Class 12 RD Sharma Solutions Chapter 1 Relations Exercise 1.2 | Set 1 (original) (raw)

Last Updated : 3 Sep, 2024

Question 1. Show that the relation R = {(a,b): a-b is divisible by 3;, a, b ∈ Z} is an equivalence relation.

**Solution:

According to question, relation R = {(a,b): a-b is divisible by 3;, a, b ∈ Z}

We have to show that R is an equivalence relation.

****(i)** reflexibity:

let a = z

=> a - a = 0

=> 0 is divisible by 3

:. a is divisible by 3

(a, a) ∈ R,

Hence, relation R is reflexive.

****(ii)** symmetry:

Let a, b∈ Z and (a, b) ∈ R

=> a - b is divisible by 3

=> a - b = 3x, for some x ∈ Z

=> b -a = 3 (-x)

here, -x ∈ Z

:. b - a is divisible by 3

Hence, (b- a) ∈ R for all a, b ∈ Z

So, relation R is symmetric.

****(iii)** transitivity:

let (a, b) and (b, c) ∈ R

a - b and b - c is divisible by 3

a - b = 3x, for some x ∈ Z & ****(eqn. 1)**

b - c = 3y, for some y ∈ Z ****(eqn. 2)**

Adding the above equations 1 and 2, we get

a - c = 3(x + y)

here, x and y ∈ Z

so, a - c is divisible by 3

Hence, a - c ∈ R for all a, c ∈ Z

thus relation R is transitive.

As we know if a relation is reflexive, symmetric and transitive at the same time, then it is called an equivalence relation.

Therefore, Relation R is an equivalence relation.

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

**Solution:

Given, R = {(a, b): 2 divides a - b} which is defined on Z

We have to show that relation R is an equivalence relation

(i) reflexibility :

let na be an arbitrary element on set Z

then a ∈ R

a - a = 0

=> 0 X 2 = 0

2 divides a - a

=> (a, a) ∈ R

****:.** Relation R is a reflexive relation.

Now, (ii) symmetric :

let (a, b) ∈ R

2 divides a - b

a - b = 2x for some x ∈ Z

b - a = 2 (- x) where - x ∈ Z

2 divides b - a

=> (b, a) ∈ R

and (a, b) ∈ R

therefore, relation R is symmetric.

(iii) transitivity :

let a, b, c ∈ Z such that (a, b) ∈ R and (b, c) ∈ R

then, (a, b) ∈ R => 2 divides b - a

b - a = 2x for some x ∈ Z ****(eqn.1)**

and(b - c) ∈ R

2 divides c - b => c - b = 2y for some y ∈ Z ****(eqn 2)**

on solving equation 1 and 2,

c - a = 2 (x + y) where x + y ∈ Z

2 divides c - a

(a, c) ∈ R

Therefore, relation R is transitive.

Hence, relation R is an equivalence relation.

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.

**Solution:

Given, relation R on Z defined by (a, b) ∈ R <=> a - b is divisible by 5

as, we have to prove it a equivalence relation, the relation R must have to be reflexive, symmetric as well as transitive.

(i) Reflexibility :

Let a be an arbitrary element of R

=> a - a = 0

=> 0 is divisible by 5

=> a - a is divisible by 5

=> (a, a) ∈ R for all a ∈ Z

****:.** relation R is reflexive.

Again, (ii) symmetry

Let (a, b) ∈ R

=> a − b is divisible by 5

=> a − b = 5x for some x ∈ Z , b − a = 5 (−x)

since (−x) ∈ Z

b − a is divisible by 5

(b, a) ∈ R for all a, b ∈ Z

So, relation R is symmetric.

(iii) transitivity ;

Let (a, b) and (b, c) ∈ R

=> a − b is divisible by 5

=>a − b = 5x for some x ∈ Z ****(eqn. 1)**

Also, b − c is divisible by 5

=> b − c = 5y for some y ∈ Z ****(eqn. 2)**

Adding the above two equations,

a −b + b − c = 5x + 5y

=> a − c = 5 (x + y)

=> a − c is divisible by 5

Since, x + y ∈ Z

=>(a, c) ∈ R for all a, c ∈ Z

So, R is transitive on Z.

As relation R is reflexive, symmetric and transitive,

Hence, R is an equivalence relation on Z.

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.

**Solution:

Given (a, b) ∈ R ⇔ a − b is divisible by n is a relation R defined on Z.

it is necessary that the relation R should be reflexive, symmetric and transitive.

