c) \in R \Rightarrow a = b and b = c \Rightarrow a = b c \Rightarrow a = c \Rightarrow (a, c) \in R So, R is transitive on A .$

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:

We observe the following properties of R.

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

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 $\in$ R}

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:

We observe the following properties on R.

$Reflexivity : Let P_{1} be an arbitrary element of A . Then, polygon P_{1} and P_{1} have the same number of sides, since they are one and the same . \Rightarrow (P_{1}, P_{1}) \in R for all P_{1} \in A So, R is reflexive on A . Symmetry : Let (P_{1}, P_{2}) \in R \Rightarrow P_{1} and P_{2} have the same number of sides . \Rightarrow P_{2} and P_{1} have the same number of sides . \Rightarrow (P_{2}, P_{1}) \in R for all P_{1}, P_{2} \in A So, R is symmetric on A . Transitivity : Let (P_{1}, P_{2}), (P_{2}, P_{3}) \in R \Rightarrow P_{1} and P_{2} have the same number of sides and P_{2} and P_{3} have the same number of sides . \Rightarrow P_{1}, P_{2} and P_{3} have the same number of sides . \Rightarrow P_{1} and P_{3} have the same number of sides . \Rightarrow (P_{1}, P_{3}) \in R for all P_{1}, P_{3} A So, R is transitive on A .$

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:

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

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

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
$\Rightarrow (P, P) \in R So, R is reflexive on A . Symmetry : Let (P, Q) \in R \Rightarrow O P = O Q \Rightarrow O Q = O P \Rightarrow (Q, P) \in R So, R is symmetric on A . Transitivity : Let (P, Q), (Q, R) \in R \Rightarrow O P = O Q and O Q = O R \Rightarrow O P = O Q = O R \Rightarrow O P = O R \Rightarrow (P, R) \in R So, R is transitive on A .$

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:

We observe the following properties of R.

Reflexivity:

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

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:

We observe the following properties of S.

$Reflexivity : Let a be an arbitrary element of R . Then, a \in R \Rightarrow a^{2} + a^{2} \neq 1 \forall a \in R \Rightarrow (a, a) \notin S So, S is not reflexive on R . Symmetry : Let (a, b) \in R \Rightarrow a^{2} + b^{2} = 1 \Rightarrow b^{2} + a^{2} = 1 \Rightarrow (b, a) \in S for all a, b \in R So, S is symmetric on R . Transitivity : Let (a, b) and (b, c) \in S \Rightarrow a^{2} + b^{2} = 1 and b^{2} + c^{2} = 1 Adding the above two, we get a^{2} + c^{2} = 2 - 2 b^{2} \neq 1 for all a, b, c \in R So, S is not transitive on R .$

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:

We observe the following properties of R.

Reflexivity:
$Let (a, b) be an arbitrary element of Z \times Z_{0} . Then, (a, b) \in Z \times Z_{0} \Rightarrow a, b \in Z, Z_{0} \Rightarrow a b = b a \Rightarrow (a, b) \in R for all (a, b) \in Z \times Z_{0} So, R is reflexive on Z \times Z_{0} .$

Symmetry:
$Let (a, b), (c, d) \in Z \times Z_{0} such that (a, b) R (c, d) . Then, (a, b) R (c, d) \Rightarrow a d = b c \Rightarrow c b = d a \Rightarrow (c, d) R (a, b) Thus, (a, b) R (c, d) \Rightarrow (c, d) R (a, b) for all (a, b), (c, d) \in Z \times Z_{0} So, R is symmetric on Z \times Z_{0} .$

