RS Aggarwal Solutions for Class 12 Chapter 01 Relation

Exercise 1A

Question 1. Find the domain and range of the relation R = {(-1, 1), (1, 1), (-2, 4), (2, 4)}.
Answer: dom(R) = {-1, 1, -2, 2} and range(R) = {1, 4}

Exam Tip: Domain consists of all first elements of ordered pairs; range contains all second elements. List them in set notation.

Question 2. Let R = {(a, a³) : a is a prime number less than 5}. Find the range of R.
Answer: range(R) = {8, 27}

Exam Tip: Prime numbers less than 5 are 2 and 3. Computing their cubes gives 2³ = 8 and 3³ = 27, which form the range.

Question 3. Let R = {(a, a³) : a is a prime number less than 10}. Find (i) R (ii) dom(R) (iii) range(R).
Answer:
(i) R = {(2, 8), (3, 27), (5, 125), (7, 343)}
(ii) dom(R) = {2, 3, 5, 7}
(iii) range(R) = {8, 27, 125, 343}

Exam Tip: Identify all primes under 10 first, then cube each to build the relation set. List domain and range separately in curly braces.

Question 4. Let R = {(x, y) : x + 2y = 6 are relation on N. Write the range of R.
Answer: {3, 2, 1}

Exam Tip: Rearrange the equation to solve for y in terms of x, then substitute natural number values to find all valid second coordinates.

Question 5. Let R = {(a, b) : a, b ∈ N and a + 3b = 12}. Find the domain and range of R.
Answer: dom(R) = {3, 6, 9} and range(R) = {3, 2, 1}

Exam Tip: Substitute natural number values for b to find corresponding a values. Both domain and range must consist only of natural numbers satisfying the equation.

Question 6. Let R = {(a, b) : b = |a - 1|, a ∈ Z and |a| < 3}. Find the domain and range of R.
Answer: dom(R) = {-2, -1, 0, 1, 2} and range(R) = {3, 2, 1, 0}

Exam Tip: The constraint |a| < 3 limits a to {-2, -1, 0, 1, 2}. Apply the absolute value function to each, then list all resulting outputs.

Question 7. Let R = {(a, b) : b = \(\frac{a}{a}\), a ∈ N and 1 < a < 5}. Find the domain and range of R.
Answer: dom(R) = {2, 3, 4} and range(R) = {1}

Exam Tip: Since \(\frac{a}{a} = 1\) for any non-zero a, the range contains only 1. The domain is determined by the inequality constraint on natural numbers.

Question 8. Let R = {(a, b) : a, b ∈ N and b = a + 5, a < 4}. Find the domain and range of R.
Answer: dom(R) = {1, 2, 3} and range(R) = {6, 7, 8}

Exam Tip: For each value of a satisfying a < 4 in the natural numbers, compute b = a + 5 to get the corresponding ordered pair.

Question 9. Let S be the set of all sets and let R = {(A, B) : A ⊂ B}, i.e., A is a proper subset of B. Show that R is (i) transitive (ii) not reflexive (iii) not symmetric.
Answer: Let R = {(A, B) : A ⊂ B}, a relation defined on S where A denotes a proper subset of B.

**Not Reflexive:**
Any set is a subset of itself, but not a proper subset.
\( \implies (A, A) \notin R \) for all \( A \in S \)
\( \implies \) R is not reflexive.

**Not Symmetric:**
Suppose (A, B) ∈ R for A, B ∈ S.
\( \implies \) A is a proper subset of B
\( \implies \) all elements of A belong to B, but B contains at least one element not in A.
\( \implies \) B cannot be a proper subset of A
\( \implies (B, A) \notin R \)
For example, if B = {1, 2, 5} and A = {1, 5}, then A is a proper subset of B. However, B is not a proper subset of A.
\( \implies \) R is not symmetric.

**Transitive:**
Suppose (A, B) ∈ R and (B, C) ∈ R for A, B, C ∈ S.
\( \implies \) A is a proper subset of B and B is a proper subset of C
\( \implies \) A is a proper subset of C
\( \implies (A, C) \in R \)
For example, A = {1, 5}, B = {1, 2, 5}, and C = {1, 2, 5, 7}. Here A ⊂ B and B ⊂ C, so A ⊂ C.
\( \implies \) R is transitive.

Thus, R is transitive but not reflexive and not symmetric.

Exam Tip: For proper subset relations, always distinguish between a subset and a proper subset - a set is never a proper subset of itself. Use concrete examples to demonstrate non-reflexivity and non-symmetry.

Question 10. Let R = {(a, \(\frac{1}{a}\)) : a ∈ N and 1 < a < 5}. Find the domain and range of R.
Answer: dom(R) = {2, 3, 4} and range(R) = \(\left\{\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right\}\)

Exam Tip: The domain consists of natural numbers between 1 and 5 (exclusive), and for each domain element, calculate its reciprocal to obtain the range.

Question 11. On the set S of all real numbers, define a relation R = {(a, b) : a ≤ b}. Show that R is (i) reflexive (ii) transitive (iii) not symmetric.
Answer: Let R = {(a, b) : a ≤ b}, a relation defined on S.

**Reflexive:**
Any element x ∈ S satisfies the property that x is less than or equal to itself.
\( \implies (x, x) \in R \) for all \( x \in S \)
\( \implies \) R is reflexive.

