NCERT Class 12 Maths Chapter 1: Exercise 1.1 (Relations & Functions)

The Intuition Behind Relations and Equivalence

To master relations and function ex 1 1 class 12 ncert, we must first build a visual intuition of what a relation actually is. Imagine a set $A$ of students. A relation $R$ establishes a mathematical connection between pairs of students $(a, b)$ based on a specific rule (like 'having the same age'). In mathematical terms, $R$ is a subset of the Cartesian product $A \times A$. For any relation to be classified as an 'Equivalence Relation', it must fulfill three strict geometric or algebraic criteria simultaneously: reflexivity, symmetry, and transitivity. If even a single ordered pair fails any of these rules, the equivalence breaks down. In CBSE Board exams, problems from Exercise 1.1 are highly structured. Scoring full marks requires writing clear, formal mathematical proofs rather than using informal explanations. This section guides you through the exact rigour expected by board evaluators.

Types of Relations: Key Definitions

Reflexive Relation

A relation $R$ on a set $A$ is reflexive if every element is related to itself. Mathematically, $(a, a) \in R$ for all $a \in A$.

Symmetric Relation

A relation $R$ on a set $A$ is symmetric if $(a, b) \in R$ implies that $(b, a) \in R$ for all $a, b \in A$.

Transitive Relation

A relation $R$ on a set $A$ is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$ for all $a, b, c \in A$.

Equivalence Relation

A relation $R$ on a set $A$ that is simultaneously reflexive, symmetric, and transitive.

How to Prove an Equivalence Relation

1. Step 1: Check Reflexivity — Substitute $a$ for $b$ in the given relation rule. Check if the resulting statement is true for all elements $a \in A$. Write down: 'Since $(a, a) \in R \forall a \in A$, $R$ is reflexive.'
2. Step 2: Check Symmetry — Assume $(a, b) \in R$ is true. This gives you a starting equation. Rearrange this equation to see if it implies $(b, a) \in R$ is also true. If yes, write: 'Since $(a, b) \in R \implies (b, a) \in R$, $R$ is symmetric.'
3. Step 3: Check Transitivity — Assume both $(a, b) \in R$ and $(b, c) \in R$ are true. Combine these two equations algebraically to eliminate $b$ and find a direct relation between $a$ and $c$. If $(a, c) \in R$, write: 'Since $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$, $R$ is transitive.'
4. Step 4: Conclude — State clearly that since the relation is reflexive, symmetric, and transitive, it is an equivalence relation.

CBSE Board Exam Traps & Tips

1. The Fractional Counterexample: When proving that a relation like $R = \{(a, b) : a \le b^2\}$ is NOT reflexive, do not use integers! For $a = 1$, $1 \le 1^2$ is true. But for fraction $a = 1/2$, $1/2 \le 1/4$ is false! Always use fractions as counterexamples.
2. Vacuous Transitivity: If $(a, b) \in R$ but there is no element $c$ such that $(b, c) \in R$, then transitivity is NOT violated. The relation is vacuously transitive. This is a common trap in MCQ questions!
3. Set Definitions: Always check the universal set $A$. A relation can be reflexive on set $A = \{1, 2\}$ but not reflexive on set $B = \{1, 2, 3\}$ if $(3,3)$ is missing.

Practice Questions with Solutions

• Q: Determine whether the relation $R$ in the set $A = \{1, 2, 3, 4, 5, 6\}$ defined as $R = \{(x, y) : y \text{ is divisible by } x\}$ is reflexive, symmetric, and transitive. A: Step 1: Reflexivity: For any $a \in A$, we know that $a$ is always divisible by itself. Thus, $(a, a) \in R$ for all $a \in A$. Therefore, $R$ is reflexive. Step 2: Symmetry: Let $(x, y) \in R$. This means $y$ is divisible by $x$. However, this does not mean $x$ is divisible by $y$. For example, $(2, 4) \in R$ because 4 is divisible by 2, but $(4, 2) \notin R$ because 2 is not divisible by 4. Thus, $R$ is not symmetric. Step 3: Transitivity: Let $(x, y) \in R$ and $(y, z) \in R$. This means $y = m \cdot x$ and $z = n \cdot y$ for some integers $m, n$. Substituting $y$, we get $z = n(m \cdot x) = (nm) \cdot x$. Since $nm$ is an integer, $z$ is divisible by $x$. Thus, $(x, z) \in R$. Therefore, $R$ is transitive. Final answer: The relation is reflexive and transitive, but not symmetric.
• Q: Show that the relation $R$ in the set $\mathbb{R}$ of real numbers, defined as $R = \{(a, b) : a \le b^2\}$ is neither reflexive nor symmetric nor transitive. A: Step 1: Reflexivity: Let us test with a fractional real number, $a = 1/2$. Here, $a^2 = 1/4$. Since $1/2 > 1/4$, $(1/2, 1/2) \notin R$. Hence, $R$ is not reflexive. Step 2: Symmetry: Let $(1, 2) \in R$ because $1 \le 2^2 = 4$. However, $2 \le 1^2$ is false because $2 > 1$. Hence, $(2, 1) \notin R$. Therefore, $R$ is not symmetric. Step 3: Transitivity: Let us choose numbers $a = 3$, $b = 2$, and $c = 1.5$. Here, $3 \le 2^2 = 4$, so $(3, 2) \in R$. Also, $2 \le (1.5)^2 = 2.25$, so $(2, 1.5) \in R$. Now check $(3, 1.5)$: $3 \le (1.5)^2 = 2.25$ is false. So $(3, 1.5) \notin R$. Therefore, $R$ is not transitive. Final answer: The relation is neither reflexive, symmetric, nor transitive.
• Q: Show that the relation $R$ in the set $\mathbb{Z}$ of integers defined as $R = \{(x, y) : x - y \text{ is an integer}\} $ is an equivalence relation. A: Step 1: Reflexivity: For any $x \in \mathbb{Z}$, $x - x = 0$, which is an integer. Thus, $(x, x) \in R$ for all $x \in \mathbb{Z}$. $R$ is reflexive. Step 2: Symmetry: Let $(x, y) \in R$. This means $x - y = k$ for some integer $k$. Then, $y - x = -k$. Since $-k$ is also an integer, $(y, x) \in R$. Hence, $R$ is symmetric. Step 3: Transitivity: Let $(x, y) \in R$ and $(y, z) \in R$. This means $x - y = p$ and $y - z = q$ for integers $p, q$. Adding these two equations gives $(x - y) + (y - z) = p + q \implies x - z = p + q$. Since the sum of two integers $p + q$ is an integer, $(x, z) \in R$. $R$ is transitive. Final answer: Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.
• Q: Show that the relation $R$ in the set of real numbers $\mathbb{R}$ defined as $R = \{(a, b) : a \le b\}$ is reflexive and transitive but not symmetric. A: Step 1: Reflexivity: For any real number $a$, $a \le a$ is always true. Thus, $(a, a) \in R$. Hence, $R$ is reflexive. Step 2: Symmetry: Let $(1, 2) \in R$ because $1 \le 2$. But $2 \le 1$ is false, so $(2, 1) \notin R$. Thus, $R$ is not symmetric. Step 3: Transitivity: Let $(a, b) \in R$ and $(b, c) \in R$. This means $a \le b$ and $b \le c$. By transitive property of inequalities, we have $a \le c$. Thus, $(a, c) \in R$. $R$ is transitive. Final answer: $R$ is reflexive and transitive, but not symmetric.

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