This MCQ module is based on: Types of Relations – Reflexive, Symmetric, Transitive, Equivalence
TOPIC 1 OF 16
Types of Relations – Reflexive, Symmetric, Transitive, Equivalence
🎓 Class 12
Mathematics
CBSE
Theory
Ch 1 — Relations and Functions
⏱ ~25 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Types of Relations – Reflexive, Symmetric, Transitive, Equivalence
Targeting Class 12 level in Functions, with Advanced difficulty.
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1.1 Introduction
In Class XI, we explored the concepts of relations, functions, domain, co-domain, and range along with various types of real-valued functions and their graphs. The term relation? in mathematics captures whether two objects or quantities share a recognisable connection. For instance, if A is the set of students in Class XII and B is the set of students in Class XI of the same school, we can form relations such as "\(a\) is a brother of \(b\)" or "\(a\) is a sister of \(b\)".
Mathematically, a relation R from set A to set B is any arbitrary subset of \(A \times B\). If \((a, b) \in R\), we say \(a\) is related to \(b\) under R and write \(a\,R\,b\). In general, we do not require a recognisable link between \(a\) and \(b\) — the pair simply belongs to R. Functions, as we recall from Class XI, are a special kind of relation.
In this chapter, we study different types of relations and functions, composition of functions, invertible functions, and binary operations.
1.2 Types of Relations
A relation in a set A is a subset of \(A \times A\). Two extreme cases arise naturally:
Definition 1 — Empty Relation
A relation R in a set A is called an empty relation if no element of A is related to any element of A, i.e., \(R = \phi \subset A \times A\).
Definition 2 — Universal Relation
A relation R in a set A is called a universal relation if every element of A is related to every element of A, i.e., \(R = A \times A\).
Both the empty relation and the universal relation are sometimes called trivial relations?.
Worked Example 1
Let A be the set of all students of a boys' school. Show that R = {(\(a, b\)) : \(a\) is sister of \(b\)} is the empty relation and R' = {(\(a, b\)) : the difference between heights of \(a\) and \(b\) is less than 3 metres} is the universal relation.
Solution
Since the school is a boys' school, no student can be a sister of any student. Hence, \(R = \phi\), showing R is the empty relation.It is also obvious that the difference between heights of any two students must be less than 3 metres. This shows that \(R' = A \times A\) is the universal relation.
One of the most important relations in mathematics is an equivalence relation?. To study it, we first consider three types of relations: reflexive, symmetric, and transitive.
Definition 3 — Reflexive, Symmetric, Transitive
A relation R in a set A is called:(i) Reflexive, if \((a, a) \in R\) for every \(a \in A\).
(ii) Symmetric, if \((a_1, a_2) \in R\) implies that \((a_2, a_1) \in R\) for all \(a_1, a_2 \in A\).
(iii) Transitive, if \((a_1, a_2) \in R\) and \((a_2, a_3) \in R\) implies that \((a_1, a_3) \in R\) for all \(a_1, a_2, a_3 \in A\).
Definition 4 — Equivalence Relation
A relation R in a set A is said to be an equivalence relation if R is reflexive, symmetric, and transitive.
Visual summary of reflexive, symmetric, and transitive properties
Worked Example 2
Let T be the set of all triangles in a plane and R be the relation defined as R = {(\(T_1, T_2\)) : \(T_1\) is congruent to \(T_2\)}. Show that R is an equivalence relation.
Solution
Reflexive: Every triangle is congruent to itself, so \((T_1, T_1) \in R\). Hence R is reflexive.Symmetric: If \(T_1\) is congruent to \(T_2\), then \(T_2\) is congruent to \(T_1\). So \((T_1, T_2) \in R \Rightarrow (T_2, T_1) \in R\). Hence R is symmetric.
Transitive: If \(T_1\) is congruent to \(T_2\) and \(T_2\) is congruent to \(T_3\), then \(T_1\) is congruent to \(T_3\). So \((T_1, T_2) \in R\) and \((T_2, T_3) \in R \Rightarrow (T_1, T_3) \in R\). Hence R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
Worked Example 3
Let L be the set of all lines in a plane and R be the relation defined as R = {(\(L_1, L_2\)) : \(L_1\) is perpendicular to \(L_2\)}. Show that R is symmetric but neither reflexive nor transitive.
Fig 1.1 — Perpendicular lines \(L_1 \perp L_2\)
Solution
Not reflexive: A line \(L_1\) cannot be perpendicular to itself, i.e., \((L_1, L_1) \notin R\).Symmetric: If \(L_1 \perp L_2\), then \(L_2 \perp L_1\). So \((L_1, L_2) \in R \Rightarrow (L_2, L_1) \in R\). R is symmetric.