**Not Symmetric:**
Suppose (x, y) ∈ R for x, y ∈ S.
\( \implies \) x is less than or equal to y
However, y cannot be less than or equal to x when x is less than or equal to y, unless they are equal.
\( \implies (y, x) \notin R \)
For example, (2, 5) ∈ R since 2 < 5, but (5, 2) is not in R because 5 is not less than or equal to 2.
\( \implies \) R is not symmetric.

**Transitive:**
Suppose (x, y) ∈ R and (y, z) ∈ R for x, y, z ∈ S.
\( \implies \) x ≤ y and y ≤ z
\( \implies \) x ≤ z
\( \implies (x, z) \in R \)
For example, (4, 5) ∈ R since 4 ≤ 5, and (5, 6) ∈ R since 5 ≤ 6. We know 4 ≤ 6, so (4, 6) ∈ R.
\( \implies \) R is transitive.

Thus, R is reflexive and transitive but not symmetric.

Exam Tip: The "less than or equal to" relation is a classic example of a reflexive and transitive but not symmetric relation. Use number examples to illustrate why symmetry fails.

Question 12. Let A = {1, 2, 3, 4, 5, 6} and let R = {(a, b) : a, b ∈ A and b = a + 1}. Show that R is (i) not reflexive, (ii) not symmetric and (iii) not transitive.
Answer: Given A = {1, 2, 3, 4, 5, 6} and R = {(a, b) : a, b ∈ A and b = a + 1}.
\( \therefore \) R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

**Not Reflexive:**
For R to be reflexive, (a, a) must belong to R for all a ∈ A. However, a = a + 1 is never true for any real number a.
Since (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) are all outside R,
\( \implies \) R is not reflexive.

**Not Symmetric:**
For R to be symmetric, if (a, b) ∈ R then (b, a) must also belong to R.
We observe that (1, 2) ∈ R, but (2, 1) is not in R because 2 ≠ 1 + 1.
\( \implies \) R is not symmetric.

**Not Transitive:**
For R to be transitive, if (a, b) ∈ R and (b, c) ∈ R then (a, c) must belong to R.
We observe that (1, 2) ∈ R and (2, 3) ∈ R, but (1, 3) is not in R because 3 ≠ 1 + 1.
\( \implies \) R is not transitive.

Exam Tip: Relations defined by strict equality conditions like b = a + 1 typically fail reflexivity and transitivity. Check each property systematically using counterexamples from the given relation.

Exercise 1B

Question 1. Define a relation on a set. What do you mean by the domain and range of a relation? Give an example.
Answer: **Relation:** Let A and B be two non-empty sets. A relation R from set A to set B is a subset of the Cartesian product A × B. Thus, R is a relation from A to B if and only if R ⊆ A × B.

If R is a relation from set A to set B and (a, b) ∈ R, we write a R b, which means "a is related to b by relation R". If (a, b) ∉ R, we write a \(\overline{R}\) b, meaning a is not related to b by R.

**Domain:** The domain of a relation R from set A to set B is the set containing all first components of the ordered pairs in R. Symbolically, dom(R) = {a : (a, b) ∈ R}. Note that dom(R) ⊆ A.

**Range:** The range of a relation R from set A to set B is the set containing all second components of the ordered pairs in R. Symbolically, range(R) = {b : (a, b) ∈ R}. Note that range(R) ⊆ B.

**Example 1:** R = {(-1, 1), (1, 1), (-2, 4), (2, 4)}
dom(R) = {-1, 1, -2, 2} and range(R) = {1, 4}

**Example 2:** R = {(a, b) : a, b ∈ N and a + 3b = 12}
dom(R) = {3, 6, 9} and range(R) = {3, 2, 1}

Exam Tip: Always remember that domain and range are subsets of the original sets involved. When listing them, extract first coordinates for domain and second coordinates for range from all ordered pairs in the relation.

Question 2. Let A be the set of all triangles in a plane. Show that the relation R = {(Δ₁, Δ₂) : Δ₁ ~ Δ₂} is an equivalence relation on A.
Answer: Let R = {(Δ₁, Δ₂) : Δ₁ ~ Δ₂}, a relation defined on A (where ~ denotes similarity).

**Reflexive:**
For R to be reflexive, (Δ, Δ) must belong to R for every Δ ∈ A. We know that every triangle is similar to itself, so Δ ~ Δ for all Δ ∈ A.
\( \implies (Δ, Δ) \in R \) for all \( Δ \in A \)
\( \implies \) R is reflexive.

**Symmetric:**
For R to be symmetric, if (Δ₁, Δ₂) ∈ R then (Δ₂, Δ₁) must also belong to R. Suppose (Δ₁, Δ₂) ∈ R, so Δ₁ ~ Δ₂. By the property of similarity, Δ₂ ~ Δ₁.
\( \implies (Δ₂, Δ₁) \in R \)
\( \implies \) R is symmetric.

**Transitive:**
For R to be transitive, if (Δ₁, Δ₂) ∈ R and (Δ₂, Δ₃) ∈ R then (Δ₁, Δ₃) must belong to R. Suppose (Δ₁, Δ₂) ∈ R and (Δ₂, Δ₃) ∈ R, so Δ₁ ~ Δ₂ and Δ₂ ~ Δ₃. By the property of similarity, Δ₁ ~ Δ₃.
\( \implies (Δ₁, Δ₃) \in R \)
\( \implies \) R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation on A.

Exam Tip: Similarity of geometric figures is always an equivalence relation. Remember that the three core properties - reflexivity, symmetry, and transitivity - must all be satisfied for equivalence. Verify each one carefully.

