NCERT Solutions · Class 12 · Chapter 1: Relations and Functions
NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.1
Exercise 1.1: Types of relations. Reflexive: \((a, a) \in R\) for every \(a\). Symmetric: \((a, b) \in R \Rightarrow (b, a) \in R\). Transitive: \((a, b), (b, c) \in R \Rightarrow (a, c) \in R\). All three: an equivalence relation, which splits the set into disjoint classes. One counterexample is enough to show a property fails.
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Exercise 1.1 questions and solutions
Exercise 1.1, Question 1
Decide whether each relation is reflexive, symmetric and transitive.
(i) On \(A = \{1, 2, \ldots, 14\}\): \(R = \{(x, y) : 3x - y = 0\}\)
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Check reflexive (every \((a, a)\)), symmetric (\((a, b) \Rightarrow (b, a)\)) and transitive (\((a, b), (b, c) \Rightarrow (a, c)\)).
(ii) On \(\mathbb N\): \(R = \{(x, y) : y = x + 5,\ x < 4\}\)
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\(R = \{(1, 6), (2, 7), (3, 8)\}\).
\((1, 1) \notin R\): not reflexive. \((6, 1) \notin R\): not symmetric.
No chain \((a, b), (b, c)\) exists in R (no pair starts with 6, 7 or 8), so the transitivity condition never has a case that could fail: R is (vacuously) transitive.
Answer: Not reflexive, not symmetric; (vacuously) transitive
(iii) On \(A = \{1, 2, 3, 4, 5, 6\}\): \(R = \{(x, y) : y \text{ is divisible by } x\}\)
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Check reflexive (every \((a, a)\)), symmetric (\((a, b) \Rightarrow (b, a)\)) and transitive (\((a, b), (b, c) \Rightarrow (a, c)\)).
Every \(x\) divides itself: reflexive.
\(1 \mid 2\) but \(2 \nmid 1\): not symmetric.
\(x \mid y\) and \(y \mid z\) give \(x \mid z\): transitive.
Answer: Reflexive and transitive, not symmetric
(iv) On \(\mathbb Z\): \(R = \{(x, y) : x - y \text{ is an integer}\}\)
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\(x - x = 0 \in \mathbb Z\); if \(x - y \in \mathbb Z\) so is \(y - x\); \((x - y) + (y - z) = x - z \in \mathbb Z\).
Answer: Reflexive, symmetric and transitive (an equivalence relation)
(v)(a) On the people in a town: \(x\) and \(y\) work at the same place.
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Everyone works where they work; 'same place' is symmetric and transitive.
Answer: Equivalence relation
(v)(b) … \(x\) and \(y\) live in the same locality.
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As in (a): a 'same attribute' relation.
Answer: Equivalence relation
(v)(c) … \(x\) is exactly 7 cm taller than \(y\).
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Nobody is 7 cm taller than themselves; if \(x\) is 7 cm taller than \(y\), \(y\) is shorter; and \(x\) would be 14 cm taller than \(z\).
Answer: None of the three
(v)(d) … \(x\) is the wife of \(y\).
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Not reflexive (no one is their own wife); not symmetric (if \(x\) is the wife of \(y\), \(y\) is not the wife of \(x\)).
A chain \((x, y), (y, z)\) would need \(y\) to be a husband and a wife at once, so it never occurs: transitivity holds vacuously.
Answer: Not reflexive, not symmetric; (vacuously) transitive
(v)(e) … \(x\) is the father of \(y\).
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No one is their own father; the father of \(y\) is not \(y\)'s child; the father of the father of \(z\) is \(z\)'s grandfather.
On \(A = \{1, 2, 3, 4, 5\}\), \(R = \{(a, b) : |a - b| \text{ is even}\}\). Show R is an equivalence relation, all elements of \(\{1, 3, 5\}\) are related to each other, all of \(\{2, 4\}\) are related to each other, and no element of the first set is related to one of the second.
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\(|a - a| = 0\) is even; \(|a - b| = |b - a|\); if \(a - b\) and \(b - c\) are even, so is \(a - c\).
\(|a - b|\) is even exactly when \(a, b\) have the same parity: odd with odd, even with even, never odd with even.
Answer: Shown: classes \(\{1, 3, 5\}\) and \(\{2, 4\}\)
On the points of a plane, P is related to Q when P and Q are the same distance from the origin. Show this is an equivalence relation, and that the points related to \(P \ne (0, 0)\) form the circle through P centred at the origin.
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'Same distance from O' is a same-attribute relation: reflexive, symmetric, transitive.
The points at distance \(OP\) from O are exactly the circle of radius \(OP\) centred at O.
On the set of triangles, \(T_1 R T_2\) if they are similar. Show R is an equivalence relation. Which of the right triangles with sides 3, 4, 5; 5, 12, 13; 6, 8, 10 are related?
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Similarity is reflexive, symmetric and transitive.
\(\tfrac63 = \tfrac84 = \tfrac{10}{5} = 2\): the 3-4-5 and 6-8-10 triangles are similar; 5-12-13 is not (\(\tfrac53 \ne \tfrac{12}{4}\)).
Answer: Equivalence; \(T_1\) (3, 4, 5) and \(T_3\) (6, 8, 10) are related
On the set of polygons, \(P_1 R P_2\) if they have the same number of sides. Show R is an equivalence relation. What is the set of polygons related to the right triangle with sides 3, 4, 5?
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'Same number of sides' is a same-attribute relation.
L is the set of lines in the XY-plane, and \(L_1 R L_2\) if \(L_1 \parallel L_2\) (a line counts as parallel to itself). Show R is an equivalence relation, and find the lines related to \(y = 2x + 4\).
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Parallelism is reflexive (by convention), symmetric and transitive.
R on \(\{1, 2, 3, 4\}\) is \(\{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\}\). Choose: (A) reflexive and symmetric, not transitive (B) reflexive and transitive, not symmetric (C) symmetric and transitive, not reflexive (D) an equivalence relation
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All \((a, a)\) are present: reflexive.
\((1, 2) \in R\), \((2, 1) \notin R\): not symmetric.
Every chain closes up (e.g. \((1, 3), (3, 2) \Rightarrow (1, 2)\)): transitive.
R on \(\mathbb N\) is \(\{(a, b) : a = b - 2,\ b > 6\}\). Which is true? (A) \((2, 4) \in R\) (B) \((3, 8) \in R\) (C) \((6, 8) \in R\) (D) \((8, 7) \in R\)
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Relations and Functions has 39 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.
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