RD Sharma Class 12 Chapter 1 Exercise 1.2 Relation Solutions Maths

RD Sharma Class 12 Chapter 1 Exercise 1.2 Relation Solutions Maths - Download PDF Free Online

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The RD Sharma Solutions for Class 12 Maths are the best to buy. The chapters and exercises in RD Sharma Class 12 are what a student might find difficult to understand. RD Sharma Class 12th Exercise 1.2 Solution will make it easy for them. Students can now solve every problem of Relation with the help of these solutions.

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RD Sharma Class 12 Solutions Chapter 1 Relations - Other Exercise
Relations Excercise: 1.2
Chapter-wise RD Sharma Class 12 Solutions

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RD Sharma Class 12 Solutions Chapter 1 Relations - Other Exercise

Chapter 1 Relations Ex 1.1

Relations Excercise: 1.2

Relation Exercise 1.2 Question 1

Answer: R is an equivalence relation on Z.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

G i v e n : R = {(a, b) : a - b i s d i v i s i b l e b y 3; a, b ϵ Z}

Explanation:
Let us check the properties on R.
Reflexivity:
Let a be an arbitrary element of R.
Then

a - a = 0 = 0 \times 3

\Rightarrow a - a i s d i v i s i b l e b y 3

\Rightarrow (a, a) ϵ R f o r a l l, a ϵ Z

So, R is reflexive on Z.
Symmetry:

L e t a, b ϵ R

\Rightarrow a - b i s d i v i s i b l e b y 3

a - b = 3 p f o r s o m e p ϵ Z

Here,

\Rightarrow b - a = 3 (- p) i s d i v i s i b l e b y 3 f o r s o m e - p ϵ Z

(b, a) ϵ R f o r a l l a, b ϵ Z

So, R is symmetric on Z.
Transitivity:

L e t (a, b) ϵ R

\Rightarrow a - b i s d i v i s i b l e b y 3 f o r s o m e p ϵ Z

Here,

\Rightarrow b - a = 3 (- p) i s d i v i s i b l e b y 3, f o r s o m e - p ϵ Z

(b, a) ϵ R f o r a l l a, b ϵ Z

So, R is symmetric on Z.

L e t (a, b) a n d (b, c) ϵ R

a - b a n d b - c a r e d i v i s i b l e b y 3 f o r s o m e p ϵ Z

. . . . (i)

b - c = 3 q f o r s o m e q ϵ Z

. . . . (i i)

Adding eq. (i) and (ii)

a - b + b - c = 3 p + 3 q

a - c = 3 (p + q)

Here,

p + q ϵ Z

\Rightarrow a - c i s d i v i s i b l e b y 3

\Rightarrow (a, c) ϵ R f o r a l l a, c ϵ Z

So, R is transitive on Z
Therefore, R is reflective, symmetric and transitive.
Hence, R is an equivalence relation on Z.

Relation Exercise1.2 Question 2

Answer: R is an equivalence relation on Z.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

G i v e n : R = {(a, b) : 2 divides a - b} i s a r e l a t i o n d e f i n e d o n Z .

Explanation:
Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of the set Z

T h e n, a - a = 0 = 0 \times 2

\Rightarrow 2 d i v i d e s a - a

\Rightarrow (a, a) ϵ R f o r a l l a ϵ Z .

So, R is reflexive on Z.
Symmetry:

L e t (a, b) ϵ R

\Rightarrow 2 d i v i d e s a - b

\Rightarrow \frac{a - b}{2} = p f o r s o m e p ϵ Z

\Rightarrow \frac{b - a}{2} = - p

H e r e, - p ϵ Z

\Rightarrow 2 d i v i d e s b - a

(b, a) ϵ R f o r a l l a, b ϵ Z

So, R is symmetric on Z
Transitivity:

L e t (a, b) a n d (b, c) ϵ R

\Rightarrow 2 d i v i d e s a - b a n d 2 d i v i d e s b - c

\Rightarrow \frac{a - b}{2} = p

. . . (i)

\Rightarrow \frac{b - c}{2} = q

. . . (i i)

f o r s o m e p, q ϵ Z

Adding eq. (i) and (ii)

\frac{a - b}{2} + \frac{b - c}{2} = p + q

\frac{a - c}{2} = p + q

H e r e, p + q ϵ Z

\Rightarrow 2 d i v i d e d a - c

(a, c) ϵ R f o r a l l a, c ϵ Z

So, R is transitive on Z
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.