Question 3. Let R = {(a, b) : a, b ∈ Z and (a + b) is even}. Show that R is an equivalence relation on Z.
Answer: To establish that R is an equivalence relation, we demonstrate that R is reflexive, symmetric, and transitive.

Given: For all a, b ∈ Z, R = {(a, b) : (a + b) is even}.

**Reflexive:**
For any a ∈ Z, we compute a + a = 2a, which is always even.
\( \implies (a, a) \in R \)
Thus, R is reflexive.

**Symmetric:**
Suppose (a, b) ∈ R for a, b ∈ Z. Then a + b is even. Since addition is commutative, b + a is also even.
\( \implies (b, a) \in R \)
Thus, R is symmetric.

**Transitive:**
Suppose (a, b) ∈ R and (b, c) ∈ R for a, b, c ∈ Z. Then a + b = 2P and b + c = 2Q for some integers P and Q. Adding these equations:
\( a + c + 2b = 2(P + Q) \)
\( a + c = 2(P + Q) - 2b \)
Since 2(P + Q) and 2b are both even, their difference is even.
\( \implies a + c \) is even
\( \implies (a, c) \in R \)
Thus, R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation on Z.

Exam Tip: For relations involving sums or differences, use parity arguments (even/odd) to prove properties efficiently. The symmetry of addition makes symmetry checks straightforward.

Question 4. Let R = {(a, b) : a, b ∈ Z and (a - b) is divisible by 5}. Show that R is an equivalence relation on Z.
Answer: To establish that R is an equivalence relation, we demonstrate that R is reflexive, symmetric, and transitive.

Given: For all a, b ∈ Z, a R b if and only if (a - b) is divisible by 5.

**Reflexive:**
For a R a to hold, (a - a) must be divisible by 5. We have a - a = 0 = 0 × 5, and since 0 is divisible by 5,
\( \implies (a, a) \in R \)
Thus, R is reflexive on Z.

**Symmetric:**
Suppose (a, b) ∈ R. Then (a - b) is divisible by 5, so a - b = 5z for some z ∈ Z. This means b - a = -5z = 5(-z). Since z ∈ Z, we have -z ∈ Z, so (b - a) is divisible by 5.
\( \implies (b, a) \in R \)
Thus, R is symmetric on Z.

**Transitive:**
Suppose (a, b) ∈ R and (b, c) ∈ R. Then (a - b) is divisible by 5 and (b - c) is divisible by 5. So a - b = 5z₁ and b - c = 5z₂ for some z₁, z₂ ∈ Z. Adding these:
\( (a - b) + (b - c) = 5z_1 + 5z_2 \)
\( a - c = 5(z_1 + z_2) \)
Since z₁ + z₂ ∈ Z, (a - c) is divisible by 5.
\( \implies (a, c) \in R \)
Thus, R is transitive on Z.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation on Z.

Exam Tip: Divisibility relations frequently generate equivalence relations. Use the definition of divisibility (a - b = kn for integer k) to verify reflexivity, symmetry, and transitivity systematically.

Question 5. Show that the relation R defined on the set A = {1, 2, 3, 4, 5}, given by R = {(a, b) : |a - b| is even} is an equivalence relation.
Answer: To establish that R is an equivalence relation, we demonstrate that R is reflexive, symmetric, and transitive.

Given: For all a, b ∈ A, R = {(a, b) : |a - b| is even}.

**Reflexive:**
For any a ∈ A, |a - a| = 0, which is even.
\( \implies (a, a) \in R \)
Thus, R is reflexive.

**Symmetric:**
Suppose (a, b) ∈ R. Then |a - b| is even. Since |a - b| = |b - a|, we have |b - a| is even.
\( \implies (b, a) \in R \)
Thus, R is symmetric.

**Transitive:**
Suppose (a, b) ∈ R and (b, c) ∈ R for a, b, c ∈ A. Then |a - b| is even and |b - c| is even. This means (a and b have the same parity) and (b and c have the same parity). Therefore, a and c have the same parity, so |a - c| is even.

Case 1: When b is even
If |a - b| is even and b is even, then a is even. If |b - c| is even and b is even, then c is even. The difference of two even numbers is even, so |a - c| is even.
\( \implies (a, c) \in R \)

Case 2: When b is odd
If |a - b| is even and b is odd, then a is odd. If |b - c| is even and b is odd, then c is odd. The difference of two odd numbers is even, so |a - c| is even.
\( \implies (a, c) \in R \)

Thus, R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation on Z.

Exam Tip: When dealing with absolute value conditions, analyze parity (even/odd) to determine relationships. Breaking transitivity into cases based on the parity of the intermediate element strengthens your proof.

Question 6. Show that the relation R on N × N, defined by (a, b) R (c, d) ⇔ a + d = b + c is an equivalent relation.
Answer: To establish that R is an equivalence relation, we demonstrate that R is reflexive, symmetric, and transitive.

Given: R is the relation in N × N defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) ∈ N × N.

**Reflexive:**
For (a, b) R (a, b) to hold, we need a + b = b + a, which is true by the commutative property of addition on N.
\( \implies \) R is reflexive.

**Symmetric:**
Suppose (a, b) R (c, d). Then a + d = b + c. By the symmetry of equality, b + c = a + d, which can be rewritten as c + b = d + a (by commutativity).
\( \implies (c, d) R (a, b) \)
Thus, R is symmetric.