Not transitive: If \(L_1 \perp L_2\) and \(L_2 \perp L_3\), then \(L_1 \parallel L_3\) (not perpendicular). So \((L_1, L_2) \in R\) and \((L_2, L_3) \in R\) but \((L_1, L_3) \notin R\). R is not transitive.
Worked Example 4
Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.
Solution
Reflexive: (1, 1), (2, 2), and (3, 3) all lie in R. So R is reflexive.Not symmetric: (1, 2) \(\in\) R but (2, 1) \(\notin\) R. So R is not symmetric.
Not transitive: (1, 2) \(\in\) R and (2, 3) \(\in\) R but (1, 3) \(\notin\) R. So R is not transitive.
Worked Example 5
Show that the relation R in the set Z of integers given by R = {(\(a, b\)) : 2 divides \(a - b\)} is an equivalence relation.
Solution
Reflexive: For all \(a \in \mathbf{Z}\), \(2\) divides \(a - a = 0\). So \((a, a) \in R\).Symmetric: If \((a, b) \in R\), then 2 divides \(a - b\). Hence 2 divides \(b - a = -(a-b)\). So \((b, a) \in R\).
Transitive: If \((a, b) \in R\) and \((b, c) \in R\), then 2 divides \(a - b\) and 2 divides \(b - c\). Now, \(a - c = (a - b) + (b - c)\). Since 2 divides both summands, 2 divides \(a - c\). So \((a, c) \in R\).
R is reflexive, symmetric, and transitive. Hence R is an equivalence relation.
In Example 5, note that all even integers are related to zero (since \(a - 0 = a\) is even), and all odd integers are related to 1 (since \(a - 1\) is even when \(a\) is odd). The set E of all even integers and the set O of all odd integers form equivalence classes? with \(\mathbf{Z} = E \cup O\) and \(E \cap O = \phi\).
Key Concept — Equivalence Classes
The subset E is called the equivalence class containing zero and is denoted by [0]. Similarly, O is the equivalence class containing 1, denoted by [1].Note that [0] = [2] = [−4] = ... (any even integer) and [1] = [3] = [−5] = ... (any odd integer).
For an arbitrary equivalence relation R in a set X, R divides X into mutually disjoint subsets \(A_i\) called partitions (or subdivisions), satisfying:
(i) all elements of \(A_i\) are related to each other, for all \(i\);
(ii) no element of \(A_i\) is related to any element of \(A_j\) when \(i \ne j\);
(iii) \(\cup A_i = X\) and \(A_i \cap A_j = \phi\) for \(i \ne j\).
Worked Example 6
Let R be the relation defined in the 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 elements of the subset {1, 3, 5, 7} are related to each other and all elements of the subset {2, 4, 6} are related to each other, but no element of {1, 3, 5, 7} is related to any element of {2, 4, 6}.
Solution
Reflexive: Given any element \(a \in A\), both \(a\) and \(a\) must be either odd or even (trivially). So \((a, a) \in R\).Symmetric: If \((a, b) \in R\), then \(a\) and \(b\) are both odd or both even. But then \(b\) and \(a\) are also both odd or both even. So \((b, a) \in R\).
Transitive: If \((a, b) \in R\) and \((b, c) \in R\), then \(a, b\) have the same parity and \(b, c\) have the same parity. Therefore, \(a\) and \(c\) share the same parity. So \((a, c) \in R\).
Hence R is an equivalence relation. The two equivalence classes are {1, 3, 5, 7} (odd) and {2, 4, 6} (even). No element of {1, 3, 5, 7} can be related to any element of {2, 4, 6} since one is odd and the other is even.
Activity: Exploring Equivalence Relations
Predict: If you define a relation on the set {1, 2, 3, 4, 5, 6, 7, 8, 9} by "a is related to b if 3 divides (a − b)", how many equivalence classes will form?
- Take A = {1, 2, 3, 4, 5, 6, 7, 8, 9} and the relation R = {(a, b) : 3 divides (a − b)}.
- Start with element 1. Find all elements related to 1: since 3 divides (a − 1), we need a − 1 = 0, ±3, ±6, ... So a = 1, 4, 7.
- Now take element 2: a − 2 divisible by 3 gives a = 2, 5, 8.
- Element 3: a − 3 divisible by 3 gives a = 3, 6, 9.
- Verify: Are there any leftover elements? 1,4,7 and 2,5,8 and 3,6,9 cover all of A.
Observation: Three equivalence classes form: [1] = {1, 4, 7}, [2] = {2, 5, 8}, [3] = {3, 6, 9}.
Explanation: Since we divide by 3, there are exactly 3 possible remainders (0, 1, 2). Elements with the same remainder when divided by 3 form one equivalence class. This is the foundation of modular arithmetic!