(i) Reflexivity:

Let a ∈ N

a − a = 0

= 0 × n

=> a − a is divisible by n

=> (a, a) ∈ R

=> (a, a) ∈ R for all a ∈ Z

****:.**R is reflexive on Z.

(ii) Symmetry:

Let (a, b) ∈ R

a − b is divisible by n

=> a − b = nx for some x ∈ Z

=> b − a = n (−x)

=> b − a is divisible by n

=> (b, a) ∈ R

So, relation R is symmetric on Z.

Similarly (iii) Transitivity:

Let (a, b) and (b, c) ∈ R

a − b is divisible by n and b − c is divisible by n.

=> a − b= n x for some x ∈ Z (eqn . 1)

And b−c = ny for some y ∈ Z (eqn. 2)

a – b + b − c = nx + n y

=> a − c = n (p + q)

=> (a, c) ∈ R for all a, c ∈ Z

So, relation R is transitive on Z.

Therefore, R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z.

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.

**Solution:

Given R = {(a, b): a, b ∈ Z and a + b is even} is a relation defined on R.

Also given that Z be the set of integers

it is necessary that the given relation should be reflexive, symmetric and transitive.

Reflexivity:

Let a be an arbitrary element of Z.

Then, a ∈ R

a + a = 2a is even for all a ∈ Z.

=> (a, a) ∈ R for all a ∈ Z

So, R is reflexive on Z.

(ii) Symmetry:

Let (a, b) ∈ R

=> a + b is even

=> b + a is even

=>(b, a) ∈ R for all a, b ∈ Z

So, R is symmetric on Z.

(iii) Transitivity:

Let (a, b) and (b, c) ∈ R

=> a + b and b + c are even

let a + b = 2x for some x ∈ Z (eqn. 1)

And b + c = 2y for some y ∈ Z ****(eqn. 2)**

Adding the above two equations, we get

A + 2b + c = 2x + 2y

=> a + c = 2 (x + y − b), which is even for all x, y, b ∈ Z

Thus, (a, c) ∈ R

So, R is transitive on Z.

Therefore, relation R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z

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?

**Solution:

Given that m is said to be related to n if m and n are integers and m − n is divisible by 13

So, we have to check whether the given relation is equivalence or not.

Let R = {(m, n): m, n ∈ Z : m − n is divisible by 13}

Reflexivity:

Let m be an arbitrary element of Z.

Then, m ∈ R

=> m − m = 0 = 0 × 13

=> m − m is divisible by 13

=> (m, m) is reflexive on Z.

Now, Symmetry:

Let (m, n) ∈ R.

Then, m − n is divisible by 13

=> m − n = 13p

Here, p ∈ Z

=>n – m = 13 (−p)

=> n − m is divisible by 13

=>(n, m) ∈ R for all m, n ∈ Z

So, R is symmetric on Z.

Transitivity:

Let (m, n) and (n, o) ∈ R

=>m − n and n − o are divisible by 13 ****(eqn. 1)**

=>m – n = 13p and n − o = 13q for some p, q ∈ Z ****(eqn.2)**

Adding the above two equations, we get

=>m – n + n − o = 13p + 13q

=> m−o = 13 (p + q)

=> m − o is divisible by 13

=>(m, o) ∈ R for all m, o ∈ Z

So, R is transitive on Z.

Therefore, R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z.

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 = y u. Show that R is an equivalence relation.

**Solution:

First let R be a relation on A

It is given that set A of ordered pair of integers defined by (x, y) R (u, v) if xv = y u

We have to check whether the given relation is equivalence or not.

Reflexivity:

Let (a, b) be an arbitrary element of the set A.

Then, (a, b) ∈ A

=> a b = b a

=>(a, b) R (a, b)

Thus, R is reflexive on A.

Again, Symmetry:

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

x v = y u

=> v x = u y

=> u y = v x

=>(u, v) R (x, y)

So, R is symmetric on A.

Transitivity:

Let (x, y), (u, v) and (p, q) ∈R such that (x, y) R (u, v) and (u, v) R (p, q)

=> x v = y u and u q = v p

Multiplying the corresponding sides, we get

x v × u q = y u × v p

=> 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.

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.

**Solution:

According to question, set A = {x ∈ Z; 0 ≤ x ≤ 12}

Also given that relation R = {(a, b): a = b} is defined on set A

We have to find whether the given relation is equivalence or not.

