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.

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