**Transitive:**
Suppose (a, b) R (c, d) and (c, d) R (e, f). Then a + d = b + c and c + f = d + e. Subtracting the second from the first:
\( (a + d) - (d + e) = (b + c) - (c + f) \)
\( a - e = b - f \)
\( a + f = b + e \)
\( \implies (a, b) R (e, f) \)
Thus, R is transitive.

Hence, R is an equivalence relation.

Exam Tip: For ordered pairs, use algebraic manipulation and the commutative property of addition carefully. When proving transitivity for composite relations, isolate terms skillfully to reveal the target equality.

Question 7. Let S be the set of all real numbers and let R = {(a, b) : a, b ∈ S and a = ± b}. Show that R is an equivalence relation on S.
Answer: To establish that R is an equivalence relation, we demonstrate that R is reflexive, symmetric, and transitive.

Given: For all a, b ∈ S, R = {(a, b) : a = ± b}.

**Reflexive:**
For any a ∈ S, a = +a (i.e., a equals a positively).
\( \implies (a, a) \in R \)
Thus, R is reflexive.

**Symmetric:**
Suppose (a, b) ∈ R. Then a = ± b, which means a = b or a = -b. If a = b, then b = a, so (b, a) ∈ R. If a = -b, then b = -a, so (b, a) ∈ R. In both cases,
\( \implies (b, a) \in R \)
Thus, R is symmetric.

**Transitive:**
Suppose (a, b) ∈ R and (b, c) ∈ R for a, b, c ∈ S. Then a = ± b and b = ± c, so a = ± c.
\( \implies (a, c) \in R \)
Thus, R is transitive.

Hence, R is an equivalence relation.

Exam Tip: The ± notation means both positive and negative cases must be covered. For transitivity, the composition of two ± relationships still yields a ± relationship between the endpoints.

Question 8. Let S be the set of all points in a plane and let R be a relation in S defined by R = {(A, B) : d(A, B) < 2 units}, where d(A, B) is the distance between the points A and B. Show that R is reflexive and symmetric but not transitive.
Answer: Given: For all A, B ∈ S, R = {(A, B) : d(A, B) < 2 units}.

**Reflexive:**
For any A ∈ S, the distance d(A, A) = 0, which is less than 2 units.
\( \implies (A, A) \in R \)
Thus, R is reflexive.

**Symmetric:**
Suppose (A, B) ∈ R. Then d(A, B) < 2 units. Since distance is symmetric (d(A, B) = d(B, A)), we have d(B, A) < 2 units.
\( \implies (B, A) \in R \)
Thus, R is symmetric.

**Not Transitive:**
Consider three points on a line: A(0, 0), B(1.5, 0), and C(3.2, 0).
d(A, B) = 1.5 units < 2 units, so (A, B) ∈ R.
d(B, C) = 1.7 units < 2 units, so (B, C) ∈ R.
However, d(A, C) = 3.2 units, which is not less than 2 units.
\( \implies (A, C) \notin R \)
Thus, (A, B) ∈ R and (B, C) ∈ R, but (A, C) ∉ R, proving R is not transitive.

Therefore, R is reflexive and symmetric but not transitive.

Exam Tip: Distance-based relations often fail transitivity because intermediate points can bridge two distant points. Provide explicit coordinate examples when disproving transitivity, using the distance formula.

Question 9. Let S be the set of all real numbers. Show that the relation R = {(a, b) : a² + b² = 1} is symmetric but neither reflexive nor transitive.
Answer: Given: For all a, b ∈ S, R = {(a, b) : a² + b² = 1}.

**Not Reflexive:**
For R to be reflexive, (a, a) must belong to R for all a ∈ S. We compute a² + a² = 2a², which equals 1 only when a = ±\(\frac{1}{\sqrt{2}}\), not for all a ∈ S.
\( \implies (a, a) \notin R \) for most a ∈ S
Thus, R is not reflexive.

**Symmetric:**
Suppose (a, b) ∈ R. Then a² + b² = 1. By the commutative property of addition, b² + a² = 1.
\( \implies (b, a) \in R \)
Thus, R is symmetric.

**Not Transitive:**
Suppose (a, b) ∈ R and (b, c) ∈ R for a, b, c ∈ S. Then a² + b² = 1 and b² + c² = 1. Adding these equations:
\( a^2 + c^2 + 2b^2 = 2 \)
\( a^2 + c^2 = 2 - 2b^2 \)
For most values of b, 2 - 2b² ≠ 1 (it only equals 1 when b² = 1/2).
\( \implies (a, c) \notin R \)
Thus, R is not transitive.

Therefore, R is symmetric but neither reflexive nor transitive.

Exam Tip: For relations defined by equations like a² + b² = 1, use algebraic manipulation to determine failures in reflexivity and transitivity. The symmetry of the equation in its variables guarantees symmetry of the relation.

Question 10. Let R = {(a, b) : a = b²} for all a, b ∈ N. Show that R satisfies none of reflexivity, symmetry and transitivity.
Answer: We have R = {(a, b) : a = b²}, a relation defined on N.

**Not Reflexive:**
For R to be reflexive, (a, a) must belong to R for all a ∈ N. We would need a = a², which holds only for a = 1, not for all natural numbers. For instance, 2 ≠ 2².
\( \implies (a, a) \notin R \) for most a ∈ N
Thus, R is not reflexive.