(i) Reflexivity:

Let a be an arbitrary element of A.

Then, a ∈ R

=>a = a ****(because, every element is equal to itself)**

=> (a, a) ∈ R for all a ∈ A

****:.** R is reflexive on A.

(ii) Symmetry:

Let (a, b) ∈ R

=>b = a

=> (b, a) ∈ R for all a, b ∈ A

So, R is symmetric on A.

(iii) Transitivity:

Let (a, b) and (b, c) ∈ R

=> a =b and b = c

=>a = b c

=>a = c

=> (a, c) ∈ R

So, R is transitive on A.

Hence, relation R is an equivalence relation on A.

Therefore, relation R is reflexive, symmetric and transitive.

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

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.

**Solution:

Given, L is the set of lines.

R = {(L1, L2) : L1 is parallel to L2} be a relation on L

Now, (i) reflexibility :

as we know, a line is always parallel to itself,

So, L1, L2 ∈ R

Therefore, R is reflexive.

Again, (ii) symmetry :

Assume, L1, L2 ∈ L and (L1, L2) ∈ R

Since, L1 is parallel to L2

:. L2 is also parallel to L1

Thus, relation R is symmetric.

(iii) transitivity :

let L1, L2, L3 ∈ R in such a way that (L1, L2) ∈ R and (L2, L3) ∈ R

L1 is parallel to L2

L2 is parallel to L3

Therefore L1 is parallel to L3

Hence, relation R is transitive.

So, relation R is an equivalence relation.

Now, the set of all lines related to line y = 2x + 4 is y = 2x + c for all c ∈ R.

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?

**Solution:

The relation R is defined as R = {(P1, P2):P1 and P2 have the same number of sides}

(i) reflexibility:

Let P be any polygon in A

Then, P and P have same number of sides

(P, P) ∈ R

Thus, relation R is reflexive.

(ii) symmetry:

Let P1 and P2 be any polygon in A such that, (P1, P2 ) ∈ R

(P1, P2) ∈ R

P1 and P2 have same number of sides

so, P2 and P1 will have also same number of sides

Hence (P1, P2) ∈ R

so, relation R is symmetric.

at last (iii) transitivity :

let P1, P2, P3 be three polygons in A in such a way that (P1 , P2) ∈ R and (P2, P3) ∈ R

then, (P1, P2) have same number of sides

(P2, P3) have same number of sides

****:.** P1 and P3 will also have same number of sides.

Hence, relation R is transitive

so, relation R is an equivalence relation

Now, let P be a polygon in A such that (P, T) ∈ R, where T is a right angle triangle with sides 3, 4 and 5

Then, (P, T) ∈ R

polygon P and triangle T have same number of sides

Hence, the set of all elements in A related to T is the set of all triangles in A.

Summary

Chapter 1 of RD Sharma's Class 12 Mathematics textbook deals with Relations. Exercise 1.2 focuses on various aspects of relations, including their properties, types, and representations. Key concepts covered include:

1. Definition of relations

2. Domain and range of relations

3. Types of relations (reflexive, symmetric, transitive, equivalence)

4. Representation of relations using sets, ordered pairs, and graphs

Practice Questions

1. Let A = {1, 2, 3, 4} and R be a relation on A defined by R = {(a, b) : a divides b}. Write R as a set of ordered pairs.

2. Define a relation R on the set of real numbers as xRy if and only if |x - y| ≤ 2. Determine if (3, 5) ∈ R.

3. Let A = {1, 2, 3, 4, 5}. Define a relation R on A as aRb if a + b is odd. Is R reflexive? Justify your answer.

4. For the relation R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3)} on the set A = {1, 2, 3}, determine if R is symmetric.

5. Let R be a relation on Z (set of integers) defined by aRb if a - b is divisible by 3. Prove that R is an equivalence relation.

6. Given the relation R = {(1, 2), (2, 3), (1, 3)} on the set A = {1, 2, 3}, find the transitive closure of R.

7. For A = {1, 2, 3, 4}, define R = {(x, y) : x < y}. Draw the graph of this relation.

8. Let R be a relation on the set of all triangles defined as TRS if triangle T is similar to triangle S. Is R an equivalence relation? Explain.

9. Given A = {1, 2, 3, 4, 5}, define R = {(x, y) : x + y is even}. Find the domain and range of R.

10. Let R be a relation on R (set of real numbers) defined by xRy if x² + y² = 25. Sketch the graph of this relation.