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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

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Definition of a Function: A function f:A→Bf: A \to B is a relation where every element in set AA (domain) is mapped to exactly one element in set BB (codomain). Visually, this is represented by an arrow diagram where every point in the left oval has exactly one arrow originating from it.

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One-to-one (Injective) Function: A function is injective if no two distinct elements in the domain have the same image in the codomain. Mathematically, f(x1)=f(x2)  ⟹  x1=x2f(x_1) = f(x_2) \implies x_1 = x_2. Visually, if you look at the codomain, no element has more than one arrow pointing to it. On a graph, this is confirmed by the Horizontal Line Test: any horizontal line must intersect the graph at most once.

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Onto (Surjective) Function: A function f:A→Bf: A \to B is surjective if every element in the codomain BB has at least one pre-image in the domain AA. This means the Range of the function is exactly equal to the Codomain (Range=BRange = B). Visually, every element in the right oval of an arrow diagram must have at least one arrow pointing to it.

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Many-to-one Function: A function is many-to-one if two or more distinct elements in the domain map to the same element in the codomain. For example, in the function f(x)=x2f(x) = x^2 defined over real numbers, both 22 and −2-2 map to 44. Visually, multiple arrows from the domain converge on a single point in the codomain.

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Into Function: A function is called an 'into' function if there is at least one element in the codomain that does not have a pre-image in the domain. In this case, the Range is a proper subset of the Codomain (Range⊂BRange \subset B). Visually, there will be 'lonely' elements in the right-hand set with no arrows pointing to them.

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Bijective Function: A function that is both one-to-one (injective) and onto (surjective) is called a bijection. Bijective functions are invertible. Visually, this creates a perfect one-to-one correspondence where every element in AA is paired with exactly one unique element in BB, leaving no elements unused in either set.

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Horizontal Line Test for Graphs: To check for injectivity visually on a coordinate plane, draw horizontal lines across the graph. If any line hits the graph in more than one place, the function is many-to-one. To check for surjectivity, ensure that every horizontal line corresponding to the yy-values in the codomain hits the graph at least once.

📐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). \n2. 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. \n2. 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.