krit.club logo

Relations and Functions - Types of Functions: One-to-one and Onto functions

Grade 12ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

A function f:A→Bf: A \to B is called One-to-One (Injective) if distinct elements in the domain AA have distinct images in the codomain BB. Mathematically, for any x1,x2∈Ax_1, x_2 \in A, if f(x1)=f(x2)f(x_1) = f(x_2), then x1=x2x_1 = x_2. Visually, no two arrows from the domain point to the same element in the codomain.

A mapping diagram showing a one-to-one function where each element of the domain maps to a unique element in the codomain.
•

A function f:A→Bf: A \to B is called Onto (Surjective) if every element in the codomain BB is the image of at least one element in the domain AA. This means the Range of ff is equal to the Codomain BB. For every y∈By \in B, there exists x∈Ax \in A such that f(x)=yf(x) = y.

A mapping diagram showing an onto function where every element in the codomain is pointed to by at least one arrow.
•

A function is Bijective if it is both one-to-one and onto. This creates a perfect pairing between elements of the domain and the codomain. If n(A)=n(B)=nn(A) = n(B) = n, the total number of such bijections is n!n!.

A mapping diagram of a bijective function showing a one-to-one correspondence.
•

The Horizontal Line Test is used to determine injectivity from a graph. If any horizontal line intersects the graph of the function at more than one point, the function is not one-to-one.

Graph of y = x^2 showing a horizontal line intersecting at two points, indicating it is not one-to-one.

📐Formulae

Injective Condition: f(x1)=f(x2)  ⟹  x1=x2f(x_1) = f(x_2) \implies x_1 = x_2

Surjective Condition: f(A)=Bf(A) = B (Range equals Codomain)

Number of one-to-one functions from AA to BB (where n(A)=m,n(B)=nn(A) = m, n(B) = n): nPm=n!(n−m)!^nP_m = \frac{n!}{(n-m)!} if n≥mn \ge m, and 00 if n<mn < m

Number of bijective functions from AA to BB (where n(A)=n(B)=nn(A) = n(B) = n): n!n!

Number of onto functions from AA to BB (where n(A)=m,n(B)=2n(A) = m, n(B) = 2): 2m−22^m - 2

💡Examples

Problem 1:

Check the injectivity and surjectivity of the function f:R→Rf: \mathbb{R} \to \mathbb{R} defined by f(x)=4x+5f(x) = 4x + 5.

Solution:

  1. Injectivity: Let f(x1)=f(x2)f(x_1) = f(x_2). Then 4x1+5=4x2+54x_1 + 5 = 4x_2 + 5. Subtracting 55 from both sides gives 4x1=4x24x_1 = 4x_2. Dividing by 44 gives x1=x2x_1 = x_2. Since f(x1)=f(x2)f(x_1) = f(x_2) implies x1=x2x_1 = x_2, the function is injective (one-to-one).
  2. Surjectivity: Let yy be an arbitrary element in the codomain R\mathbb{R}. We set y=f(x)=4x+5y = f(x) = 4x + 5 and solve for xx: x=y−54x = \frac{y - 5}{4}. For any real value of yy, xx will also be a real number. Since x∈Rx \in \mathbb{R} (the domain) exists for every y∈Ry \in \mathbb{R} (the codomain), the function is surjective (onto). \nConclusion: The function is bijective.

Explanation:

We use the standard algebraic definitions: for injectivity, we prove that equal outputs imply equal inputs; for surjectivity, we prove that any output yy can be produced by a valid input xx.

Problem 2:

Show that the function f:N→Nf: \mathbb{N} \to \mathbb{N} defined by f(x)=x2f(x) = x^2 is injective but not surjective.

Solution:

  1. Injectivity: Let f(x1)=f(x2)  ⟹  x12=x22f(x_1) = f(x_2) \implies x_1^2 = x_2^2. Since the domain is natural numbers (N={1,2,3,...}\mathbb{N} = \{1, 2, 3, ...\}), x1x_1 and x2x_2 must be positive. Therefore, x12=x22  ⟹  x1=x2x_1^2 = x_2^2 \implies x_1 = x_2. The function is injective.
  2. Surjectivity: The codomain is N\mathbb{N}. Consider an element in the codomain, say 2∈N2 \in \mathbb{N}. If f(x)=2f(x) = 2, then x2=2x^2 = 2, which means x=2x = \sqrt{2}. However, 2\sqrt{2} is not a natural number (x∉Nx \notin \mathbb{N}). Since there are elements in the codomain (like 2,3,52, 3, 5) that have no pre-image in the domain, the function is not surjective.

Explanation:

Injectivity holds because natural numbers are always positive, eliminating the negative root. Surjectivity fails because only perfect squares in the codomain have pre-images in the domain.

Problem 3:

Determine if the function f:R→Rf: \mathbb{R} \to \mathbb{R} defined by f(x)=x3f(x) = x^3 is a bijection.

Graph of y = x^3 which passes both the horizontal line test and covers the entire y-axis.

Solution:

Injectivity: Let f(x1)=f(x2)f(x_1) = f(x_2). Then x13=x23x_1^3 = x_2^3. Taking the cube root of both sides, we get x1=x2x_1 = x_2. Thus, ff is injective.

Surjectivity: Let y∈Ry \in \mathbb{R}. We seek xx such that x3=yx^3 = y. This implies x=y1/3x = y^{1/3}. Since every real number has a unique real cube root, x∈Rx \in \mathbb{R} for every yy. Thus, the range is R\mathbb{R} and ff is surjective.

Since ff is both injective and surjective, it is a bijection.

Explanation:

A cubic function x3x^3 is strictly increasing over the set of real numbers, ensuring each output is unique and every real value is eventually reached.

Problem 4:

Check the surjectivity of the function f:R→[0,∞)f: \mathbb{R} \to [0, \infty) defined by f(x)=∣x∣f(x) = |x|. Is it injective?

V-shaped graph of y = |x| showing that a horizontal line intersects the graph at two points.

Solution:

Injectivity: Let f(x)=∣x∣f(x) = |x|. For x1=2x_1 = 2 and x2=−2x_2 = -2, we have f(2)=∣2∣=2f(2) = |2| = 2 and f(−2)=∣−2∣=2f(-2) = |-2| = 2. Since f(2)=f(−2)f(2) = f(-2) but 2≠−22 \neq -2, the function is not injective.

Surjectivity: The codomain is given as [0,∞)[0, \infty). For any y∈[0,∞)y \in [0, \infty), we can find x=yx = y such that f(x)=∣y∣=yf(x) = |y| = y. Thus, the range is [0,∞)[0, \infty), which equals the codomain. The function is surjective.

Explanation:

The absolute value function maps both positive and negative values to the same positive result, failing injectivity, but it covers all non-negative real numbers.