Relation Exercise 1 Question 3

Answer:R is an equivalence relation on Z.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive,symmetric and transitive.

G i v e n : R = {(a, b) : (a - b) i s d i v i s i b l e b y 5} i s a r e l a t i o n d e f i n e d o n Z

.
Explanation:
Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of the set Z

\Rightarrow a - a = 0 = 0 \times 5

\Rightarrow a - a i s d i v i s i b l e b y 5

(a, a) ϵ R f o r a l l a ϵ Z .

So, R is reflexive on Z.
Symmetry:

L e t (a, b) ϵ R

\Rightarrow a - b i s d i v i s i b l e b y 5

a - b = 5 p f o r s o m e p ϵ Z

t h e n b - a = 5 (- p)

H e r e, - p ϵ Z [∵ p ϵ Z]

b - a i s d i v i s i b l e b y 5

(b, a) ϵ R f o r a l l a, b ϵ Z

So, R is symmetric on Z
Transitivity:

Let (a, b) and (b, c) ϵ R

\Rightarrow a - b i s d i v i s i b l e b y 5

a - b = 5 p f o r s o m e p ϵ Z

. . . (i)

A l s o, b - c i s d i v i s i b l e b y 5

. . . (i i)

b - c = 5 q f o r s o m e q ϵ Z

Adding eq. (i) and (ii)

a - b + b - c = 5 p + 5 q

a - c = 5 (p + q)

a - c i s d i v i s i b l e b y 5

H e r e, p + q ϵ Z

(

a, c) ϵ R i s t r a n s i t i v e o n Z

Therefore R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.

Relation Exercise1.2 Question 4

Answer: R is an equivalence relation on Z.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

G i v e n : R = {(a, b) : a - b i s d i v i s i b l e b y n} i s a r e l a t i o n d e f i n e d o n Z

Explanation:
Let us check these properties on R
Reflexivity:

L e t a ϵ N

H e r e, a - a = 0 \times n

a - a i s d i v i s i b l e b y n

(a, a) ϵ R f o r a l l a ϵ Z

So, R is reflexive on Z
Symmetry:

L e t (a, b) ϵ R

H e r e, a - b i s d i v i s i b l e b y n

a - b = n p f o r s o m e p ϵ Z

b - a = n (- p)

b - a i s d i v i s i b l e b y n [i f p ϵ Z, t h e n - p ϵ Z]

(b, a) ϵ R

So, R is symmetric on Z.
Transitivity:

L e t (a, b) a n d (b, c) ϵ R

H e r e, a - b i s d i v i s i b l e b y n a n d b - c i s d i v i s i b l e b y n

a - b = n p f o r s o m e p ϵ Z . . . (i)

b - c = n q f o r s o m e q ϵ Z . . . . (i i)

Adding eq. (i) and (ii)

a - b + b - c = n p + n q

a - c = n p + q [H e r e, p + q ϵ Z]

(a, c) ϵ R f o r a l l a, c ϵ Z

So, R is transitive on Z.
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.

Relation Exercise1.2 Question 5

Answer: R is an equivalence relation on Z
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetry and transitive.
Explanation:

G i v e n : Z b e s e t o f i n t e g e r s a n d R = {(a, b) : a, b ϵ Z a n d a + b i s e v e n}

Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of Z.

T h e n, a + a = 2 a i s e v e n f o r a l l a ϵ Z .

(a, a) ϵ R f o r a l l a ϵ Z .

So, R is reflexive on Z.
Symmetry:

L e t (a, b) ϵ R

t h e n, a + b i s e v e n

\Rightarrow b + a i s e v e n

(b, a) ϵ R f o r a l l a, b ϵ Z

So, R is symmetric on Z.
Transitivity:

L e t (a, b) a n d (b, c) ϵ R

t h e n, a + b a n d b + c a r e e v e n

N o w, l e t a + b = 2 x f o r s o m e x ϵ Z . . . (i)

b + c = 2 y f o r s o m e y ϵ Z . . . (i i)

Adding (i) and (ii)

\Rightarrow a + b + b + c = 2 x + 2 y

\Rightarrow a + 2 b + c = 2 x + 2 y

\Rightarrow a + c = 2 x + 2 y - 2 b

\Rightarrow a + c = 2 (x + y - b), w h i c h i s e v e n f o r a l l x, y, b ϵ Z

T h u s, (a, c) ϵ R

So, R is transitive on Z
Therefore, R is reflexive, symmetry and transitive.
Hence, R is an equivalence relation on Z.

