Class 12 RD Sharma Solutions - Chapter 1 Relations - Exercise 1.2 | Set 2

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.

Solution:

Let A be the set of points on plane

and let R = {(P, Q): OP = OQ} be a relation on A where O is the origin.

now (i) reflexibility:

let P ∈ A

since, OP = OP

:. P, P ∈ R

so, relation R is reflexive

(ii) symmetry:

let (P, Q) ∈ R for P, Q ∈ A

=>OQ = OP

:. (Q, P) ∈ R

therefore, relation R is symmetric .

Similarly (iii) transitive:

let (P,Q) ∈ R and (Q,S) ∈ R

OP = OS

(P, S) ∈ R

hence, relation R is transitive.

so, relation R is an equivalence relation on A .

Question 12. Let R be the relation defined on 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 element 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}

Solution:

Given, A={1,2,3,4,5,6,7}

R={(a,b): both a and b are either odd or even}

clearly, (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7) ∈ R

so, relation R is reflexive.

Also, for symmetry,

a, b ∈ A such that (a, b) ∈ R

both a and b are either odd or even

both b and a are either odd or even

=>(b, a) ∈ R

so, relation R is also symmetry

Similarly, for transitivity,

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

both a and b are either odd or even

both b and c are either odd or even

if both a and b are even then,

(b,c) ∈ R => both b and c are even

:. both a and c are even

and if both a and b are odd, then

(b,c) ∈ R =>both b and c are odd

both a and c are odd

Thus, both a and c are even and odd

therefore, (a,c) ∈ R

so, (a,b) ∈ R and (b,c) ∈ R => (a,c) ∈ R

so, relation R is transitive

it proves that R is an equivalence relation.

We observe that {1,3,5,7} are related with each other only and {2,4,6} are related with each other.

Question 13. let set S be a relation on the set of all real numbers defined as S = {(a,b) ∈ R x R: a² + b² = 1}

Solution:

Let us observe the following properties of S

(i) Reflexibility: let a be an arbitrary element of R

a ∈ R

=> a² + a² ≠ 1 for all a ∈ R

so, S is not reflexive on R .

(ii) symmetry:

let (a, b) ∈ R

a² + b² = 1

b² + a² = 1

=> (b, a) ∈ S for all a, b ∈ R

So, S is symmetric on R

(iii) transitivity:

Let (a,b) and (b,c) ∈ S

a² + b² =1 and b² + c² = 1

Adding both the equations we will get,

a² + c² = 2 - 2b² ≠ 1 for all a, b, c ∈ R

So, S is not transitive on R

Hence, S is not an equivalence relation on R

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 x Z₀ be defined as (a,b) R (c,d) <=>ad =bc for all (a,b), (c,d) ∈ Z x Z₀. Prove that R is an equivalence relation on Z x Z₀.

Solution:

Let us observe the properties of R

(i) reflexibility:

let (a,b) be an arbitrary element of Z x Z₀

(a,b) ∈ ZxZ₀

a,b ∈ Z,Z₀

ab = ba

(a,b) ∈ R for all (a,b) ∈ ZxZ₀

R is reflexive

(ii) symmetry:

Let (a,b), (c,d) ∈ ZxZ₀ such that (a,b) R (c,d)

=> ad = bc

=> cb = da

(c,d) R (a,b)

Thus, (a,b) R (c,d) => (c,d) R (a,b) for all (a,b), (c,d) ∈ ZxZ₀

so, R is symmetric.

(iii) transitivity:

Let (a,b), (c,d), (e,f) ∈ NxN₀ such that (a,b) R (c,d) and (c,d) R (e,f)

(a,b) R (c,d) => ad = bc

(c,d) R (e,f) => cf = de

from this, we get (ad) (cf) = (bc) (de)

=> af = be

(a,b) R (e,f)

so, R is transitive.

Hence it is proved that relation R is an equivalence relation.

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.

Solution:

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

=> R ⊂ A x A and S ⊂ A x A

=> R ∩ S ⊂ A x A

thus, R ∩ S is a relation on A

let a, b ∈ A such that (a,b) ∈ R ∩ S

(a,b) ∈ R ∩ S

=> (a,b) ∈ R and (a,b) ∈ S

=> (b,a) ∈ R and (b,a) ∈ S

thus, (a,b) ∈ R ∩ S

=> (b,a) ∈ R ∩ S for all a, b ∈ A

so, R ∩ S is symmetric on A

Also, let a, b ∈ R such that (a,b) ∈ R ∪ S

=> (a,b) ∈ R or (a,b) ∈ S

= (b,a) ∈ R or (b,a) ∈ S [since R and S are symmetric]

=> (b,a) ∈ R ∪ S

hence, R ∪ S is symmetric on A

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

Suppose a ∈ A

then, (a,a) ∈ R [since R is reflexive]

=> (a,a) ∈ R ∪ S

=> R ∪ S is reflexive on A .