Transitivity:
$Let (a, b), (c, d), (e, f) \in N \times N_{0} such that (a, b) R (c, d) and (c, d) R (e, f) . Then, \begin{array}{r} (a, b) R (c, d) \Rightarrow a d = b c \\ (c, d) R (e, f) \Rightarrow c f = d e \end{array}\} \Rightarrow (a d) (c f) = (b c) (d e) \Rightarrow a f = b e \Rightarrow (a, b) R (e, f) Thus, (a, b) R (c, d) a n d (c, d) R (e, f) \Rightarrow (a, b) R (e, f) \Rightarrow (a, b) R (e, f) for all values (a, b), (c, d), (e, f) \in N \times N_{0} So, R is transitive on N \times N_{0} .$

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:

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

$\Rightarrow R \subset A \times A and S \subset A \times A \Rightarrow R \cap S \subset A \times A Thus, R \cap S is a relation on A . Let a, b \in A such that (a, b) \in R \cap S . Then, (a, b) \in R \cap S \Rightarrow (a, b) \in R and (a, b) \in S \Rightarrow (b, a) \in R and (b, a) \in S [Since R and S are symmetric] \Rightarrow (b, a) \in R \cap S Thus, (a, b) \in R \cap S \Rightarrow (b, a) \in R \cap S for all a, b \in A So, R \cap S is symmetric on A .$

Also,
$Let a, b \in A such that (a, b) \in R \cup S \Rightarrow (a, b) \in R or (a, b) \in S \Rightarrow (b, a) \in R or (b, a) \in S [Since R and S are symmetric] \Rightarrow (b, a) \in R \cup S So, R \cup S is symmetric on A .$

(ii) R is reflexive and S is any relation.
$Suppose a \in A . Then, (a, a) \in R [Since R is reflexive] \Rightarrow (a, a) \in R \cup S \Rightarrow R \cup S is reflexive on A .$

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:

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.

$(a, b) \in R \cup S and (b, c) \in R \cup S But (a, c) \notin R \cup S$

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 C₀be the set of all no-zero complex numbers. Let a relation R on C₀be defined as

$z_{1}$ R $z_{2} \Leftrightarrow \frac{z_{1} - z_{2}}{z_{1} + z_{2}}$ is real for all $z_{1}, z_{2} \in$ C₀.

Show that R is an equivalence relation.

Answer:

(i) Test for reflexivity:

Since, $\frac{z_{1} - z_{1}}{z_{1} + z_{1}} = 0$ , which is a real number.

So, $(z_{1}, z_{1}) \in R$

Hence, R is relexive relation.

(ii) Test for symmetric:

Let $(z_{1}, z_{2}) \in R$ .

Then, $\frac{z_{1} - z_{2}}{z_{1} + z_{2}} = x$ , where x is real

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

So, $(z_{2}, z_{1}) \in R$

Hence, R is symmetric relation.

(iii) Test for transivity:

Let $(z_{1}, z_{2}) \in R and (z_{2}, z_{3}) \in R$ .

Then,

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

Also,

$\frac{z_{2} - z_{3}}{z_{2} + z_{3}} = y, where y is a real number . \Rightarrow z_{2} - z_{3} = y z_{2} + Y z_{3} \Rightarrow z_{2} - y z_{2} = z_{3} + y z_{3} \Rightarrow z_{2} (1 - y) = z_{3} (1 + y) \Rightarrow \frac{z_{2}}{z_{3}} = \frac{(1 + y)}{(1 - y)} . . . (2)$

Dividing (1) and (2), we get

$\frac{z_{1}}{z_{3}} = (\frac{1 + x}{1 - x}) \times (\frac{1 - y}{1 + y}) = z, where z is a real number . \Rightarrow \frac{z_{1} - z_{3}}{z_{1} + z_{3}} = \frac{z - 1}{z + 1}, which is real \Rightarrow (z_{1}, z_{3}) \in R$

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:

(c) (6, 8) ∈ R

$(6, 8) \in R Then, a = b - 2 \Rightarrow 6 = 8 - 2 and b = 8 > 6 Henc e, (6, 8) \in R$

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:

(c) {0, ± 3, ± 4, ± 5}

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

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

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:

