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:
Aspect | Relation | Function |
|---|---|---|
Definition | A relation is a set of ordered pairs, where each pair consists of two elements, establishing a relationship between them. | A function is a special type of relation where each input value (domain) is associated with exactly unique one output value (range). |
Input-Output Mapping | A single input can be related to multiple outputs. | Each input is associated with only one output. |
Relationship | The relationship between elements doesn't guarantee a unique output for each input. | Every input has a precisely defined output. |
Vertical Line Test | Fails the vertical line test if a vertical line intersects the graph in more than one point. | Passes the vertical line test as the graph intersects the vertical line at only one point. |
Representation | Can be represented as a set of ordered pairs, table, or graph. | Represented similarly but adheres to the "unique output" rule. |
General Notation | Often denoted as R, where R ⊆ A × B, with A and B being sets | Denoted as f: A → B, where f is the function, A is the domain, and B is the range. |
Examples | If R = {(1, 2), (2, 3), (3, 4)}, it represents a relation between elements where each element is related to the next one. | If f(x) = x2, it represents a function where each input x is associated with its square as the output. |
Real-world Example | A person and their phone numbers (one person can have multiple numbers). | A student and their unique roll number (one student has exactly one roll number). |
Also Read,
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.
- 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.
- 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:
- 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.
- 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.