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.

Solution:

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) ∈ R ∪ S and (b,c) ∈ R ∪ S

but (a,c) ∉ R ∪ S

Hence, R ∪ S is not a transitive relation on A.

Question 17. Let C be the set of all complex numbers and C₀ be the set of all non - zero complex numbers. Let a relation R and C₀ be defined as Z₁ R Z₂ <=> z₁ - z₂ / z₁ + z₂ is real for all Z₁, Z₂ ∈ C_0. Show that R is an equivalence relation.

Solution:

(i) reflexibility:

since, z₁ - z₂ / z₁ + z₂ = 0 which is a real number

so, (z₁, z₁) ∈ R so,

relation R is reflexive.

(ii)symmetry;

z₁ - z₂ / z₁ + z₂ = x, where x is real number

=> - (z₁ - z₂ / z₁ + z₂ ) = -x

so, (z₂, z₁) ∈ R

Hence, R is symmetric.

(iii) transitivity:

Let (z₁,z₂) ∈ R and (z₂,z₃) ∈ R

then,

z₁-z₂ / z₁+z₂ = x where x is a real number

=> z₁ - z₂ = xz₁ + xz₂

=> z₁ - xz₁ = z₂ + xz₂

=> z₁ (1 - x) = z₂ (1+x)

=> z₁/z₂ = (1+x)/(1-x) .... eqn. (1)

also, z₂-z₃/z₂+z₃ = y where y is a real number

=> z₂ - z₃ = yz₂ + yz₃

=> z₂ - yz₂ = z₃ + yz₃

=> z₂(1-y) = z₃ (1+y)

=> z₂/z₃ = (1+y)/(1-y)....eqn.(2)

dividing (1) and (2) we will get

z₁/z₃ = (1+x / 1-x) X (1-y / 1+y) = z where z is a real number

=> z₁-z₃ / z₁+z₃ = z-1/z+1 which is real

=> (z₁, z₃) ∈ R

Hence, R is transitive

Therefore, it is proved that relation R is an equivalence relation.

Summary

Exercise 1.2 in Chapter 1 of RD Sharma's Class 12 Mathematics textbook continues to explore the concept of Relations. This set focuses on:

1. Identifying and analyzing different types of relations

2. Applying properties of relations (reflexive, symmetric, transitive, equivalence)

3. Working with composite relations

4. Representing relations using matrices and graphs

5. Solving problems involving real-world applications of relations

Practice Questions (Set 2):

1. Let A = {1, 2, 3, 4, 5} and R be a relation on A defined by R = {(a, b) : |a - b| ≤ 1}. Is R an equivalence relation? Justify your answer.

2. Define a relation R on the set of all lines in a plane as follows: l₁Rl₂ if and only if l₁ is parallel to l₂. Prove that R is an equivalence relation.

3. Let R be a relation on Z (set of integers) defined by aRb if a² ≡ b² (mod 3). Determine if R is reflexive, symmetric, and transitive.

4. Given relations R = {(1, 2), (2, 3), (3, 4)} and S = {(1, 1), (2, 2), (3, 3), (4, 4)} on the set A = {1, 2, 3, 4}, find R ∘ S and S ∘ R.

5. Let A = {1, 2, 3, 4} and R be a relation on A defined by the matrix:

[1 0 1 0]

[0 1 0 1]

[1 0 1 0]

[0 1 0 1]

Is R symmetric? Justify your answer.

6. Define a relation R on the set of all triangles as follows: T₁RT₂ if and only if T₁ and T₂ have the same perimeter. Is R an equivalence relation? Explain.

7. Let R be a relation on R (set of real numbers) defined by xRy if x² + y² < 1. Sketch the graph of this relation.

8. Given A = {a, b, c, d} and R = {(a, b), (b, c), (c, d), (d, a)}, find R², R³, and R⁴. What do you observe?

9. Let R be a relation on the set of all people defined as xRy if x and y have at least one grandparent in common. Is R transitive? Explain your reasoning.

10. Define a relation R on the set of all subsets of a given set A as follows: For any two subsets X and Y of A, XRY if X ⊆ Y. Prove that R is reflexive and transitive, but not symmetric unless A has at most one element.