Relation Exercise 1.2 Question 6

Answer: R is an equivalence relation on Z
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetry and transitive.

G i v e n : R = {(m, n) : m - n i s d i v i s i b l e b y 13} b e a r e l a t i o n o n Z

Explanation:
Let us check these properties on R.
Reflexivity:
Let m be an arbitrary element of Z.

T h e n, m - m = 0 = 0 \times 13

\Rightarrow m - m i s d i v i s i b l e b y 13.

\Rightarrow (m, m) ϵ R

Hence, R is reflexive on Z.
Symmetry:

L e t (m, n) ϵ R

T h e n, m - n i s d i v i s i b l e b y 13

m - n = 13 p

H e r e, p ϵ Z

n - m = 13 (- p)

H e r e, - p ϵ Z

n - m i s d i v i s i b l e b y 13

(n, m) ϵ R f o r a l l m, n ϵ Z

So, R is symmetric on Z.
Transitivity:

L e t (m, n) a n d (n, o) ϵ R

m - n a n d n - o a r e d i v i s i b l e b y 13

m - n = 13 p f o r s o m e p ϵ Z . . . (i)

n - o = 13 q f o r s o m e q ϵ Z . . . (i i)

Adding (i) and (ii)

m - n + n - o = 13 p + 13 q

m - o = 13 (p + q)

H e r e, p + q ϵ Z

m - o i s d i v i s i b l e b y 13

(m, o) ϵ R f o r a l l m, o ϵ Z

So, R is transitive on Z
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.

Relation Exercise 1.2 Question 7

Answer: R is an equivalence relation.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
Explanation:
Given: R be a relation on A

S e t A o f o r d e r p a i r o f i n t e g e r s d e f i n e d b y (x, y) R (u, v) i f x v = y u

Reflexivity:

L e t (a, b) b e a n a r b i t r a r y e l e m e n t o f t h e s e t A .

T h e n, (a, b) ϵ A

a b = b a

(a, b) R (a, b)

Thus, R is reflexive on A.
Symmetry:

Let (x, y) and (u, v) ϵ A such that (x, y) R(u, v), Then

x v = y u

\Rightarrow v x = u y

\Rightarrow u y = v x

(u, v) R (x, y)

So, R is symmetric on A.
Transitivity:

L e t (x, y), (u, v) a n d (p, q) ϵ R s u c h t h a t (x, y) R (u, v) a n d (u, v) R (p, q)

x v = y u . . . (i)

u q = v p . . . (i i)

Multiply eq. (i) and (ii)

x v \times u q = y u \times v p

x v u q = y u v p

cancelling out vu it is common on both sides

x q = y p

(x, y) R (p, q)

So, R is transitive on A.
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on A.

Relation Exercise 1.2 Question 8

Answer:R is an equivalence relation and the set of all elements related to 1 is 1.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

G i v e n : S e t A = {x ϵ Z; 0 \leq x \leq 12}

A l s o g i v e n t h a t r e l a t i o n R = {(a, b) : a = b} i s d e f i n e d o n s e t A

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

a = a [S i n c e, e v e r y e l e m e n t i s e q u a l t o i t s e l f]

(a, a) ϵ R f o r a l l a ϵ A

So, R is reflexive on A
Symmetry:

L e t (a, b) ϵ R

a = b

b = a

(b, a) ϵ R f o r a l l a, b ϵ A

So, R is symmetric on A.
Transitivity:

L e t a, b a n d (b, c) ϵ R

a = b . . . (i) a n d b = c . . . (i i)

multiplying eqn (i) and (ii), we get

a b = b c

a = c

t h e r e f o r e, (a, c) ϵ R

So, R is transitive on A.
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on A.

R = {(a, b) : a = b} a n d 1 i s o n e l e m e n t o f A .

R = {(1, 1) : 1 = 1}

Thus, the set of all elements related to 1 is 1.

Relation Exercise 1.2 Question 9

Answer: R is an equivalence relation.

T h e s e t o f l i n e p a r a l l e l t o t h e l i n e y = 2 x + 4

y = 2 x + C f o r a l l C ϵ R w h e r e R i s t h e s e t o f r e a l n u m b e r s .