Interactive: Relation Property Checker
Enter pairs of a relation on set {1, 2, 3} and check its properties
Exercise 1.1
Q1. Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x − y = 0}
(ii) Relation R in N defined as R = {(x, y) : y = x + 5 and x < 4}
(iii) Relation R in A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x}
(iv) Relation R in Z defined as R = {(x, y) : x − y is an integer}
(v) Relation R in the set A of human beings at a particular time given by
(a) R = {(x, y) : x and y work at the same place}
(b) R = {(x, y) : x and y live in the same locality}
(c) R = {(x, y) : x is exactly 7 cm taller than y}
(d) R = {(x, y) : x is wife of y}
(e) R = {(x, y) : x is father of y}
(i) Relation R in A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x − y = 0}
(ii) Relation R in N defined as R = {(x, y) : y = x + 5 and x < 4}
(iii) Relation R in A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x}
(iv) Relation R in Z defined as R = {(x, y) : x − y is an integer}
(v) Relation R in the set A of human beings at a particular time given by
(a) R = {(x, y) : x and y work at the same place}
(b) R = {(x, y) : x and y live in the same locality}
(c) R = {(x, y) : x is exactly 7 cm taller than y}
(d) R = {(x, y) : x is wife of y}
(e) R = {(x, y) : x is father of y}
(i) R = {(1,3),(2,6),(3,9),(4,12)}.
Reflexive: (1,1) \(\notin\) R. Not reflexive.
Symmetric: (1,3) \(\in\) R but (3,1) \(\notin\) R. Not symmetric.
Transitive: (1,3) and (3,9) \(\in\) R but (1,9) \(\notin\) R. Not transitive.
(ii) R = {(1,6),(2,7),(3,8)}.
Not reflexive: (1,1) \(\notin\) R. Not symmetric: (1,6) \(\in\) R but (6,1) \(\notin\) R. Not transitive: no pair (a,b) and (b,c) both in R.
Neither reflexive, nor symmetric, nor transitive.
(iii) Reflexive: \(x\) divides \(x\) for all \(x \in A\). Yes.
Symmetric: (1,2) \(\in\) R (2 is divisible by 1) but (2,1) \(\notin\) R (1 not divisible by 2). Not symmetric.
Transitive: If \(y\) is divisible by \(x\) and \(z\) is divisible by \(y\), then \(z\) is divisible by \(x\). Yes.
Reflexive and transitive but not symmetric.
(iv) Reflexive: \(x - x = 0\) is an integer. Yes.
Symmetric: If \(x - y\) is an integer, then \(y - x = -(x-y)\) is also an integer. Yes.
Transitive: If \(x - y\) and \(y - z\) are integers, then \(x - z = (x-y)+(y-z)\) is an integer. Yes.
Equivalence relation.
(v)(a) Reflexive: x works at the same place as x. Yes. Symmetric: obvious. Transitive: obvious. Equivalence relation.
(v)(b) Same reasoning as (a). Equivalence relation.
(v)(c) Not reflexive: x is not 7 cm taller than x. Not symmetric: if x is 7 cm taller than y, then y is NOT 7 cm taller than x. Not transitive: if x is 7 cm taller than y and y is 7 cm taller than z, then x is 14 cm taller than z, not 7. None of the three.
(v)(d) Not reflexive: x is not wife of x. Not symmetric: if x is wife of y, y is husband of x (not wife). Not transitive: no chain possible. None of the three.
(v)(e) Not reflexive: x is not father of x. Not symmetric: if x is father of y, y is not father of x. Not transitive: if x is father of y and y is father of z, then x is grandfather of z, not father. None of the three.
Reflexive: (1,1) \(\notin\) R. Not reflexive.
Symmetric: (1,3) \(\in\) R but (3,1) \(\notin\) R. Not symmetric.
Transitive: (1,3) and (3,9) \(\in\) R but (1,9) \(\notin\) R. Not transitive.
(ii) R = {(1,6),(2,7),(3,8)}.
Not reflexive: (1,1) \(\notin\) R. Not symmetric: (1,6) \(\in\) R but (6,1) \(\notin\) R. Not transitive: no pair (a,b) and (b,c) both in R.
Neither reflexive, nor symmetric, nor transitive.
(iii) Reflexive: \(x\) divides \(x\) for all \(x \in A\). Yes.
Symmetric: (1,2) \(\in\) R (2 is divisible by 1) but (2,1) \(\notin\) R (1 not divisible by 2). Not symmetric.
Transitive: If \(y\) is divisible by \(x\) and \(z\) is divisible by \(y\), then \(z\) is divisible by \(x\). Yes.
Reflexive and transitive but not symmetric.
(iv) Reflexive: \(x - x = 0\) is an integer. Yes.
Symmetric: If \(x - y\) is an integer, then \(y - x = -(x-y)\) is also an integer. Yes.