(b) reflexive and symmetric

$Reflexivity : Let x \in R . Then, x - x = 0 < 1 \Rightarrow |x - x| \leq 1 \Rightarrow (x, x) \in R for all x \in Z So, R is reflexive on Z . Symmetry : Let (x, y) \in R . Then, |x - y| \leq 0 \Rightarrow |- (y - x)| \leq 1 \Rightarrow |y - x| \leq 1 [Since |x - y| = |y - x|] \Rightarrow (y, x) \in R for all x, y \in Z So, R is symmetri c on Z . Transitivity : Let (x, y) \in R and (y, z) \in R . Then, |x - y| \leq 1 and |y - z| \leq 1 \Rightarrow It is not always true that |x - y| \leq 1 . \Rightarrow (x, z) \notin R So, R is not transitive on Z .$

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:

(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 l₁R l₂ ⇔ l ₁⊥ l₂. Then, R is
(a) symmetric
(b) reflexive
(c) transitive
(d) an equivalence relation

Answer:

(a) symmetric

A = Set of all straight lines in the plane

$R = \{(l_{1,} l_{2}) : l_{1}, l_{2} \in A : l_{1} ⊥ l_{2}\} Reflexivity : l_{1} is not ⊥ l_{1} \Rightarrow (l_{1}, l_{1}) \notin R So, R is not reflexive on A . Symmetry : Let (l_{1}, l_{2}) \in R \Rightarrow l_{1} ⊥ l_{2} \Rightarrow l_{2} ⊥ l_{1} \Rightarrow (l_{2}, l_{1}) \in R So, R is symmetric on A . Transitivity : Let (l_{1}, l_{2}) \in R, (l_{2}, l_{3}) \in R \Rightarrow l_{1} ⊥ l_{2} and l_{2} ⊥ l_{3} But l_{1} is not ⊥ l_{3} \Rightarrow (l_{1}, l_{3}) \notin R So, R is not transitive on A .$

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:

(c) transitive only

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:

(c) 6

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:

(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.

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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:

(c) an equivalence relation

We observe the following properties of relation R.

$Reflexivity : Let (a, b) \in N \times N \Rightarrow a, b \in N \Rightarrow a + b = b + a \Rightarrow (a, b) \in R So, R is reflexive on N \times N . Symmetry : Let (a, b), (c, d) \in N \times N such that (a, b) R (c, d) \Rightarrow a + d = b + c \Rightarrow d + a = c + b \Rightarrow (d, c), (b, a) \in R So, R is symmetric on N \times N . Transitivity : Let (a, b), (c, d), (e, f) \in N \times N such that (a, b) R (c, d) and (c, d) R (e, f) \Rightarrow a + d = b + c and c + f = d + e \Rightarrow a + d + c + f = b + c + d + e \Rightarrow a + f = b + e \Rightarrow (a, b) R (e, f) So, R is transitive on N \times N .$

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:

(c) {1}

Here,
$R = \{(x, y) : x \in A and y \in B : x > y\} \Rightarrow R = \{(2, 1), (3, 1)\}$

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:

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

The relation R is defined as
$R = \{(x, y) : x \in \{2, 3, 4, 5\}, y \in \{3, 6, 7, 10\} : x is relatively prime to y\} \Rightarrow R = \{(2, 3), (2, 7), (3, 7), (3, 10), (4, 7), (5, 3), (5, 7)\}$

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

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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:

(d) i ϕ 1

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

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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:

(c) {2,4,6}

The relation R is defined as
$R = \{(x, y) : x, y \in N and x + 2 y = 8\} \Rightarrow R = \{(x, y) : x, y \in N and y = \frac{(8 - x)}{2}\}$

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:

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

The relation R is defined by
$R = \{(x, y) : x \in \{11, 12, 13\}, y \in \{8, 10, 12\} : y = x - 3\} \Rightarrow R = \{(11, 8), (13, 10)\} So, R^{- 1} = \{(8, 11), (10, 13)\}$

