Difference between Relation and Function

A relation in mathematics is simply a connection or a rule that links one set of things (called the domain) to another set of things (called the range). You can think of it as a collection of pairs that shows how items from one group are connected to items in another group.

For example, Imagine you have a list of students and their favourite subjects.

• Student A → Math
• Student B → Science
• Student A → English

Here, the relation connects students (from the domain) to their favourite subjects (from the range).

A function in mathematics is a special type of relation where each input (from the domain) is connected to exactly one output (from the range). In simple terms, a function makes sure that one thing in the first group is linked to only one thing in the second group.

For example, Imagine you have a vending machine.

• You press Button A → It gives you chips.
• You press Button B → It gives you chocolate.

Here, each button (input) is connected to only one snack (output). You won't press Button A and get both chips and soda—just chips.

The difference between relation and function is given below:

Also Read,

• Relations and Functions
• Types of Relations
• Domain and Range of Relation
• Domain and Range of a Function

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.