Transitive: If \(x - y\) and \(y - z\) are integers, then \(x - z = (x-y)+(y-z)\) is an integer. Yes.
Equivalence relation.
(v)(a) Reflexive: x works at the same place as x. Yes. Symmetric: obvious. Transitive: obvious. Equivalence relation.
(v)(b) Same reasoning as (a). Equivalence relation.
(v)(c) Not reflexive: x is not 7 cm taller than x. Not symmetric: if x is 7 cm taller than y, then y is NOT 7 cm taller than x. Not transitive: if x is 7 cm taller than y and y is 7 cm taller than z, then x is 14 cm taller than z, not 7. None of the three.
(v)(d) Not reflexive: x is not wife of x. Not symmetric: if x is wife of y, y is husband of x (not wife). Not transitive: no chain possible. None of the three.
(v)(e) Not reflexive: x is not father of x. Not symmetric: if x is father of y, y is not father of x. Not transitive: if x is father of y and y is father of z, then x is grandfather of z, not father. None of the three.
Q2. Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b²} is neither reflexive nor symmetric nor transitive.
Not reflexive: Take \(a = \frac{1}{2}\). Then \(a \leq a^2\) means \(\frac{1}{2} \leq \frac{1}{4}\), which is false. So \((\frac{1}{2}, \frac{1}{2}) \notin R\).
Not symmetric: \((1, 4) \in R\) since \(1 \leq 16\), but \((4, 1) \notin R\) since \(4 \leq 1\) is false.
Not transitive: \((3, 2) \in R\) since \(3 \leq 4\), and \((2, 1.5) \in R\) since \(2 \leq 2.25\), but \((3, 1.5) \notin R\) since \(3 \leq 2.25\) is false.
Not symmetric: \((1, 4) \in R\) since \(1 \leq 16\), but \((4, 1) \notin R\) since \(4 \leq 1\) is false.
Not transitive: \((3, 2) \in R\) since \(3 \leq 4\), and \((2, 1.5) \in R\) since \(2 \leq 2.25\), but \((3, 1.5) \notin R\) since \(3 \leq 2.25\) is false.
Q3. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.
R = {(1,2),(2,3),(3,4),(4,5),(5,6)}.
Not reflexive: (1,1) \(\notin\) R since \(1 \neq 1+1\).
Not symmetric: (1,2) \(\in\) R but (2,1) \(\notin\) R.
Not transitive: (1,2) \(\in\) R and (2,3) \(\in\) R but (1,3) \(\notin\) R (since \(3 \neq 1+1\)).
R is neither reflexive, nor symmetric, nor transitive.
Not reflexive: (1,1) \(\notin\) R since \(1 \neq 1+1\).
Not symmetric: (1,2) \(\in\) R but (2,1) \(\notin\) R.
Not transitive: (1,2) \(\in\) R and (2,3) \(\in\) R but (1,3) \(\notin\) R (since \(3 \neq 1+1\)).
R is neither reflexive, nor symmetric, nor transitive.
Q4. Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.
Reflexive: \(a \leq a\) for all \(a \in \mathbf{R}\). Yes.
Not symmetric: \((1, 2) \in R\) since \(1 \leq 2\), but \((2, 1) \notin R\) since \(2 \leq 1\) is false.
Transitive: If \(a \leq b\) and \(b \leq c\), then \(a \leq c\). Yes.
Not symmetric: \((1, 2) \in R\) since \(1 \leq 2\), but \((2, 1) \notin R\) since \(2 \leq 1\) is false.
Transitive: If \(a \leq b\) and \(b \leq c\), then \(a \leq c\). Yes.
Q5. Check whether the relation R in R defined by R = {(a, b) : a ≤ b³} is reflexive, symmetric or transitive.
Not reflexive: Take \(a = -2\). Then \(a \leq a^3 \Rightarrow -2 \leq -8\), false. So \((-2, -2) \notin R\).
Not symmetric: Take \(a = 1, b = 2\). \(1 \leq 8\) is true, so \((1, 2) \in R\). But \(2 \leq 1\) is false, so \((2, 1) \notin R\).
Not transitive: \((25, 3) \in R\) since \(25 \leq 27\), \((3, \frac{3}{2}) \in R\) since \(3 \leq \frac{27}{8} = 3.375\), but \((25, \frac{3}{2}) \notin R\) since \(25 \leq \frac{27}{8}\) is false.
Not symmetric: Take \(a = 1, b = 2\). \(1 \leq 8\) is true, so \((1, 2) \in R\). But \(2 \leq 1\) is false, so \((2, 1) \notin R\).
Not transitive: \((25, 3) \in R\) since \(25 \leq 27\), \((3, \frac{3}{2}) \in R\) since \(3 \leq \frac{27}{8} = 3.375\), but \((25, \frac{3}{2}) \notin R\) since \(25 \leq \frac{27}{8}\) is false.