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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:

(b) reflexive

$Reflexivity : Since (a, a) \in R \forall a \in A, R is reflexive on A . Symmetry : Since (a, b) \in R but (b, a) \notin R, R is not symmetric on A . \Rightarrow R is not antisymmetric on A . Also, R is not an identity relation on A .$

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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:

(c) transitive

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

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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:

(b) S ⊂ R

Since R is the largest equivalence relation on set A,
$R \subseteq A \times A$

Since S is any relation on A,
$S \subset A \times A$

So, S ⊂ R

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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:

(d) none of these

The relation R is defined as
$R = \{(x, y) : x, y \in A : y = 3 x\} \Rightarrow R = \{(1, 3), (2, 6), (3, 9)\}$

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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:

(d) all the three options

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

Hence, R is an equivalence relation on A.

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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:

(a) symmetric and transitive only

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

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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:

(c) transitive only

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.

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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:

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

$Reflexivity : Clearly, (a, a) \in R \forall a \in A So, R is reflexive on A . Symmetry : Since (1, 2) \in R, but (2, 1) \notin R, R is not symmetric on A . Transitivity : S ince, (1, 3), (3, 2) \in R and (1, 2) \in R, R is transitive on A .$

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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:

(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:

(c) an equivalence relation

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

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:

(d) an equivalence relation

Reflexivity: Let a $\in$ R
Then,

$a a = a^{2} > 0 \Rightarrow (a, a) \in R \forall a \in R$

So, S is reflexive on R.

Symmetry: Let (a, b) $\in$ S
Then,

$(a, b) \in S \Rightarrow a b \geq 0 \Rightarrow b a \geq 0 \Rightarrow (b, a) \in S \forall a, b \in R$

So, S is symmetric on R.

Transitivity:

$If (a, b), (b, c) \in S \Rightarrow a b \geq 0 and b c \geq 0 \Rightarrow a b \times b c \geq 0 \Rightarrow a c \geq 0 [∵ b^{2} \geq 0] \Rightarrow (a, c) \in S for all a, b, c \in set R$

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:

(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:

We have,

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

$As, (a, a) \in R \forall a \in A So, R is reflexive relation Also, (1, 2) \in R but (2, 1) \notin R So, R is not symmetric relation And, (1, 2) \in R, (2, 3) \in R and (1, 3) \in R So, R is transitive relation$

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:

We have,

S = {(a, b) : a $\geq$ b; a, b $\in$ R}

$As, a = a \forall a \in R \Rightarrow (a, a) \in S So, S is reflexive relation Let (a, b) \in S \Rightarrow a \geq b But b \leq a \Rightarrow (b, a) \notin S So, S is not symmetric relation Let (a, b) \in S and (b, c) \in S \Rightarrow a \geq b and b \geq c \Rightarrow a \geq c \Rightarrow (a, c) \in S So, S is transitive relation$

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:

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

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:

We have,

R = {(m, n) : n divides m; m, n $\in$ N}

$As, m divides m \Rightarrow (m, m) \in R \forall m \in N So, R is reflexive Since, (2, 1) \in R i . e . 1 divides 2 but 2 cannot divide 1 i . e . (2, 1) \notin R So, R is not symmetric Let (m, n) \in R and (n, p) \in R . Then, n divides m and p divides n \Rightarrow p divides m \Rightarrow (m, p) \in R So, R is transitive$

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:

$We have, R = \{(l, m) : l is perpendicular to m; l, m \in L\} As, l is not perpencular to l \Rightarrow (l, l) \notin R So, R is not reflexive relation Let (l, m) \in R \Rightarrow l is perpendicular to m \Rightarrow m is also perpendicular to l \Rightarrow (m, l) \in R So, R is symmetric relation Let (l, m) \in R and (m, n) \in R \Rightarrow l is perpendicular to m and m is perpendicular to n \Rightarrow l is parallel to n (Lines perpendicular to same line are parallel) \Rightarrow (m, l) \notin R So, R is not transitive relation$

Hence, the correct alternative is option (b).

