NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.4

Understanding Binary Operations

A binary operation on a non-empty set $A$ is a rule that takes any two elements of $A$ and combines them to produce another element of the same set $A$. Formally, a binary operation $$ on a set $A$ is a function $: A \times A \to A$. We write $(a, b)$ as $a b$. The most critical feature is closure: the result of $a * b$ must always belong to $A$ for all $a, b \in A$. For instance, addition is a binary operation on natural numbers because the sum of any two natural numbers is always a natural number. However, subtraction is not a binary operation on natural numbers because $2 - 5 = -3$, which is not a natural number.

Key Properties of Binary Operations

Commutative Property

A binary operation $$ on a set $A$ is commutative if $a b = b * a$ for all $a, b \in A$.

Associative Property

A binary operation $$ on a set $A$ is associative if $(a b) c = a (b * c)$ for all $a, b, c \in A$.

Identity Element

An element $e \in A$ is an identity element for $$ if $a e = e * a = a$ for all $a \in A$.

Inverse Element

For an identity element $e \in A$, an element $b \in A$ is the inverse of $a$ if $a b = b a = e$. It is written as $a^{-1}$.

How to Find Identity and Inverse Elements

1. Define the Operation — Identify the binary operation definition $a * b$ and the specified set $A$.
2. Find the Identity Element (e) — Set up the equation $a e = a$. Solve for $e$. Verify that $e a = a$ and ensure $e$ belongs to the set $A$.
3. Find the Inverse Element — If a valid identity element $e$ exists, set up the equation $a * b = e$ and solve for $b$ in terms of $a$. This $b$ is the inverse $a^{-1}$.
4. Verify Domain Restraints — Double-check that the calculated inverse element $b$ belongs to the set $A$ for all $a \in A$.

Detailed Solved Examples

• Example 1: Let $$ be a binary operation on the set of rational numbers $\mathbb{Q}$ defined by $a b = \frac{ab}{3}$. Determine if $$ is commutative and associative. Step 1: Check Commutativity. $a b = \frac{ab}{3}$ and $b a = \frac{ba}{3}$. Since multiplication of rational numbers is commutative, $ab = ba$, which means $a b = b a$. Thus, $$ is commutative. Step 2: Check Associativity. $(a b) c = \left(\frac{ab}{3}\right) c = \frac{(\frac{ab}{3})c}{3} = \frac{abc}{9}$. $a (b c) = a \left(\frac{bc}{3}\right) = \frac{a(\frac{bc}{3})}{3} = \frac{abc}{9}$. Since $(a b) c = a (b c)$, the operation is associative.
• Example 2: For the binary operation $$ defined on $\mathbb{Q}$ as $a b = \frac{ab}{3}$, find the identity element and the inverse of any non-zero element $a$. Step 1: Find Identity $e$. We need $a e = a \implies \frac{ae}{3} = a$. Since $a \neq 0$, we can divide both sides by $a$ to get $\frac{e}{3} = 1 \implies e = 3$. Since $3 \in \mathbb{Q}$, $e = 3$ is the identity element. Step 2: Find Inverse of $a$. Let $b$ be the inverse of $a$. Then $a b = e \implies \frac{ab}{3} = 3$. Solving for $b$, we get $ab = 9 \implies b = \frac{9}{a}$. Thus, the inverse of $a$ is $\frac{9}{a}$, which is rational for all non-zero $a \in \mathbb{Q}$.

Exam Trap: Always Check the Set Domain!

A very common mistake CBSE students make is calculating an identity element or inverse correctly but forgetting to check if it belongs to the given set. For example, if the operation $a b = a + b - 1$ is defined on the set of natural numbers $\mathbb{N}$, solving $a e = a$ gives $e = 1 \in \mathbb{N}$. But when finding the inverse, $a * b = 1 \implies a + b - 1 = 1 \implies b = 2 - a$. For $a = 3$, the inverse would be $2 - 3 = -1$. Since $-1$ is not a natural number, the inverse does not exist for all elements in $\mathbb{N}$. Always write down the domain restriction explicitly!

Practice Questions with Solutions

• Q: Let $$ be a binary operation on $\mathbb{Z}$ defined by $a b = a + b - 4$. Is $$ associative? A: Step 1: Write down the LHS of the associative equation: $(a b) c$. $(a b) c = (a + b - 4) c = (a + b - 4) + c - 4 = a + b + c - 8$. Step 2: Write down the RHS: $a (b c)$. $a (b c) = a (b + c - 4) = a + (b + c - 4) - 4 = a + b + c - 8$. Step 3: Compare both sides. Since $(a b) c = a (b c) = a + b + c - 8$, the operation is associative. Final answer: Yes, the operation $$ is associative.
• Q: Determine whether the operation $$ defined on $\mathbb{R}^+$ by $a b = a^b$ is commutative. A: Step 1: Calculate $a b$. By definition, $a b = a^b$. Step 2: Calculate $b a$. By definition, $b a = b^a$. Step 3: Test with counter-examples. Let $a = 2$ and $b = 3$. Then $2 3 = 2^3 = 8$. But $3 2 = 3^2 = 9$. Since $8 \neq 9$, $a b \neq b a$ in general. Final answer: No, the operation is not commutative.
• Q: Find the identity element for the binary operation $$ defined on $\mathbb{Q} \setminus \{0\}$ by $a b = 5ab$. A: Step 1: Apply the identity property condition: $a e = a$. $5ae = a$. Step 2: Solve for $e$. Since $a \in \mathbb{Q} \setminus \{0\}$, we can divide by $a$ on both sides: $5e = 1 \implies e = \frac{1}{5}$. Step 3: Verify the condition $e a = a$. $e * a = 5(\frac{1}{5})a = a$. Since $\frac{1}{5} \in \mathbb{Q} \setminus \{0\}$, it is a valid identity element. Final answer: The identity element is $e = \frac{1}{5}$.
• Q: Let $$ be a binary operation on $\mathbb{R} \setminus \{-1\}$ defined by $a b = a + b + ab$. Find the inverse of $a$. A: Step 1: Find the identity element $e$ first. $a e = a \implies a + e + ae = a \implies e(1 + a) = 0$. Since $a \neq -1$, $1 + a \neq 0$, which gives $e = 0$. Step 2: Use the identity to find the inverse $b$. $a b = e \implies a + b + ab = 0$. Step 3: Solve for $b$. $b(1 + a) = -a \implies b = \frac{-a}{1 + a}$. Since $a \neq -1$, this denominator is non-zero, and the value belongs to $\mathbb{R} \setminus \{-1\}$. Final answer: The inverse of $a$ is $a^{-1} = \frac{-a}{1 + a}$.

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