Q6. Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
Symmetric: (1,2) \(\in\) R and (2,1) \(\in\) R. Yes.
Not reflexive: (1,1), (2,2), (3,3) all \(\notin\) R.
Not transitive: (1,2) \(\in\) R and (2,1) \(\in\) R but (1,1) \(\notin\) R.
Not reflexive: (1,1), (2,2), (3,3) all \(\notin\) R.
Not transitive: (1,2) \(\in\) R and (2,1) \(\in\) R but (1,1) \(\notin\) R.
Q7. Show that the relation R in the set A of all books in a library, given by R = {(x, y) : x and y have the same number of pages} is an equivalence relation.
Reflexive: Every book has the same number of pages as itself. So (x, x) \(\in\) R.
Symmetric: If x and y have the same number of pages, then y and x also have the same number of pages. So (x, y) \(\in\) R \(\Rightarrow\) (y, x) \(\in\) R.
Transitive: If x and y have the same page count, and y and z have the same page count, then x and z have the same page count.
Hence R is an equivalence relation.
Symmetric: If x and y have the same number of pages, then y and x also have the same number of pages. So (x, y) \(\in\) R \(\Rightarrow\) (y, x) \(\in\) R.
Transitive: If x and y have the same page count, and y and z have the same page count, then x and z have the same page count.
Hence R is an equivalence relation.
Q8. Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a − b| is even} is an equivalence relation. Show that all elements of {1, 3, 5} are related to each other and all elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
Reflexive: \(|a - a| = 0\) is even for all \(a\). Yes.
Symmetric: \(|a - b| = |b - a|\). If one is even, so is the other. Yes.
Transitive: If \(|a - b|\) is even and \(|b - c|\) is even, then \(a - b\) is even and \(b - c\) is even, so \(a - c = (a-b) + (b-c)\) is even. Hence \(|a - c|\) is even. Yes.
R is an equivalence relation. The equivalence classes are {1, 3, 5} (odd numbers) and {2, 4} (even numbers). Since odd minus even is always odd (not even), no element of {1, 3, 5} is related to any element of {2, 4}.
Symmetric: \(|a - b| = |b - a|\). If one is even, so is the other. Yes.
Transitive: If \(|a - b|\) is even and \(|b - c|\) is even, then \(a - b\) is even and \(b - c\) is even, so \(a - c = (a-b) + (b-c)\) is even. Hence \(|a - c|\) is even. Yes.
R is an equivalence relation. The equivalence classes are {1, 3, 5} (odd numbers) and {2, 4} (even numbers). Since odd minus even is always odd (not even), no element of {1, 3, 5} is related to any element of {2, 4}.
Q9. Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by
(i) R = {(a, b) : |a − b| is a multiple of 4}
(ii) R = {(a, b) : a = b}
is an equivalence relation. Find the set of all elements related to 1 in each case.
(i) R = {(a, b) : |a − b| is a multiple of 4}
(ii) R = {(a, b) : a = b}
is an equivalence relation. Find the set of all elements related to 1 in each case.
(i) Reflexive: \(|a-a| = 0\) is a multiple of 4. Yes.
Symmetric: \(|a-b| = |b-a|\). Yes.
Transitive: If 4 divides \(|a-b|\) and 4 divides \(|b-c|\), then 4 divides \(|a-c|\) (since \(a-c = (a-b) + (b-c)\)). Yes.
Set of elements related to 1: \(|a - 1|\) is a multiple of 4. So \(a = 1, 5, 9\). {1, 5, 9}.
(ii) The identity relation. Reflexive: \(a = a\). Symmetric: \(a = b \Rightarrow b = a\). Transitive: \(a = b\) and \(b = c \Rightarrow a = c\).
Set of elements related to 1: only \(a = 1\). {1}.
Symmetric: \(|a-b| = |b-a|\). Yes.
Transitive: If 4 divides \(|a-b|\) and 4 divides \(|b-c|\), then 4 divides \(|a-c|\) (since \(a-c = (a-b) + (b-c)\)). Yes.
Set of elements related to 1: \(|a - 1|\) is a multiple of 4. So \(a = 1, 5, 9\). {1, 5, 9}.
(ii) The identity relation. Reflexive: \(a = a\). Symmetric: \(a = b \Rightarrow b = a\). Transitive: \(a = b\) and \(b = c \Rightarrow a = c\).
Set of elements related to 1: only \(a = 1\). {1}.
Q10. Give an example of a relation. Which is:
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.
Let A = {1, 2, 3}.
(i) R = {(1, 2), (2, 1)}. Symmetric: yes. Not reflexive: (1,1) \(\notin\) R. Not transitive: (1,2) and (2,1) \(\in\) R but (1,1) \(\notin\) R.