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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:

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

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:

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

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:

$We have, R = \{(x, y) : x - y + \sqrt{2} is an irrational number; x, y \in R\} As, x - x + \sqrt{2} = \sqrt{2}, which is an irrational number \Rightarrow (x, x) \in R So, R is reflexive relation Since, (\sqrt{2}, 2) \in R i . e . \sqrt{2} - 2 + \sqrt{2} = 2 \sqrt{2} - 2, which is an irrational number but 2 - \sqrt{2} + \sqrt{2} = 2, which is a rational number \Rightarrow (2, \sqrt{2}) \notin R So, R is not symmetric relation Also, (\sqrt{2}, 2) \in R and (2, 2 \sqrt{2}) \in R i . e . \sqrt{2} - 2 + \sqrt{2} = 2 \sqrt{2} - 2, which is an irrational number and 2 - 2 \sqrt{2} + \sqrt{2} = 2 - \sqrt{2}, which is also an irrational number But \sqrt{2} - 2 \sqrt{2} + \sqrt{2} = 0, which is a rational number \Rightarrow (\sqrt{2}, 2 \sqrt{2}) \notin R So, R is not transitive relation$

Hence, the correct alternative is option (a).

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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:

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).

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Question 1:

If R = $\{(x, y) : x^{2} + y^{2} \leq 4, x, y \in Z\}$ is a relation in Z, then the domain of R is ______________________.

Answer:

Given: R = $\{(x, y) : x^{2} + y^{2} \leq 4, x, y \in Z\}$

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 = $\{(x, y) : x^{2} + y^{2} \leq 4, x, y \in Z\}$ is a relation in Z, then the domain of R is {−2, −1, 0, 1, 2}.

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Question 2:

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

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}.

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Question 3:

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

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.

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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:

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}.

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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:

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}.

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Question 6:

If R s a relation defined on set A ={1, 2, 3} by the rule (a,b) $\in R \Leftrightarrow |a^{2} - b^{2}| \leq 5,$ then R^-1=____________________.

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) $\in R \Leftrightarrow |a^{2} - b^{2}| \leq 5,$ then R^-1= R.

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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:

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)}.

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Question 8:

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

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:

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)}.

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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:

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, ....}.

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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:

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, ....}.

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Question 12:

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

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:

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:

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

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Question 2:

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

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:

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

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Question 3:

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

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:

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)}

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Question 5:

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

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}.

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Question 6:

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

Answer:

Here, R is symmetric on the set A.

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

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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:

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)}

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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:

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:

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:

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

i.e. R is reflexive $\Leftrightarrow (a, a) \in R for all a \in A$

Page No 1.33:

Question 11:

Define a symmetric relation.

Answer:

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

$(a, b) \in R \Rightarrow (b, a) \in R for all a, b \in A i . e . a R b \Rightarrow b R a for all a, b \in A$

Page No 1.33:

Question 12:

Define a transitive relation.

Answer:

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

$(a, b) \in R and (b, c) \in R \Rightarrow (a, c) \in R for all a, b, c \in R i . e . a R b and b R c \Rightarrow a R c for all a, b, c \in R$

Page No 1.33:

Question 13:

Define an equivalence relation.

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:

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

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:

Since R = {(x, y) : y is one half of x; x, y $\in$ A}

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:

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

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:

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

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:

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:

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

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

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:

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:

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

$As, (a, a) \in R \forall a \in A So, R is a reflexive relation Also, (a, b) \in R and (b, a) \in R So, R is a symmetric relation as well And, (0, 1) \in R but (1, 2) \notin R and (2, 3) \notin R So, R is not a transitive relation$

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:

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:

As, R = {(a, b) : 2a + 3b = 30; a, b $\in$ N}

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:

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

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