Hint:To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
Given: We have, L is the set of lines

R = {(L 1, L 2) : L 1 i s p a r a l l e l t o L 2} b e a r e l a t i o n o n L .

Explanation:
Now,
Reflexivity:

L e t L 1 ϵ L

Since, one line is always parallel to itself.

(L 1, L 1) ϵ R

R is reflexive.
Symmetric:

L e t, L 1, L 2 ϵ L a n d (L 1, L 2) ϵ R

L 1 i s p a r a l l e l t o L 2

L 2 i s p a r a l l e l t o L 1

(L 2, L 1) ϵ R

R is symmetric
Transitive:

L e t, L 1, L 2 a n d L 3 ϵ L s u c h t h a t (L 1, L 2) ϵ R a n d (L 2, L 3) ϵ R

L 1 i s p a r a l l e l t o L 3 a n d L 2 i s p a r a l l e l t o L 3

L 1 i s p a r a l l e l t o L 3

(L 1, L 3) ϵ R

R is transitive
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.
The set of lines parallel to the line

y = 2 x + 4

y = 2 x + C f o r a l l C ϵ R

where R is the set of real numbers.

Relation Exercise 1.2 Question 10

Answer: R is an equivalence relation.
The set of all elements in A related to triangle T is the set of all triangles.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

G i v e n : R = {(P_{1}, P_{2}) : P_{1} a n d P_{2} h a v e s a m e n u m b e r o f s i d e s}

Explanation:
Reflexive:

R i s r e f l e x i v e since (P_{1}, P_{1}) ϵ R is t h e s a m e p o l y g o n a s i t s e l f .

Symmetric:

L e t (P_{1}, P_{2}) ϵ R

P_{1} and P_{2} h a v e t h e s a m e n u m b e r s o f s i d e s .

P_{2} and P_{1} have t h e s a m e n u m b e r o f s i d e .

(P_{2}, P_{1}) ϵ R

R is symmetric
Transitive:

L e t (P_{1}, P_{2}), (P_{2}, P_{3}) ϵ R

P_{1} and P_{2} h a v e t h e s a m e n u m b e r o f s i d e s

A l s o P_{2} a n d P_{3} h a v e t h e s a m e n u m b e r o f s i d e s .

P_{1} a n d P_{3} h a v e t h e s a m e n u m b e r o f s i d e s

(P_{1}, P_{3}) ϵ R

R is transitive.
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.
The elements in A related to the right-angle triangle (T) with sides 3, 4, and 5 are those polygons that have 3 sides (since T is a polygon with 3 sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.

Relation Exercise 1.2 Question 11

Answer: R is an equivalence relation on A
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

G i v e n : O b e t h e o r i g i n a n d R = {(P, Q) : O P = O Q} b e a r e l a t i o n o n A w h e r e O i s t h e o r i g i n

.
Explanation:
Let A be a set of points on a plane.
Reflexivity:

L e t P ϵ A

S i n c e, O P = O P = (P, P) ϵ R

R is reflexive.
Symmetric:

L e t (P, Q) ϵ R f o r P, Q ϵ A

T h e n O P = O Q

O Q = O P

(Q, P) ϵ R

R is symmetric
Transitive:

L e t (P, Q) ϵ R a n d (Q, S) ϵ R

O P = O Q . . . (i) a n d O Q = O S . . . . (i i)

Putting (ii) in (i), we get

O P = O S

(P, S) ϵ R

R is transitive
Therefore, R is reflexive, symmetric and transitive.
Thus, R is an equivalence relation on A.

Relation Exercise 1.2 Question 12

Answer: R is an equivalence relation on A.
No number of the subset {1, 3, 5, 7} is related to any number of the subset {2, 4, 6}.
Hint: To prove an equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

G i v e n : {A = 1, 2, 3, 4, 5, 6, 7}

R = {(a, b) : b o t h a a n d b a r e e i t h e r o d d o r e v e n n u m b e r}

Explanation:

R = {(1, 1), (1, 3), (1, 5), (1, 7), (3, 1), (3, 3), (3, 5), (3, 7), (5, 1), (5, 3), (5, 5), (5, 7), (7, 1), (7, 3), (7, 5) (7, 7), (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2) (6, 4), (6, 6)}

We observe the following properties of R on A.
Reflexivity:

C l e a r l y (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7) ϵ R

.
So, R is a reflexive relation in A
Symmetric:

L e t a, b ϵ A be such that (a,b) ϵ R .