(ii) R = {(1, 2)}. Transitive: vacuously true (no chain to check). Not reflexive: (1,1) \(\notin\) R. Not symmetric: (1,2) \(\in\) R but (2,1) \(\notin\) R.
(iii) R = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}. Reflexive: yes. Symmetric: yes. Not transitive: (1,2) and (2,3) \(\in\) R but (1,3) \(\notin\) R.
(iv) R = {(1,1),(2,2),(3,3),(1,2)}. Reflexive and transitive: yes. Not symmetric: (1,2) \(\in\) R but (2,1) \(\notin\) R.
(v) R = {(1,1),(1,2),(2,1),(2,2)}. Symmetric: yes. Transitive: yes. Not reflexive: (3,3) \(\notin\) R.
(i) R = {(1, 2), (2, 1)}. Symmetric: yes. Not reflexive: (1,1) \(\notin\) R. Not transitive: (1,2) and (2,1) \(\in\) R but (1,1) \(\notin\) R.
(ii) R = {(1, 2)}. Transitive: vacuously true (no chain to check). Not reflexive: (1,1) \(\notin\) R. Not symmetric: (1,2) \(\in\) R but (2,1) \(\notin\) R.
(iii) R = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}. Reflexive: yes. Symmetric: yes. Not transitive: (1,2) and (2,3) \(\in\) R but (1,3) \(\notin\) R.
(iv) R = {(1,1),(2,2),(3,3),(1,2)}. Reflexive and transitive: yes. Not symmetric: (1,2) \(\in\) R but (2,1) \(\notin\) R.
(v) R = {(1,1),(1,2),(2,1),(2,2)}. Symmetric: yes. Transitive: yes. Not reflexive: (3,3) \(\notin\) R.
Q11. Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of P from origin = distance of Q from origin} is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle through P with origin as centre.
Reflexive: dist(P, O) = dist(P, O) for every point P. So (P, P) \(\in\) R.
Symmetric: If dist(P, O) = dist(Q, O), then dist(Q, O) = dist(P, O). So (P, Q) \(\in\) R \(\Rightarrow\) (Q, P) \(\in\) R.
Transitive: If dist(P, O) = dist(Q, O) and dist(Q, O) = dist(R, O), then dist(P, O) = dist(R, O).
The set of all points related to P = {Q : dist(Q, O) = dist(P, O)} = all points at distance \(r = \sqrt{x_P^2 + y_P^2}\) from origin = the circle through P with centre at origin.
Symmetric: If dist(P, O) = dist(Q, O), then dist(Q, O) = dist(P, O). So (P, Q) \(\in\) R \(\Rightarrow\) (Q, P) \(\in\) R.
Transitive: If dist(P, O) = dist(Q, O) and dist(Q, O) = dist(R, O), then dist(P, O) = dist(R, O).
The set of all points related to P = {Q : dist(Q, O) = dist(P, O)} = all points at distance \(r = \sqrt{x_P^2 + y_P^2}\) from origin = the circle through P with centre at origin.
Q12. Show that the relation R defined in the set A of all triangles as R = {(T₁, T₂) : T₁ is similar to T₂} is an equivalence relation. Consider three right-angle triangles T₁ with sides 3, 4, 5; T₂ with sides 5, 12, 13; and T₃ with sides 6, 8, 10. Which triangles among T₁, T₂, T₃ are related?
Reflexive: Every triangle is similar to itself. Symmetric: If T₁ ~ T₂, then T₂ ~ T₁. Transitive: If T₁ ~ T₂ and T₂ ~ T₃, then T₁ ~ T₃.
Hence R is an equivalence relation.
T₁ has sides 3, 4, 5. T₃ has sides 6, 8, 10 = 2(3, 4, 5). So T₁ ~ T₃.
T₂ has sides 5, 12, 13 which are not proportional to 3, 4, 5.
T₁ and T₃ are related. T₂ is not related to T₁ or T₃.
Hence R is an equivalence relation.
T₁ has sides 3, 4, 5. T₃ has sides 6, 8, 10 = 2(3, 4, 5). So T₁ ~ T₃.
T₂ has sides 5, 12, 13 which are not proportional to 3, 4, 5.
T₁ and T₃ are related. T₂ is not related to T₁ or T₃.
Q13. Show that the relation R defined in the set A of all polygons as R = {(P₁, P₂) : P₁ and P₂ have the 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?
Reflexive: Every polygon has the same number of sides as itself.
Symmetric: If P₁ has the same number of sides as P₂, then P₂ has the same as P₁.
Transitive: obvious chain of equalities.
Hence R is an equivalence relation.
The triangle T has 3 sides. All elements related to T = the set of all triangles in A.
Symmetric: If P₁ has the same number of sides as P₂, then P₂ has the same as P₁.
