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• RD Sharma Class 12 Solutions for Maths

# 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)

=> 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 ∈ C0. 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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