T h e n (a, b) ϵ R

Both a and b are either odd or even.
Both b and a are either odd or even.

(b, a) ϵ R

T h u s, (a, b) ϵ R

\Rightarrow (b, a) ϵ R f o r a l l a, b ϵ A

So, R is a symmetric relation on A
Transitivity:

L e t a, b, c ϵ A b e s u c h t h a t (a, b) ϵ R, (b, c) ϵ R

T h e n (a, b) ϵ R

Both a and b are either odd or even.

(b, c) ϵ R

⇒ Both b and c are odd or even.

If both a and b are even, then (b, c) ϵ R both b and c are even

If both a and b are odd, then (b, c) ϵ R both b and c are odd

⇒ Both a and c are even or odd.

∴ (a, c) ϵ R

S o, (a, b) ϵ R and (b, c) ϵ R

(a, c) ϵ R

R is a transitive relation on A
Thus, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on A.
We observe that two numbers in A are related if both are odd or both are even.
Since, {1, 3, 5, 7} has all odd numbers of A
So, all the numbers of {1, 3, 5, 7} are related to each other
Similarly, all the numbers of {2, 4, 6} are related to each other as it contains all even numbers of set A.
An even, odd number in A is related to an even, odd number in A respectively.
So, no number of the subset {1, 3, 5, 7} is related to any number of the subset {2, 4, 6}

Relation Exercise 1.2 Question 13

Answer: Hence prove, S is not an equivalence relation on R.
Hint: To prove an equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
Explanation:

G i v e n : S {(a, b) : a^{2} + b^{2} = 1}

Now,
Reflexivity:

L e t a = \frac{1}{2} ϵ R

T h e n, a^{2} + a^{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \neq 1

(a, a) \notin S

S is not reflexive.
When one the three properties are not valid, relation not an equivalence.
Hence, S is not an equivalence relation on R.

Relation Exercise 1.2 Question 14

Answer: R is an equivalence relation on

Z \times Z_{0}

Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

G i v e n : Z b e s e t o f i n t e g e r s a n d Z_{0} b e t h e s e t o f n o n - z e r o i n t e g e r s .

R = {((a, b), (c, d)) : a d = b c} b e a r e l a t i o n o n Z \times Z_{0}

Explanation:
Reflexivity:

(a, b) ϵ Z \times Z_{0}

a b = b a

(a, b), (a, b)) ϵ R

R is reflexive
Symmetric:

L e t ((a, b), (c, d)) ϵ R

a d = b c

c b = d a

((c, d), (a, b)) ϵ R

R is symmetric
Transitive:

L e t ((a, b), (c, d)) ϵ R a n d (c, d), (e, f)) ϵ R

a d = b c a n d c f = d e

\frac{a}{b} = \frac{c}{d} . . . (i) a n d \frac{c}{d} = \frac{e}{f} . . . . (i i)

Using (ii) in (i), we get

\frac{a}{b} = \frac{e}{f}

a f = b c

((a, b), (e, f)) ϵ R

R is transitive
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on

Z \times Z_{0}

Relation Exercise 1.2 Question 15 (i)

Given:

H e n c e p r o v e d, R \cap S a n d R \cup S a r e s y m m e t r i c .

Hint: For symmetric relation.

a=b is true then b=a is also true.

Given: R and S are relations on a set A.
R and S are two symmetric relations on set A.

T o P r o v e : R \cap S i s s y m m e t r i c

Explanation:

L e t (a, b) ϵ R \cap S

(a, b) ϵ R a n d (a, b) ϵ S

(b, a) ϵ R a n d (b, a) ϵ S

(b, a) ϵ R \cap S ∵ [R a n d S a r e s y m m e t r i c]

(b, a) ϵ R \cap S

R \cap S i s s y m m e t r i c

T o p r o v e : R \cup S i s s y m m e t r i c

L e t (a, b) ϵ R \cup S

(a, b) ϵ R o r (a, b) ϵ S

(b, a) ϵ R o r (b, a) ϵ S

(b, a) ϵ R \cup S ∵ [R a n d S a r e s y m m e t r i c]

R \cup S i s s y m m e t r i c

H e n c e p r o v e d, R \cap S a n d R \cup S a r e s y m m e t r i c .

