RD Sharma Class 12 Solutions Chapter 3 - Binary Operations (Ex 3.2) Exercise 3.2 - Free PDF
FAQs on RD Sharma Class 12 Solutions Chapter 3 - Binary Operations (Ex 3.2) Exercise 3.2
1. How do you solve problems in RD Sharma Ex 3.2 that ask to check if a binary operation * is commutative?
To check if a binary operation *, defined on a set S, is commutative, you must verify if the condition `a * b = b * a` holds true for all elements `a` and `b` in S. The standard method is:
Calculate the expression for `a * b` using the given definition.
Calculate the expression for `b * a` using the same definition.
Compare the results. If they are identical for all a, b ∈ S, the operation is commutative. Otherwise, it is not.
2. What is the correct method to test for associativity for a binary operation in RD Sharma Chapter 3?
To test if a binary operation * is associative on a set S, you need to check if `(a * b) * c = a * (b * c)` for all `a, b, c ∈ S`. The step-by-step process is:
First, calculate the Left Hand Side (LHS): Evaluate the operation inside the parentheses `(a * b)` first, and then apply the operation with `c`.
Next, calculate the Right Hand Side (RHS): Evaluate `(b * c)` first, and then apply the operation of `a` with that result.
If the final expressions for the LHS and RHS are equal, the operation is associative. If they differ, it is not.
3. How do you find the identity element for a binary operation in Exercise 3.2?
To find the identity element, denoted by `e`, for a binary operation * on a set S, you need to find an element `e ∈ S` such that `a * e = e * a = a` for every element `a ∈ S`.
Assume `e` is the identity element.
Solve the equation `a * e = a` for `e`.
The value you get for `e` must be a constant, independent of `a`, and must belong to the set S.
Verify that `e * a = a` also holds true with this value of `e`. If both conditions are met, then `e` is the identity element.
4. What are the steps to find the inverse of an element for a binary operation in this chapter?
To find the inverse of an element `a` in a set S, you must first confirm the existence of an identity element, `e`. If `e` exists, you then find an element `b` (often denoted `a⁻¹`) such that `a * b = b * a = e`.
Identify the identity element `e` of the operation.
Let `b` be the inverse of `a`.
Set up the equation `a * b = e`.
Solve this equation for `b` in terms of `a`.
The resulting expression for `b` is the inverse. Crucially, you must ensure that for any `a` in the set, its inverse `b` is also an element of the set S.
5. If an identity element 'e' does not exist for a binary operation, what does this imply about finding inverse elements?
The concept of an inverse element is fundamentally dependent on the existence of an identity element. The inverse of an element `a` is defined as another element `b` such that `a * b = b * a = e`. If there is no identity element `e`, the very definition of an inverse cannot be satisfied. Therefore, if a binary operation does not have an identity element, no element in the set can have an inverse.
6. Why is checking the underlying set (e.g., N, Z, Q) crucial when solving problems in Ex 3.2?
The underlying set is crucial because it determines two key factors:
Closure Property: A binary operation * on a set S is only valid if for every `a, b ∈ S`, the result `a * b` is also in S. For instance, subtraction is a binary operation on integers (Z) but not on natural numbers (N), as `3 - 5 = -2`, which is not in N.
Existence of Identity/Inverse: The identity element `e` and the inverse `a⁻¹` must belong to the given set. For example, for multiplication on Z, the inverse of 2 is 1/2, which is not in Z. Always check if your results belong to the specified set.
7. How can you definitively prove a binary operation is *not* associative or *not* commutative?
To prove that a binary operation is not associative or not commutative, you do not need a general algebraic proof. You simply need to provide a single, valid counterexample.
For non-commutativity: Find one specific pair of elements `a` and `b` from the given set for which `a * b ≠ b * a`.
For non-associativity: Find one specific set of three elements `a`, `b`, and `c` from the given set for which `(a * b) * c ≠ a * (b * c)`.
Showing that the property fails for even one case is sufficient proof that it does not hold universally for the operation.
