RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2: Students will mostly gain conceptual knowledge of equivalence relations through this practice. If a relation is reflexive, symmetric, and transitive, it is said to be an equivalence relationship. Class 12 is a crucial year for students in terms of shaping and achieving their long-term objectives. These RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2 problems’ solutions are created in a form of comprehensive manner, which develops students’ problem-solving abilities.
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RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2
Access answers to Maths RD Sharma Solutions For Class 12 Chapter 1 – Relations Exercise 1.2 Question
1. Show that the relation R defined by R = {(a, b): a – b is divisible by 3; a, b ∈ Z} is an equivalence relation.
Solution:
Given R = {(a, b): a – b is divisible by 3; a, b ∈ Z} is a relation
To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.
Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of R.
Then, a – a = 0 = 0 × 3
⇒ a − a is divisible by 3
⇒ (a, a) ∈ R for all a ∈ Z
So, R is reflexive on Z.
Symmetry:
Let (a, b) ∈ R
⇒ a − b is divisible by 3
⇒ a − b = 3p for some p ∈ Z
⇒ b − a = 3 (−p)
Here, −p ∈ Z
⇒ b − a is divisible by 3
⇒ (b, a) ∈ R for all a, b ∈ Z
So, R is symmetric on Z.
Transitivity:
Let (a, b) and (b, c) ∈ R
⇒ a − b and b − c is divisible by 3
⇒ a – b = 3p for some p ∈ Z
And b − c = 3q for some q ∈ Z
Adding the above two equations, we get
a − b + b – c = 3p + 3q
⇒ a − c = 3 (p + q)
Here, p + q ∈ Z
⇒ a − c is divisible by 3
⇒ (a, c) ∈ R for all a, c ∈ Z
So, R is transitive on Z.
Therefore R is reflexive, symmetric, and transitive.
Hence, R is an equivalence relation to Z.
2. Show that the relation R on the set Z of integers, given by
R = {(a, b): 2 divides a – b}, is an equivalence relation.
Solution:
Given R = {(a, b): 2 divides a – b} is a relation defined on Z.
To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.
Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of the set Z.
Then, a ∈ R
⇒ a − a = 0 = 0 × 2
⇒ 2 divides a − a
⇒ (a, a) ∈ R for all a ∈ Z
So, R is reflexive on Z.
Symmetry:
Let (a, b) ∈ R
⇒ 2 divides a − b
⇒ (a-b)/2 = p for some p ∈ Z
⇒ (b-a)/2 = – p
Here, −p ∈ Z
⇒ 2 divides b − a
⇒ (b, a) ∈ R for all a, b ∈ Z
So, R is symmetric on Z
Transitivity:
Let (a, b) and (b, c) ∈ R
⇒ 2 divides a−b and 2 divides b−c
⇒ (a-b)/2 = p and (b-c)/2 = q for some p, q ∈ Z
Adding the above two equations, we get
(a – b)/2 + (b – c)/2 = p + q
⇒ (a – c)/2 = p +q
Here, p+ q ∈ Z
⇒ 2 divides a − c
⇒ (a, c) ∈ R for all a, c ∈ Z
So, R is transitive on Z.
Therefore R is reflexive, symmetric, and transitive.
Hence, R is an equivalence relation on Z.
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
To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.
Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of R. Then,
⇒ a − a = 0 = 0 × 5
⇒ a − a is divisible by 5
⇒ (a, a) ∈ R for all a ∈ Z
So, R is reflexive on Z.
Symmetry:
Let (a, b) ∈ R
⇒ a − b is divisible by 5
⇒ a − b = 5p for some p ∈ Z
⇒ b − a = 5 (−p)
Here, −p ∈ Z [Since p ∈ Z]
⇒ b − a is divisible by 5
⇒ (b, a) ∈ R for all a, b ∈ Z
So, R is symmetric on Z.
Transitivity:
Let (a, b) and (b, c) ∈ R
⇒ a − b is divisible by 5
⇒ a − b = 5p for some Z
Also, b − c is divisible by 5
⇒ b − c = 5q for some Z
Adding the above two equations, we get
a −b + b − c = 5p + 5q
⇒ a − c = 5 ( p + q )
⇒ a − c is divisible by 5
Here, p + q ∈ Z
⇒ (a, c) ∈ R for all a, c ∈ Z
So, R is transitive on Z.