**Not Symmetric:**
Suppose (a, b) ∈ R. Then a = b². For (b, a) to be in R, we would need b = a². If a = b² and b = a², then substituting the first into the second gives b = (b²)² = b⁴, which means b⁴ = b. This holds only for b = 1, not universally. For example, (4, 2) ∈ R since 4 = 2², but (2, 4) ∉ R because 2 ≠ 4².
\( \implies \) R is not symmetric.

**Not Transitive:**
Suppose (a, b) ∈ R and (b, c) ∈ R. Then a = b² and b = c². From these, a = (c²)² = c⁴. For (a, c) to be in R, we would need a = c², but we have a = c⁴. These are equal only when c = 1. For example, (16, 4) ∈ R since 16 = 4², and (4, 2) ∈ R since 4 = 2². However, (16, 2) ∉ R because 16 ≠ 2² (we have 16 = 2⁴ instead).
\( \implies \) R is not transitive.

Thus, R satisfies none of reflexivity, symmetry, and transitivity.

Exam Tip: For exponential relations like a = b², provide concrete numerical counterexamples to disprove each property. The exponential nature prevents the relation from behaving reflexively or transitively on the full domain.

Question 11. Show that the relation R = {(a, b) : a > b} on N is transitive but neither reflexive nor symmetric.
Answer: We have R = {(a, b) : a > b}, a relation defined on N.

**Not Reflexive:**
For R to be reflexive, (a, a) must belong to R for all a ∈ N. We would need a > a, which is never true for any real number a.
\( \implies (a, a) \notin R \) for all a ∈ N
Thus, R is not reflexive.

**Not Symmetric:**
Suppose (a, b) ∈ R. Then a > b. For (b, a) to be in R, we would need b > a. Since a > b and b > a cannot both be true simultaneously,
\( \implies (b, a) \notin R \)
For example, (5, 2) ∈ R since 5 > 2, but (2, 5) ∉ R because 2 is not greater than 5.
Thus, R is not symmetric.

**Transitive:**
Suppose (a, b) ∈ R and (b, c) ∈ R for a, b, c ∈ N. Then a > b and b > c. By the transitive property of the "greater than" relation, a > c.
\( \implies (a, c) \in R \)
For example, (5, 4) ∈ R since 5 > 4, and (4, 3) ∈ R since 4 > 3. We know 5 > 3, so (5, 3) ∈ R.
Thus, R is transitive.

Therefore, R is transitive but not reflexive or symmetric.

Exam Tip: Order relations like "greater than" are classically transitive but never reflexive or symmetric. Use the algebraic properties of inequalities to justify transitivity.

Question 12. Let A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}. Show that R is reflexive but neither symmetric nor transitive.
Answer: Given: A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}.

**Reflexive:**
R is reflexive because (1, 1), (2, 2), and (3, 3) all belong to R. All diagonal elements from A × A are present in R.
\( \implies \) R is reflexive.

**Not Symmetric:**
For R to be symmetric, if (a, b) ∈ R then (b, a) must also belong to R. We observe that (1, 2) ∈ R, but (2, 1) ∉ R. Similarly, (2, 3) ∈ R, but (3, 2) ∉ R.
\( \implies \) R is not symmetric.

**Not Transitive:**
For R to be transitive, if (a, b) ∈ R and (b, c) ∈ R then (a, c) must belong to R. We have (1, 2) ∈ R and (2, 3) ∈ R, but (1, 3) ∉ R.
\( \implies \) R is not transitive.

Thus, R is reflexive but neither symmetric nor transitive.

Exam Tip: When checking properties of finite relations on small sets, list all ordered pairs explicitly and verify each property systematically by checking all relevant pairs.

Question 13. Let A = {1, 2, 3, 4} and R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (3, 2)}. Show that R is reflexive and transitive but not symmetric.
Answer: Given: A = {1, 2, 3, 4} and R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (3, 2)}.

**Reflexive:**
R is reflexive because (1, 1), (2, 2), (3, 3), and (4, 4) all belong to R. All diagonal elements from A × A are present in R.
\( \implies \) R is reflexive.

**Not Symmetric:**
For R to be symmetric, if (a, b) ∈ R then (b, a) must also belong to R. We observe that (1, 2) ∈ R, but (2, 1) ∉ R. Similarly, (1, 3) ∈ R, but (3, 1) ∉ R. Also, (3, 2) ∈ R, but (2, 3) ∉ R.
\( \implies \) R is not symmetric.

**Transitive:**
We verify transitivity by checking all pairs (a, b) and (b, c) in R. Since (1, 3) ∈ R and (3, 2) ∈ R, we check that (1, 2) ∈ R, which is true. No other non-trivial transitive pairs violate the condition. All other pairs either involve an element's relation to itself (reflexive pairs) or cannot be chained.
\( \implies \) R is transitive.

Thus, R is reflexive and transitive but not symmetric.

Exam Tip: Always verify transitivity by examining all possible chains: when both (a, b) and (b, c) exist in R, ensure (a, c) also exists. List checks explicitly to avoid missing problematic pairs.

Objective Questions

Question 1. Mark the tick against the correct answer in the following: Let A = {1, 2, 3} and let R = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 2), (1, 2)}. Then, R is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation
Answer: (b) reflexive and transitive but not symmetric
In simple words: R contains all diagonal pairs (1,1), (2,2), (3,3), making it reflexive. It fails symmetry because (1,3) ∈ R but (3,1) ∉ R. It is transitive since (1,3) ∈ R and (3,2) ∈ R implies (1,2) ∈ R, and we can verify no other violations occur.

