Learn Bijective Function on Vedantu
Bijection, or bijective function, is a one-to-one correspondence function between the elements of two sets. In such a function, each element of one set pairs with exactly one element of the other set, and each element of the other set has exactly one paired partner in the first set. While understanding bijective mapping, it is important to not confuse such functions with one-to-one correspondence. This article will help you in understanding what a bijective function is, it’s examples, properties, and how to prove that a function is bijective.
Surjective, Injective and Bijective Functions
Functions can be one-to-one functions (injections), onto functions (surjections), or both one-to-one and onto functions (bijections).
Here is a brief overview of surjective, injective and bijective functions:
Surjective: If f: P → Q is a surjective function, for every element in Q, there is at least one element in P, that is, f (p) = q.
Injective: If f: P → Q is an injective function, then distinct elements of P will be mapped to distinct elements of Q, such that p=q whenever f (p) = f (q).
Bijective: If f: P → Q is a bijective function, for every element in Q, there is exactly one element in P, that is, f (p) = q.
Bijective Function Definition
What is a bijective function?
A bijective function has no unpaired elements and satisfies both injective (one-to-one) and surjective (onto) mapping of a set P to a set Q. Thus, bijective functions satisfy injective as well as surjective function properties and have both conditions to be true. In mathematical terms, let f: P → Q is a function; then, f will be bijective if every element ‘q’ in the co-domain Q, has exactly one element ‘p’ in the domain P, such that f (p) =q.
Bijective Function Properties
As we have understood what a bijective mapping means, let us understand the properties that are the characteristic of bijective functions. We know the function f: P → Q is bijective if every element q ∈ Q is the image of only one element p ∈ P, where element ‘q’ is the image of element ‘p,’ and element ‘p’ is the preimage of element ‘q’.
Also,
Each element of P should be paired with at least one element of Q.
No element of P must be paired with more than one element of Q.
Each element of Q must be paired with at least one element of P, and
No element of Q must be paired with more than one element of P.
Examples of Bijective Function
Example 1: The function f (x) = x2 from the set of positive real numbers to positive real numbers is injective as well as surjective. Thus, it is also bijective. However, the same function from the set of all real numbers R is not bijective since we also have the possibilities f (2)=4 and f (-2)=4.
Example 2: The function f: {months of a year} {1,2,3,4,5,6,7,8,9,10,11,12} is a bijection if the function is defined as f (M)= the number ‘n’ such that M is the nth month.
How to prove that a Function is Bijective?
If we have defined a map f: P → Q and we have to prove that the function f is a bijection, we have to satisfy two conditions. First of all, we have to prove that f is injective, and secondly, we have to show that f is surjective. Only when we have established that the elements of domain P perfectly pair with the elements of co-domain Q, such that, |P|=|Q|=n, we can conveniently say that there are n bijections between P and Q. Let us understand the proof with the following example:
Example: Show that the function f (x) = 5x+2 is a bijective function from R to R.
Solution:
Given function: f (x) = 5x+2.
Step 1: To prove that the given function is injective.
To prove injection, we have to show that f (p) = z and f (q) = z, and then p = q.
Say, f (p) = z and f (q) = z.
Therefore, we can write z = 5p+2 and z = 5q+2 which can be thus written as: 5p+2 = 5q+2. Simplifying the equation, we get p =q, thus proving that the function f is injective.
Step 2: To prove that the given function is surjective.
To prove surjection, we have to show that for any point “c” in the range, there is a point “d” in the domain so that f (q) = p.
Let, c = 5x+2.
Therefore, d will be (c-2)/5. Since this number is real and in the domain, f is a surjective function.
Therefore, since the given function satisfies the one-to-one (injective) as well as the onto (surjective) conditions, it is proved that the given function is bijective.
Important Points to Remember for Bijective Function:
Bijective functions only when the given function is said to be both Injective function as well as surjective function. If the function is not an injective function but a surjective function or a surjective function but not an injective function, then the function is not a Bijective function. This is a very basic concept to keep in mind.
An injective function is one of the easiest concepts to understand from the topic function. It is defined as a function in which a distinct element in the domain of the function maps with a distinct element in its codomain or range. Injective function is also referred to as one to one function.
In surjective function, one element in a codomain can be mapped by one or more than one element in the domain. We can say that in a surjective function, more than one preimage is possible. Onto function is the other name of surjective function.
Hence, we can say that a bijective function carries the properties of both an injective or one to one function and surjective or a onto function.
The domain and the codomain in a bijective function has equal number of elements and each element in the domain will have a certain image.
The Bijective function can have an inverse function. However, a constant function can never be a bijective function.
When a bijective function is represented with the help of a graph by plotting down the elements on the graph, the figure obtained by doing so is always a straight line.
A bijective function is also reflexive, symmetric and transitive.
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FAQs on Bijective Function
1. What are Some Examples of Surjective and Injective Functions?
The function f: {Indian cricket players’ jersey} N defined as f (W) = the jersey number of W is injective, that is, no two players are allowed to wear the same jersey number.
f (x) = x2 from a set of real numbers R to R is not an injective function. This is because: f (2) = 4 and f (-2) = 4. So, even if f (2) = f (-2), 2 and the definition f (x) = f (y), x = y is not satisfied.
The function f (x) = 2x from the set of natural numbers N to a set of positive even numbers is a surjection.
The function f: {Lok Sabha seats} → {Indian states} defined by f (L) = the state that L represents is surjective since every Indian state has at least one Lok Sabha seat.
2. What are the Fundamental Differences Between Injective, Surjective and Bijective Functions?
Injective: In this function, a distinct element of the domain always maps to a distinct element of its co-domain.
Surjective: In this function, one or more elements of the domain map to the same element in the co-domain.
Bijective: These functions follow both injective and surjective conditions.
3. What is a bijective function?
Bijection, or bijective function, is a one-to-one correspondence function between the elements of two sets. In such a function, each element of one set pairs with exactly one element of the other set, and each element of the other set has exactly one paired partner in the first set.
4. How to prove that a function is an injective function?
Let's say we have a function called f(x) such that f(x)=10 x + 2. Now consider two cases where x is replaced by two variables p and q such that f(p) and f(q) are the functions given below
f(p) = 10 p + 2 and f(q) = 10 q + 2. In injective function, it is important that f(p) = f(q) = m. Hence,
f(p) = 10 p + 2 = m and f(q) = 10 q + 2 = m.
That means 10 p + 2 = 10 q + 2 = m.
Therefore, f(p) = f(q). Hence, the function is said to be an injective function.
5. How to prove that a function is a surjective function?
We will again take a simple equation. Let's say f (x) = 10 x + 2. We have to check whether the given function is a surjective function or not. For this we will consider f (x) = m, where m is variable. We can further infer this as
f (x) = m
10 x + 2 = m
10 x = m - 2
x = \[\frac{(m - 2)}{10}\]
In the above equation we can infer that x is a real number that means all the real numbers can satisfy the above equation. This means that f (x) will have real values satisfying it. That means for all the elements in the codomain of this function f (x), there will be some element in its domain as its preimage. Hence, we can say that this function f (x) = 10 x + 2 is a surjective function.