Transitive: obvious chain of equalities.
Hence R is an equivalence relation.
The triangle T has 3 sides. All elements related to T = the set of all triangles in A.
Q14. Let L be the set of all lines in XY plane and R be the relation defined as R = {(L₁, L₂) : L₁ is parallel to L₂}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
Reflexive: Every line is parallel to itself. Symmetric: If L₁ ∥ L₂, then L₂ ∥ L₁. Transitive: If L₁ ∥ L₂ and L₂ ∥ L₃, then L₁ ∥ L₃.
R is an equivalence relation.
Lines related to y = 2x + 4: all lines with slope 2, i.e., the set {y = 2x + c : c ∈ R}.
R is an equivalence relation.
Lines related to y = 2x + 4: all lines with slope 2, i.e., the set {y = 2x + c : c ∈ R}.
Q15. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
Reflexive: (1,1), (2,2), (3,3), (4,4) all \(\in\) R. Yes.
Symmetric: (1,2) \(\in\) R but (2,1) \(\notin\) R. Not symmetric.
Transitive: (1,3) \(\in\) R, (3,2) \(\in\) R, (1,2) \(\in\) R. Check all chains — none violate transitivity.
Answer: (B) Reflexive and transitive but not symmetric.
Symmetric: (1,2) \(\in\) R but (2,1) \(\notin\) R. Not symmetric.
Transitive: (1,3) \(\in\) R, (3,2) \(\in\) R, (1,2) \(\in\) R. Check all chains — none violate transitivity.
Answer: (B) Reflexive and transitive but not symmetric.
Q16. Let R be the relation in the set N given by R = {(a, b) : a = b − 2, b > 6}. Choose the correct answer.
(A) (2, 4) ∈ R (B) (3, 8) ∈ R (C) (6, 8) ∈ R (D) (8, 7) ∈ R
(A) (2, 4) ∈ R (B) (3, 8) ∈ R (C) (6, 8) ∈ R (D) (8, 7) ∈ R
R = {(a, b) : a = b − 2, b > 6}. So we need b > 6 AND a = b − 2.
(A) (2, 4): b = 4, not > 6. No.
(B) (3, 8): a = 3, b = 8 > 6, but 8 − 2 = 6 ≠ 3. No.
(C) (6, 8): a = 6, b = 8 > 6, and 8 − 2 = 6 = a. Yes!
(D) (8, 7): a = 8, b = 7, 7 − 2 = 5 ≠ 8. No.
Answer: (C)
(A) (2, 4): b = 4, not > 6. No.
(B) (3, 8): a = 3, b = 8 > 6, but 8 − 2 = 6 ≠ 3. No.
(C) (6, 8): a = 6, b = 8 > 6, and 8 − 2 = 6 = a. Yes!
(D) (8, 7): a = 8, b = 7, 7 − 2 = 5 ≠ 8. No.
Answer: (C)
Competency-Based Questions
A mathematics teacher defines a relation R on the set of integers Z by the rule: "a is related to b if and only if \(a \equiv b \pmod{3}\)", meaning 3 divides \((a - b)\). She asks the class to explore properties of this relation and determine the equivalence classes formed.
Q1. Which of the following pairs belongs to R?
(a) (7, 4) (b) (5, 9) (c) (12, 8) (d) (15, 6)
L1 Remember(a) (7, 4) (b) (5, 9) (c) (12, 8) (d) (15, 6)
Answer: (a) and (d).
(a) 7 − 4 = 3, divisible by 3. Yes.
(b) 5 − 9 = −4, not divisible by 3. No.
(c) 12 − 8 = 4, not divisible by 3. No.
(d) 15 − 6 = 9, divisible by 3. Yes.
(a) 7 − 4 = 3, divisible by 3. Yes.
(b) 5 − 9 = −4, not divisible by 3. No.
(c) 12 − 8 = 4, not divisible by 3. No.
(d) 15 − 6 = 9, divisible by 3. Yes.
Q2. Prove that this relation R is an equivalence relation.
L2 UnderstandReflexive: \(a - a = 0\) is divisible by 3 for all \(a \in \mathbf{Z}\). So \((a, a) \in R\).
Symmetric: If 3 divides \(a - b\), then 3 divides \(-(a-b) = b - a\). So \((a, b) \in R \Rightarrow (b, a) \in R\).
Transitive: If 3 divides \(a - b\) and 3 divides \(b - c\), then 3 divides \((a - b) + (b - c) = a - c\).
Hence R is an equivalence relation.
Symmetric: If 3 divides \(a - b\), then 3 divides \(-(a-b) = b - a\). So \((a, b) \in R \Rightarrow (b, a) \in R\).
Transitive: If 3 divides \(a - b\) and 3 divides \(b - c\), then 3 divides \((a - b) + (b - c) = a - c\).