Therefore R is reflexive, symmetric, and transitive.
Hence, R is an equivalence relation to Z.
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.
Show that R is an equivalence relation on Z.
Solution:
Given (a, b) ∈ R ⇔ a − b is divisible by n is a relation R defined on Z.
To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.
Let us check these properties on R.
Reflexivity:
Let a ∈ N
Here, a − a = 0 = 0 × n
⇒ a − a is divisible by n
⇒ (a, a) ∈ R
⇒ (a, a) ∈ R for all a ∈ Z
So, R is reflexive on Z.
Symmetry:
Let (a, b) ∈ R
Here, a − b is divisible by n
⇒ a − b = n p for some p ∈ Z
⇒ b − a = n (−p)
⇒ b − a is divisible by n [ p ∈ Z⇒ − p ∈ Z]
⇒ (b, a) ∈ R
So, R is symmetric on Z.
Transitivity:
Let (a, b) and (b, c) ∈ R
Here, a − b is divisible by n, and b − c is divisible by n.
⇒ a − b= n p for some p ∈ Z
And b−c = n q for some q ∈ Z
a – b + b − c = n p + n q
⇒ a − c = n (p + q)
⇒ (a, c) ∈ R for all a, c ∈ Z
So, R is transitive on Z.
Therefore R is reflexive, symmetric, and transitive.
Hence, R is an equivalence relation to Z.
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
To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.
Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of Z.
Then, a ∈ R
Clearly, a + a = 2a is even for all a ∈ Z.
⇒ (a, a) ∈ R for all a ∈ Z
So, R is reflexive on Z.
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.
Transitivity:
Let (a, b) and (b, c) ∈ R
⇒ a + b and b + c are even
Now, let a + b = 2x for some x ∈ Z
And b + c = 2y for some y ∈ Z
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 R is reflexive, symmetric, and transitive.
Hence, R is an equivalence relation to 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
Now we have to check whether the given relation is equivalence or not.
To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.
Let R = {(m, n): m, n ∈ Z : m − n is divisible by 13}
Let us check these properties on R.
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.
Symmetry:
Let (m, n) ∈ R.
Then, m − n is divisible by 13
⇒ m − n = 13p
Here, p ∈ Z
⇒ n – m = 13 (−p)
Here, −p ∈ Z
⇒ 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
⇒ m – n = 13p and n − o = 13q for some p, q ∈ Z
Adding the above two equations, we get
m – n + n − o = 13p + 13q
⇒ m−o = 13 (p + q)
Here, p + q ∈ Z
⇒ 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 to Z.
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
Now we have to check whether the given relation is equivalence or not.
To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.
Reflexivity:
Let (a, b) be an arbitrary element of set A.
Then, (a, b) ∈ A
⇒ a b = b a
⇒ (a, b) R (a, b)
Thus, R is reflexive on A.
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 to A.
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:
Given set A = {x ∈ Z; 0 ≤ x ≤ 12}
Also given that relation R = {(a, b): a = b} is defined on set A
Now we have to check whether the given relation is equivalence or not.
To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.
Reflexivity:
Let a be an arbitrary element of A.
Then, a ∈ R
⇒ a = a [Since every element is equal to itself]
⇒ (a, a) ∈ R for all a ∈ A
So, R is reflexive on A.
Symmetry:
Let (a, b) ∈ R
⇒ a b
⇒ b = a
⇒ (b, a) ∈ R for all a, b ∈ A
So, R is symmetric on A.
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, R is an equivalence relation to A.
Therefore R is reflexive, symmetric, and transitive.
The set of all elements related to 1 is {1}.
RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2: Important Topics From The Chapter
Let us have a look at the important topics covered in this exercise.
- Types of relations
- Void relation
- Universal relation
- Identity relation
- Reflexive relation
- Symmetric relation
- Transitive relation
- Antisymmetric relation
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