Relation Exercise 1.2 Question 15 (ii)

Given: $H e n c e p r o v e d, R \cup S i s r e f l e x i v e .$
Hint: $(a, a) ϵ R \forall a ϵ X o r a s I \subseteq R, where, I, is the identity relation on A.$
Explanation:
Given: R and S are two relations on A such that R is reflexive.
To Prove: $R \cup S i s r e f l e x i v e .$
$S u p p o s e R \cup S i s n o t r e f l e x i v e .$
$This means that there is and R \cup S such that (a, a) \notin R \cup S$
$S i n c e, a ϵ R \cup S,$
$a ϵ R o r a ϵ S$
$I f a ϵ R, t h e n (a, a) ϵ R [R i s r e f l e x i v e]$
$(a, a) ϵ R \cup S$
$H e n c e, R \cup S i s r e f l e x i v e .$

Relation Exercise 1.2 Question 16

Answer:

R \cup S i s n o t t r a n s i t i v e

Hint: For transitive relation

x = y a n d y = z, t h e n a l s o x = z .

Given:

R and S are transitive relation on a set A

To Prove:

R \cup S may not be a transitive relation on A

Explanation:
We will prove this by means of an example.

L e t A = {a, b, c} b e a s e t

R = {(a, a), (b, b), (c, c), (a, b), (b, a)} a n d

S = {(a, a), (b, b), (c, c), (b, c), (c, b)} a r e t w o r e l a t i o n s o n A

Clearly, R and S are transitive relation on A.

N o w, R \cup S = {(a, a), (b, b), (c, c), (a, b), (b, a), (b, c), (c, b)}

H e r e, (a, b) ϵ R \cup S a n d (b, c) ϵ R \cup S

b u t (a, c) \notin R \cup S

R \cup S i s n o t t r a n s i t i v e

Relation Exercise 1.2 Question 17

Answer: Hence prove, R is an equivalence relation.
Hint: To prove an equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
Explanation:
Given:

C\: be\: the\: set\: o\! f\: all\: complex \: numbers.

C_{0} be\: the\: set\: o\! f\: all\: non-zero\: complex\: numbers.

Relation R on C_{0} be 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} ϵ C_{0}

(i) Test for reflexivity

S i n c e, \frac{z_{1} - z_{1}}{z_{1} + z_{1}} = 0, w h i c h i s a r e a l n u m b e r

S o, (z_{1}, z_{1}) ϵ R

Hence, R is a reflexive relation.
(ii) Test for symmetric

\frac{z_{1} - z_{2}}{z_{1} + z_{2}} = x, w h e r e x i s r e a l

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

\frac{z_{2} - z_{1}}{z_{2} + z_{1}} = - x i s a l s o a r e a l n u m b e r

S o, (z_{2}, z_{1}) ϵ R

Hence, R is a symmetric relation.
(iii) Test for transitivity

L e t (z_{1}, z_{2}) ϵ R a n d (z_{2}, z_{3}) ϵ R

T h e n \frac{z_{1} - z_{2}}{z_{1} + z_{2}} = x

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

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

. . . (i)

A l s o, \frac{z_{2} - z_{3}}{z_{2} + z_{3}} = y

z_{2} - z_{3} = y z_{2} + y z_{3}

z_{2} (1 - y) = z_{3} (1 + y)

\frac{z_{2}}{z_{3}} = \frac{1 + y}{1 - y}

. . . (i i)

Dividing (i) and (ii), we get

\frac{z_{1}}{z_{3}} = \frac{1 + x}{1 - x} \times \frac{1 + y}{1 - y}

= z w h e r e z i s a r e a l n u m b e r .

\frac{z_{1} - z_{3}}{z_{1} + z_{3}} = \frac{z - 1}{z + 1}, w h i c h i s r e a l

(z_{1}, z_{3}) ϵ R

. . . (i i i)

From (i), (ii) & (iii)

Hence, R\: is\: transitivity\: relation.

Hence\: proved, R \: is\: an\: equivalence \: relation.

Chapter-wise RD Sharma Class 12 Solutions

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5. What are the different types of relations?

There are three types of relations, and these are void, universal, and identity relations.

You can learn more about these relations by RD Sharma Class 12 Chapter 1 Exercise 1.2.