Solved Examples on Relation and Function

Question 1: Given the set A = {1, 2, 3, 4} and set B = {a, b, c}, define a relation from set A to set B where each element of set A is related to each element of set B.

Solution:

To define a relation from set A to set B, we can create a relation where each element of set A is related to each element of set B. This is essentially a Cartesian product of A and B. So, the relation R can be defined as follows:

R = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c), (3, a), (3, b), (3, c), (4, a), (4, b), (4, c)}

Question 2: Let f: ℝ → ℝ be defined as f(x) = x^2 + 1. Determine whether the function f is injective, surjective, or bijective.

Solution:

To determine if f is injective, we need to check if every element in the co-domain has at most one pre-image. To check if f is surjective, we need to verify if every element in the co-domain has at least one pre-image. Finally, if f is both injective and surjective, it’s bijective.

1. Injective (One-to-One): Assume f(x₁) = f(x₂) for some x₁, x₂ in the domain. f(x₁) = x₁² + 1 and f(x₂) = x₂² + 1. If x₁ ≠ x₂, then f(x₁) ≠ f(x₂), as squaring a real number always results in a non-negative value and adding 1 makes it strictly greater. So, f is not injective.
2. Surjective (Onto): To check if f is surjective, we need to verify if for every y in the co-domain, there exists an x in the domain such that f(x) = y. Let’s take y = 0. Solving x² + 1 = 0 does not yield any real solutions. Hence, f is not surjective.

Question 3: Given the set A = {1, 2, 3} and set B = {x, y, z}, define a relation from set A to set B where each element of set A is related to exactly one element of set B.

Solution:

To define a relation from set A to set B where each element of set A is related to exactly one element of set B, we can simply pair each element of A with an element of B in a one-to-one manner. So, the relation R can be defined as follows:

R = {(1, x), (2, y), (3, z)}

Question 4: Let g: ℝ → ℝ be defined as g(x) = 2x – 3. Determine whether the function g is injective, surjective, or bijective.

Solution:

1. Injective (One-to-One): Assume g(x₁) = g(x₂) for some x₁, x₂ in the domain. 2x₁ – 3 = 2x₂ – 3 2x₁ = 2x₂ Dividing by 2, x₁ = x₂. Since every element in the co-domain has at most one pre-image, g is injective.
2. Surjective (Onto): To check if g is surjective, we need to verify if for every y in the co-domain, there exists an x in the domain such that g(x) = y. Let’s take any y in ℝ, say y = 0. Solving 2x – 3 = 0, we get x = 3/2. So, g(3/2) = 0. Since for any y in ℝ, there exists x = 3/2 such that g(x) = y, g is surjective.

Since g is both injective and surjective, it’s bijective.

Relation defines how elements of one set relate to elements of another set whereas a Function is a special type of relation in which each element in the domain (input) is related to exactly one element in the codomain (output).

This article explores relations and functions, highlighting their definitions, properties, differences, and applications in mathematics.