One of the fundamental concepts in mathematics, particularly with regard to relations and mapping, is the concept of bijective functions. A function is said to be bijective if it is both surjective (or onto) and injective (or one-to-one). By guaranteeing that every element in the domain maps uniquely to an element in the codomain and vice versa, these features make it possible for the elements of two sets to be precisely matched.
A function f: A → B is considered bijective if every element in set B corresponds uniquely to an element in set A, and vice versa. This means that for every b ∈ B, there exists a unique a ∈ A such that f (a) = b. Here, b is referred to as the image of a, while a is the preimage of b.
The key characteristics of bijective functions can be outlined as follows:
This one-to-one mapping ensures that all elements in both sets are perfectly paired without overlap or omission.
To grasp the idea of bijective functions, it's essential to understand injective, surjective, bijective functions:
To establish that a function f: A → B is bijective, you must show that it satisfies both injectivity and surjectivity. Start by defining the mapping f, ensuring that every element in A has a corresponding element in B.
If the sets A and B differ in size (∣A∣ ≠ ∣B∣), a bijective mapping is impossible. When ∣A∣ = ∣B∣ = n, n! unique bijections exist, pairing elements from A to B in all possible one-to-one correspondences.
This approach highlights bijections as precise pairings between sets.
Problem 1: Prove f (x) = 2x + 3 is Bijective
Thus, f (x) is injective.
Since x ∈ R, every y has a corresponding x. Hence, f (x) is surjective.
Therefore, f (x) = 2x + 3 is bijective.
Problem 2: Inverse of a Bijective Function
Solution: Find the inverse of f (x) =
Let y =
, then rearrange to solve for x:
y (x+2) = x − 1 ⟹ xy +2y = x−1
x (y−1) = −1 −2y ⟹
Thus, the inverse function is:
f−1 (y) =
Problem 3: Let g:{1, 2, 3} → {a, b, c} be defined as g(1) = a, g(2) = b and g(3) = c. Is g a bijective function?
Considering these factors, g unquestionably qualifies as a bijective function.
Bijective functions have broad applications in various fields:
(Session 2025 - 26)