Exam Tip: Quickly check reflexivity by confirming diagonal elements (a, a). Check symmetry by finding one counterexample. Verify transitivity systematically by examining all pairs (a, b) and (b, c) and ensuring (a, c) appears.

Question 2. Mark the tick against the correct answer in the following: Let A = {a, b, c} and let R = {(a, a), (a, b), (b, a)}. Then, R is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation
Answer: (c) symmetric and transitive but not reflexive
In simple words: R is not reflexive because (b, b) and (c, c) are missing from R. It is symmetric since (a, b) ∈ R and (b, a) ∈ R are both present. It is transitive: the pairs (a, b) and (b, a) both map to (a, a), which is included in R.

Exam Tip: Missing reflexive elements are easy to spot - look for any diagonal pair (x, x) absent from R. Use the symmetry of given pairs to establish symmetry, and check transitivity carefully for all compositional chains.

Question 3. Mark the tick against the correct answer in the following: Let A = {a, b, c} and let R = {(a, a), (a, b), (b, a)}. Then, R is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation
Answer: (c) symmetric and transitive but not reflexive
In simple words: Since (b, b) and (c, c) do not belong to R, the relation fails reflexivity. Both (a, b) and (b, a) appear in R, confirming symmetry. For transitivity, (a, b) ∈ R and (b, a) ∈ R yield (a, a) ∈ R, which holds, so it is transitive.

Exam Tip: This question duplicates the previous one - confirm by counting set elements and listed pairs. When verifying transitivity in small finite relations, enumerate all possible two-step paths and confirm their endpoints exist in R.

Question 1. Let A = {1, 2, 3} and let R = {(1, 1),(2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}. Then, R is
(a) reflexive and symmetric but not transitive
(b) symmetric and transitive but not reflexive
(c) reflexive and transitive but not symmetric
(d) an equivalence relation
Answer: (a) reflexive and symmetric but not transitive
In simple words: A relation shows three main properties. Reflexive means every element relates to itself. Symmetric means if one element relates to another, then the second relates back to the first. Transitive means if the first relates to the second, and the second relates to a third, then the first must relate to the third. Here the relation has the first two properties but fails the third because (1,2) and (2,3) are in R, yet (1,3) is missing.

Exam Tip: Always verify all three properties - check reflexivity by confirming (a,a) is in R for every element, symmetry by matching pairs in both directions, and transitivity by testing the chain rule with actual ordered pairs.

Question 2. Let S be the set of all straight lines in a plane. Let R be a relation on S defined by a R b - a ⊥ b. Then, R is
(a) reflexive but neither symmetric nor transitive
(b) symmetric but neither reflexive nor transitive
(c) transitive but neither reflexive nor symmetric
(d) an equivalence relation
Answer: (b) symmetric but neither reflexive nor transitive
In simple words: When two lines are perpendicular (one is at right angles to the other), the relationship works both ways - if line a is perpendicular to line b, then line b is also perpendicular to line a. However, a line cannot be perpendicular to itself, and if line a is perpendicular to line b and line b is perpendicular to line c, then line a and line c will be parallel, not perpendicular.

Exam Tip: Geometric relations often fail reflexivity - a line cannot be perpendicular or parallel to itself. Recognize that perpendicularity is symmetric but breaks transitivity when you chain the relationship.

Question 3. Let S be the set of all straight lines in a plane. Let R be a relation on S defined by a R b - a || b. Then, R is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation
Answer: (d) an equivalence relation
In simple words: When two lines are parallel, several things hold true. Every line is parallel to itself (a line runs in the same direction as itself). If line a is parallel to line b, then line b is also parallel to line a. If line a is parallel to line b, and line b is parallel to line c, then line a must be parallel to line c. These three conditions make parallelism an equivalence relation.

Exam Tip: Parallelism is one of the classic equivalence relations in geometry - it partitions all lines into equivalence classes where lines in the same class share the same slope or direction.

Question 4. Let Z be the set of all integers and let R be a relation on Z defined by a R b - (a - b) is divisible by 3. Then, R is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation
Answer: (d) an equivalence relation
In simple words: Two integers are related if their difference is a multiple of 3. When you subtract any number from itself, you get 0, which is divisible by 3. If a minus b is divisible by 3, then b minus a is also divisible by 3 (negative multiples of 3 are still multiples of 3). If a minus b and b minus c are both divisible by 3, then when you add these differences, a minus c will also be divisible by 3. All three conditions are satisfied, making this an equivalence relation that groups integers into three classes based on remainders.

Exam Tip: Divisibility relations are equivalence relations - they partition the set into remainder classes. Always verify transivity by adding two conditions algebraically.

Question 5. Let R be a relation on the set N of all natural numbers, defined by a R b - a is a factor of b. Then, R is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation
Answer: (b) reflexive and transitive but not symmetric
In simple words: A number is always a factor of itself, so every element relates to itself. If one number divides another, the reverse is not always true - for example, 2 divides 6, but 6 does not divide 2. If one number is a factor of a second, and the second is a factor of a third, then the first must be a factor of the third. The relation is reflexive and transitive but fails symmetry.

Exam Tip: Factor relations are rarely symmetric because divisibility is directional - a smaller number divides a larger one, not the reverse. Always test with small numbers like (2,6) to disprove symmetry.

