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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 1 |

Exercise | 1.2 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Class 12 Ex 1.2 Solutions Chapter 1 Relations**

### 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.

lly (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?

### Question 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 = {(L_{1, }L_{2}) : L_{1} is parallel to L_{2}}. 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 = {(L

_{1}, L_{2}) : L_{1}is parallel to L_{2}} be a relation on LNow, (i) reflexibility :

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

So, L

_{1}, L_{2 }∈ RTherefore, R is reflexive.

Again, (ii) symmetry :

Assume, L

_{1}, L_{2}∈ L and (L1, L2) ∈ RSince, L

_{1}is parallel to L_{2}:. L

_{2}is also parallel to L_{1}Thus, relation R is symmetric.

(iii) transitivity :

let L

_{1}, L_{2}, L_{3}∈ R in such a way that (L_{1}, L_{2}) ∈ R and (L_{2}, L_{3}) ∈ RL

_{1}is parallel to L_{2}L

_{2}is parallel to L_{3 }Therefore L

_{1}is parallel to L_{3}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 = {(P

_{1}, P_{2}):P_{1}and P_{2}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 P

_{1}and P_{2}be any polygon in A such that, (P_{1}, P_{2 }) ∈ R(P

_{1}, P_{2}) ∈ RP

_{1}and P_{2}have same number of sidesso, P

_{2}and P_{1}will have also same number of sidesHence (P

_{1}, P_{2}) ∈ Rso, relation R is symmetric.

at last (iii) transitivity :

let P

_{1}, P_{2}, P_{3}be three polygons in A in such a way that (P_{1 }, P_{2}) ∈ R and (P_{2}, P_{3}) ∈ Rthen, (P

_{1}, P_{2)}have same number of sides(P

_{2}, P_{3}) have same number of sides

:.P_{1}and P_{3}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.

### 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 ∈ Rso, relation R is reflexive

(ii) symmetry:

let (P, Q) ∈ R for P, Q ∈ A

=>OQ = OP

:.(Q, P) ∈ Rtherefore, relation R is symmetric .

lly (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

lly, 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: a^{2 }+ b^{2 }= 1}

**Solution:**

Let us observe the following properties of S

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

a ∈ R

=> a

^{2}+ a^{2}≠ 1 for all a ∈ Rso, S is not reflexive on R .

(ii) symmetry:

let (a, b) ∈ R

a

^{2}+ b^{2}= 1b

^{2}+ a^{2}= 1=> (b, a) ∈ S for all a, b ∈ R

So, S is symmetric on R

(iii) transitivity:

Let (a,b) and (b,c) ∈ S

a

^{2 }+ b^{2}=1 and b^{2 }+ c^{2}= 1Adding both the equations we will get,

a

^{2}+ c^{2 }= 2 – 2b^{2}≠ 1 for all a, b, c ∈ RSo, 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 Z_{0} be the set of all non – zero integers. Let a relation R on Z x Z_{0 }be defined as (a,b) R (c,d) <=>ad =bc for all (a,b), (c,d) ∈ Z x Z_{0}. Prove that R is an equivalence relation on Z x Z_{0}.

**Solution:**

Let us observe the properties of R

(i) reflexibility:

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

_{0}(a,b) ∈ ZxZ

_{0}a,b ∈ Z,Z

_{0}ab = ba

(a,b) ∈ R for all (a,b) ∈ ZxZ

_{0}R is reflexive

(ii) symmetry:

Let (a,b), (c,d) ∈ ZxZ

_{0}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) ∈ ZxZ

_{0}so, R is symmetric.

(iii) transitivity:

Let (a,b), (c,d), (e,f) ∈ NxN

_{0}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 auch 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 C_{0} be the set of all non – zero complex numbers. Let a relation R and C_{0} be defined as Z_{1} R Z_{2} <=> z_{1} – z_{2} / z_{1} + z_{2} is real for all Z_{1}, Z_{2} ∈ C_{0. }Show that R is an equivalence relation.

**Solution:**

(i) reflexibility:

since, z

_{1 }– z_{2}/ z_{1}+ z_{2}= 0 which is a real numberso, (z

_{1}, z_{1}) ∈ R so,relation R is reflexive.

(ii)symmetry;

z

_{1}– z_{2}/ z_{1}+ z_{2}= x, where x is real number=> – (z

_{1}– z_{2}/ z_{1}+ z_{2 }) = -xso, (z

_{2}, z_{1}) ∈ RHence, R is symmetric.

(iii) transitivity:

Let (z

_{1},z_{2}) ∈ R and (z_{2},z_{3}) ∈ Rthen,

z

_{1}-z_{2}/ z_{1}+z_{2}= x where x is a real number=> z

_{1}– z_{2}= xz_{1}+ xz_{2}=> z

_{1}– xz_{1}= z_{2}+ xz_{2}=> z

_{1}(1 – x) = z_{2}(1+x)=> z

_{1}/z_{2}= (1+x)/(1-x) …. eqn. (1)also, z

_{2}-z_{3}/z_{2}+z_{3}= y where y is a real number=> z

_{2}– z_{3}= yz_{2}+ yz_{3}=> z

_{2}– yz_{2}= z_{3}+ yz_{3}=> z

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

_{2}/z_{3}= (1+y)/(1-y)….eqn.(2)dividing (1) and (2) we will get

z

_{1}/z_{3}= (1+x / 1-x) X (1-y / 1+y) = z where z is a real number=> z

_{1}-z_{3}/ z_{1}+z_{3}= z-1/z+1 which is real=> (z

_{1}, z_{3}) ∈ RHence, R is transitive

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

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