Hence R is an equivalence relation.
Q3. List all elements of the equivalence class [2] within the set {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6, 7, 8}.
L3 ApplyAnswer: [2] = {a : 3 divides (a − 2)} = {a : a − 2 = 0, ±3, ±6, ...}
= {2, 5, 8, −1, −4}.
So [2] within the given set = {−4, −1, 2, 5, 8}.
= {2, 5, 8, −1, −4}.
So [2] within the given set = {−4, −1, 2, 5, 8}.
Q4. A student claims: "If a relation is symmetric and transitive, it must be reflexive." Evaluate this claim with a counterexample.
L5 EvaluateThe claim is false. Counterexample: Let A = {1, 2, 3} and R = {(1, 2), (2, 1), (1, 1), (2, 2)}.
R is symmetric: (1,2) and (2,1) both present. R is transitive: all chains close.
But R is not reflexive because (3, 3) ∉ R. Element 3 is not related to any element, so we cannot use symmetry and transitivity to derive (3, 3).
R is symmetric: (1,2) and (2,1) both present. R is transitive: all chains close.
But R is not reflexive because (3, 3) ∉ R. Element 3 is not related to any element, so we cannot use symmetry and transitivity to derive (3, 3).
Assertion–Reason Questions
Assertion (A): The relation R = {(1, 1), (2, 2), (3, 3)} on {1, 2, 3} is an equivalence relation.
Reason (R): A relation that is reflexive, symmetric, and transitive is an equivalence relation.
Reason (R): A relation that is reflexive, symmetric, and transitive is an equivalence relation.
Answer: (a) — The identity relation is reflexive (all diagonal pairs present), symmetric (vacuously, since no off-diagonal pairs), and transitive (vacuously). Both A and R are true, and R correctly explains why A is an equivalence relation.
Assertion (A): The relation "is perpendicular to" on the set of lines in a plane is an equivalence relation.
Reason (R): An equivalence relation must be reflexive, symmetric, and transitive.
Reason (R): An equivalence relation must be reflexive, symmetric, and transitive.
Answer: (d) — A is false: "is perpendicular to" is not reflexive (a line is not perpendicular to itself) and not transitive. R is true: it correctly states the definition of an equivalence relation. R explains why A fails.
Assertion (A): The empty relation on a non-empty set A is symmetric and transitive.
Reason (R): Symmetric and transitive properties are satisfied vacuously when no pairs exist in R.
Reason (R): Symmetric and transitive properties are satisfied vacuously when no pairs exist in R.
Answer: (a) — Both are true. The empty relation has no pairs, so there are no counterexamples to symmetry or transitivity. These properties hold vacuously. R correctly explains why the empty relation satisfies them.
Frequently Asked Questions
What is a reflexive relation?
A relation R in a set A is reflexive if every element is related to itself, i.e., (a, a) belongs to R for every a in A.
What is an equivalence relation?
A relation R in a set A is an equivalence relation if it is reflexive, symmetric, and transitive simultaneously.
What is the difference between empty and universal relations?
An empty relation has no ordered pairs, while a universal relation contains all possible ordered pairs. Both are trivial relations.
How to check if a relation is an equivalence relation?
Verify three properties: reflexive (a,a) in R for all a; symmetric if (a,b) implies (b,a); transitive if (a,b) and (b,c) imply (a,c). All three must hold.
What are equivalence classes?
An equivalence class of element a under equivalence relation R is the set of all elements related to a. Equivalence classes partition the set into non-overlapping subsets.
Frequently Asked Questions — Relations and Functions
What is Types of Relations - Reflexive, Symmetric, Transitive, Equivalence in NCERT Class 12 Mathematics?
Types of Relations - Reflexive, Symmetric, Transitive, Equivalence is a key concept covered in NCERT Class 12 Mathematics, Chapter 1: Relations and Functions. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Types of Relations - Reflexive, Symmetric, Transitive, Equivalence step by step?
To solve problems on Types of Relations - Reflexive, Symmetric, Transitive, Equivalence, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 1: Relations and Functions?
The essential formulas of Chapter 1 (Relations and Functions) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Types of Relations - Reflexive, Symmetric, Transitive, Equivalence important for the Class 12 board exam?
Types of Relations - Reflexive, Symmetric, Transitive, Equivalence is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Types of Relations - Reflexive, Symmetric, Transitive, Equivalence?
Common mistakes in Types of Relations - Reflexive, Symmetric, Transitive, Equivalence include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Types of Relations - Reflexive, Symmetric, Transitive, Equivalence?
End-of-chapter NCERT exercises for Types of Relations - Reflexive, Symmetric, Transitive, Equivalence cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 1, and solve at least one previous-year board paper to consolidate your understanding.
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