Question 6. Let Z be the set of all integers and let R be a relation on Z defined by a R b - a ≥ b. Then, R is
(a) symmetric and transitive but not reflexive
(b) reflexive and symmetric but not transitive
(c) reflexive and transitive but not symmetric
(d) an equivalence relation
Answer: (c) reflexive and transitive but not symmetric
In simple words: Any number is greater than or equal to itself, so the relation is reflexive. If a is greater than or equal to b, and b is greater than or equal to c, then a must be greater than or equal to c - this shows transitivity. However, for symmetry to hold, both a ≥ b and b ≥ a would need to be true, which happens only when a equals b, not in general. The relation is reflexive and transitive but not symmetric.

Exam Tip: Order relations like ≥ or > are never symmetric unless equality is involved - use counterexamples like a=2, b=1 to quickly disprove symmetry.

Question 7. Let S be the set of all real numbers and let R be a relation on S defined by a R b - |a| ≤ b. Then, R is
(a) reflexive but neither symmetric nor transitive
(b) symmetric but neither reflexive nor transitive
(c) transitive but neither reflexive nor symmetric
(d) none of these
Answer: (c) transitive but neither reflexive nor symmetric
In simple words: The absolute value of a number is not always less than or equal to the number itself - for negative numbers like -2, the absolute value is 2, which is not ≤ -2. If the absolute value of a is ≤ b, the reverse (absolute value of b ≤ a) may not hold. However, if |a| ≤ b and |b| ≤ c, then |a| ≤ c, showing that the relation is transitive even though it lacks the other two properties.

Exam Tip: Always test reflexivity with negative numbers - many relations break down when a is negative. Remember that absolute value creates asymmetry in most relations.

Question 8. Let S be the set of all real numbers and let R be a relation on S, defined by a R b - |a - b| ≤ 1. Then, R is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation
Answer: (a) reflexive and symmetric but not transitive
In simple words: When you subtract any number from itself, the result is 0, which is ≤ 1, making the relation reflexive. If the distance between a and b is at most 1, then the distance between b and a is also at most 1, showing symmetry. However, if a and b are within distance 1, and b and c are within distance 1, then a and c could be up to 2 units apart - for instance, a = -5, b = -6, c = -7 gives differences of 1 and 1, but a and c are 2 units apart, so transitivity fails.

Exam Tip: Distance-based relations often fail transitivity when you chain them - the error accumulates. Always test with numbers spread across the threshold (like -5, -6, -7) to find a counterexample.

Question 9. Let S be the set of all real numbers and let R be a relation on S, defined by a R b - (1 + ab) > 0. Then, R is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) none of these
Answer: (a) reflexive and symmetric but not transitive
In simple words: When you multiply any number by itself, the product is never negative, so 1 plus this product is always positive - making the relation reflexive. Multiplication is commutative, meaning a times b equals b times a, so if (1 + ab) > 0, then (1 + ba) > 0 - the relation is symmetric. For transitivity, if a = -1, b = 0, c = 2, then (1 + (-1)(0)) = 1 > 0 and (1 + (0)(2)) = 1 > 0 are both true, but (1 + (-1)(2)) = -1 < 0 is false, so transitivity breaks down.

Exam Tip: When testing transitive relations involving products, use integers with different signs - negative times positive creates the critical failure cases.

Question 10. Let S be the set of all triangles in a plane and let R be a relation on S defined by Δ₁ R Δ₂ - Δ₁ ≡ Δ₂. Then, R is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation
Answer: (d) an equivalence relation
In simple words: Congruence among triangles is reflexive because every triangle is congruent to itself. If one triangle is congruent to another, then the second is congruent to the first, so the relation is symmetric. If triangle 1 is congruent to triangle 2, and triangle 2 is congruent to triangle 3, then triangle 1 must be congruent to triangle 3 - the relation is transitive. Since all three properties hold, congruence is an equivalence relation that groups triangles by shape and size.

Exam Tip: Congruence and similarity relations in geometry are equivalence relations - they partition geometric objects into classes of equivalent figures. Use this to quickly identify the answer when congruence or equality is involved.

Question 11. Let S be the set of all real numbers and let R be a relation on S defined by a R b - a² + b² = 1. Then, R is
(a) symmetric but neither reflexive nor transitive
(b) reflexive but neither symmetric nor transitive
(c) transitive but neither reflexive nor symmetric
(d) none of these
Answer: (a) symmetric but neither reflexive nor transitive
In simple words: For reflexivity, when you add the square of any number to itself, you get 2a², which equals 1 only in rare cases like a = ±1/√2 - not for all numbers, so the relation is not reflexive. If a² + b² = 1, then b² + a² = 1 (addition is commutative), so the relation is symmetric. For transitivity, if a² + b² = 1 and b² + c² = 1, it does not follow that a² + c² = 1 - for example, a = -1, b = 0, c = 1 gives (-1)² + 0² = 1 and 0² + 1² = 1, but (-1)² + 1² = 2 ≠ 1.

Exam Tip: Relations involving sums of squares often fail reflexivity and transitivity but maintain symmetry because squares are commutative. Test with specific values that satisfy the equation to find counterexamples quickly.

Question 12. Let R be a relation on N × N, defined by(a, b) R (c, d) - a + d = b + c. Then, R is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation
Answer: (d) an equivalence relation
In simple words: For reflexivity, when you pair (a,b) with itself, you need a + b = b + a, which is always true by the commutative property of addition. For symmetry, if a + d = b + c, then rearranging gives c + b = d + a, so the condition is satisfied both ways. For transitivity, if a + d = b + c and c + f = d + e, adding these equations yields a + f = b + e, which is exactly the condition for (a,b) R (e,f). All three properties hold, making this an equivalence relation on ordered pairs.

Exam Tip: Relations on Cartesian products often become equivalence relations when based on algebraic identities involving sums or differences. Verify transitivity by adding or manipulating the given equations.

Question 13. Let A be the set of all points in a plane and let O be the origin. Let R = {(P, Q) : OP = OQ}. Then, R is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation
Answer: (d) an equivalence relation
In simple words: The relation connects two points if they are the same distance from the origin. Every point is the same distance from the origin as itself, so the relation is reflexive. If point P is the same distance from O as point Q, then point Q is the same distance from O as point P, making the relation symmetric. If point P has the same distance from O as point Q, and point Q has the same distance from O as point R, then point P must have the same distance from O as point R - this is transitive. All conditions are met, making this an equivalence relation that groups points into circles around the origin.

Exam Tip: Relations based on equal distances or equal measurements are typically equivalence relations. Recognize that transitivity follows directly from the symmetry of the equality sign in the distance equation.

Question 16. Mark the tick against the correct answer in the following: Let Q be the set of all rational numbers, and * be the binary operation, defined by a * b = a + 2b, then
(A) * is commutative but not associative
(B) * is associative but not commutative
(C) * is neither commutative nor associative
(D) * is both commutative and associative
Answer: (C) * is neither commutative nor associative
In simple words: An operation is commutative when switching the order of two numbers gives the same result. It is associative when grouping three numbers differently still produces the same result. For this operation, neither property holds true.

Exam Tip: Always test commutativity by computing a * b and b * a with specific values - if they differ, the operation fails this property. Similarly, verify associativity by comparing (a * b) * c with a * (b * c).

Question 17. Mark the tick against the correct answer in the following: Let a * b = a + ab for all a, b ∈ Q. Then,
(A) * is not a binary composition
(B) * is not commutative
(C) * is commutative but not associative
(D) * is both commutative and associative
Answer: (B) * is not commutative
In simple words: When you swap the positions of a and b in this formula, you get a different result. For example, a + ab does not equal b + ba in general, so the operation fails the commutative property.

Exam Tip: For non-commutative operations, construct a quick numerical counterexample using simple values like a = 2, b = 3 to prove the operation fails commutativity.

Question 18. Mark the tick against the correct answer in the following: Let Q+ be the set of all positive rationals. Then, the operation * on Q+ defined by for all a, b ∈ Q+ is
(A) commutative but not associative
(B) associative but not commutative
(C) neither commutative nor associative
(D) both commutative and associative
Answer: (D) both commutative and associative
In simple words: When you multiply two positive rationals and divide by 2, swapping their order gives the same result - that is commutativity. Grouping three of them in different ways also yields identical outcomes - that is associativity.

Exam Tip: For operations involving multiplication and division, commutativity often holds because the basic operations are commutative; verify associativity by carefully expanding both (a * b) * c and a * (b * c) with the operation definition.

Question 19. Mark the tick against the correct answer in the following: let Z be the set of all integers and let a * b = a - b + ab. Then, * is
(A) commutative but not associative
(B) associative but not commutative
(C) neither commutative nor associative
(D) both commutative and associative
Answer: (C) neither commutative nor associative
In simple words: The subtraction part (a - b versus b - a) makes this operation non-commutative. The algebra also shows that grouping changes the outcome, so it is not associative either.

Exam Tip: When an operation involves subtraction (which is not inherently commutative), suspect failure of commutativity and test it first with concrete integers before moving to the associativity check.

Question 20. Mark the tick against the correct answer in the following: Let Z be the set of all integers. Then, the operation * on Z defined by a * b = a + b - ab is
(A) commutative but not associative
(B) associative but not commutative
(C) neither commutative nor associative
(D) both commutative and associative
Answer: (D) both commutative and associative
In simple words: Reordering a and b in the formula a + b - ab gives b + a - ba, which simplifies to the same thing. Grouping three values in different ways also produces identical results in both cases.

Exam Tip: Notice that addition and subtraction (when written as addition of a negative term) together can form associative operations; verify by expanding both grouped forms completely and comparing all terms.

Question 21. Mark the tick against the correct answer in the following: Let Z+ be the set of all positive integers. Then, the operation * on Z+ defined by a * b = ab is
(A) commutative but not associative
(B) associative but not commutative
(C) neither commutative nor associative
(D) both commutative and associative
Answer: (C) neither commutative nor associative
In simple words: Raising a to the power b is not the same as raising b to the power a (for example, 2³ = 8 but 3² = 9). Similarly, grouping changes the outcome: (2³)⁴ = 2¹² while 2^(3⁴) = 2⁸¹, which are entirely different.

Exam Tip: Exponentiation is notoriously non-commutative and non-associative; always test with small integer values to expose how dramatically the results differ under reordering or regrouping.

Question 22. Mark the tick against the correct answer in the following: Define * on Q - {-1} by a * b = a + b + ab. Then, * on Q - {-1} is
(A) commutative but not associative
(B) associative but not commutative
(C) neither commutative nor associative
(D) both commutative and associative
Answer: (D) both commutative and associative
In simple words: Swapping a and b yields a + b + ba, which is identical to a + b + ab (since multiplication of numbers is commutative). Regrouping three elements and expanding fully also produces the same final expression, confirming associativity.

Exam Tip: When an operation is built from commutative building blocks (addition and multiplication), it often inherits commutativity; verify associativity by fully expanding both (a * b) * c and a * (b * c) and matching every term.