Relations and Functions | ISC Class 12 Maths Notes

This note covers relations on a set, empty, universal and identity relations, reflexive, symmetric and transitive relations, equivalence classes, functions, domain and range, one-one and many-one functions, into and onto functions, composition, invertibility, and graphs of a function and its inverse.

What is a relation on a set?

A set is a well-defined collection of objects, called its elements. Let A and B denote sets. An ordered pair (a, b) records an element a first and an element b second. Reversing their positions can change the pair.

The notation a ∈ A means that a belongs to A; a ∉ A means that it does not. Braces enclose a set's elements. The Cartesian product A × B contains all ordered pairs (a, b) with a ∈ A and b ∈ B.

Definition: A relation R from A to B is a subset of A × B. A subset contains only elements of the set to which it belongs. A relation on A is a subset of A × A.

The symbol ⊆ means “is a subset of”. Thus R ⊆ A × A says that both entries of every pair in R belong to A. Writing a R b is another way of saying (a, b) ∈ R.

How can the same relation be written in different ways?

The roster form lists the pairs. The set-builder form gives the condition selecting them; the colon inside braces means “such that”. The underlying set matters because a pair is admitted only when its entries belong to the specified sets.

For A = {1, 2, 3, 4}, the rule b = a + 1 selects the relation R = {(1, 2), (2, 3), (3, 4)}. In set-builder notation, this is R = {(a, b) ∈ A × A : b = a + 1}.

Checking the rule without checking membership would incorrectly admit further pairs. The rule and the sets together determine the relation. To compare ordered pairs, compare their first entries and then their second entries: both corresponding entries must agree.

How do empty, universal and identity relations differ?

Three useful relations are defined by how many pairs they admit and which pairs they contain. The empty set, denoted by ∅, contains no elements. The empty relation on a set therefore contains no ordered pairs at all.

The universal relation on A contains every pair in A × A. The identity relation on A contains exactly the pairs (a, a), where a belongs to A. Each element is related to itself, with no pair joining distinct elements.

How does a condition produce an extreme relation?

In the following condition, the symbol − denotes subtraction.

Worked example 1. On A = {1, 2, 3, 4}, classify R = {(a, b) ∈ A × A : a − b = 10}.

Answer: No two elements of A have difference 10. Hence no ordered pair satisfies the defining condition, so R = ∅. This is an empty relation, although A itself is not empty.

For the same A, the condition |a − b| ≥ 0 selects every pair. The modulus |a − b| is the non-negative absolute value of the difference; ≥ means “greater than or equal to”. Thus this relation is universal.

Identity and universal relations must not be confused. The identity rule requires equality of the entries; the universal rule imposes no further restriction once both entries belong to A. A relation can contain all identity pairs and still contain additional pairs.

How are reflexivity, symmetry and transitivity tested?

A reflexive relation R on A contains (a, a) for every element a of A. A symmetric relation contains (b, a) whenever it contains (a, b). A transitive relation contains (a, c) whenever it contains both (a, b) and (b, c).

In the transitivity test, a, b and c denote elements of A, and b is the shared middle element. They need not be distinct. Symmetry reverses a pair; transitivity links two pairs. Neither operation is a substitute for the other.

How do you give a complete classification?

Worked example 2. On A = {1, 2, 3}, classify R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}.

Answer: R is reflexive because all three self-pairs occur. It is not symmetric because (1, 2) occurs but (2, 1) does not. It is not transitive because (1, 2) and (2, 3) occur but (1, 3) does not.

A counterexample is a particular case that disproves a general claim. Each negative conclusion above gives one. A positive conclusion needs the property for every relevant element or pair, rather than just a successful sample.

For a finite relation presented as a list, inspect that complete list. For a relation given by an algebraic condition, work with arbitrary elements satisfying the condition. The proof must establish exactly the property being claimed.

When is a relation an equivalence relation?

Definition: An equivalence relation is a relation that is reflexive, symmetric and transitive. All three requirements must hold on the stated underlying set.

Congruent triangles have the same shape and size. Let T be the set of triangles in a plane, and relate two triangles when they are congruent. The following proof checks the three requirements separately, using arbitrary triangles from T.

Worked example 3. Show that congruence is an equivalence relation on T.

Answer: Every triangle is congruent to itself, giving reflexivity. If a triangle is congruent to another, the second is congruent to the first, giving symmetry. If the first is congruent to a second and the second to a third, the first is congruent to the third, giving transitivity.

Result: the intersection of equivalence relations is an equivalence relation

Let R₁ and R₂ be equivalence relations on the same set A. Subscripts distinguish the two relations. Their intersection, written R₁ ∩ R₂, consists of pairs belonging to both relations. Membership in just one relation is insufficient.

1. For each a ∈ A, the pair (a, a) belongs to both R₁ and R₂, so it belongs to their intersection.
2. If (a, b) belongs to the intersection, it belongs to both relations. Symmetry in each puts (b, a) in both.
3. If (a, b) and (b, c) belong to the intersection, both chains occur in each relation. Transitivity in each puts (a, c) in both.
4. The intersection is therefore reflexive, symmetric and transitive, establishing the result.

Notice the importance of “the same set A”. The reflexivity statement refers to every element of that set. Carrying the underlying set through the proof prevents a change of meaning halfway through the argument.

How do equivalence classes divide a set?

An equivalence class groups elements related by an equivalence relation. For an element a of the underlying set, [a] denotes the class containing a. Square brackets here name the class containing the indicated element.

Let ℤ denote the set of integers, including negative integers, zero and positive integers. Say that 2 divides a − b when a − b is an integer multiple of 2. This gives a relation on ℤ.

Worked example 4. Prove that R = {(a, b) ∈ ℤ × ℤ : 2 divides a − b} is an equivalence relation.

Answer: For every integer a, a − a = 0 is divisible by 2, so R is reflexive. If a − b is divisible by 2, its negative b − a is also divisible by 2, giving symmetry.

If a − b and b − c are divisible by 2, their sum a − c = (a − b) + (b − c) is divisible by 2. Hence R is transitive and therefore an equivalence relation.

Which classes arise in this example?

An even integer is divisible by 2; an odd integer is not. The class [0] contains all even integers, while [1] contains all odd integers. Every two even integers are related, as are every two odd integers.

No even integer is related to an odd integer under this rule. The two classes are disjoint, meaning they have no common element, and together they contain every integer. Their union, the set containing everything in either class, is ℤ.

More generally, an equivalence relation separates its underlying set into mutually disjoint classes whose union is the entire set. Such a division is a partition. Elements inside one class are related; elements in different classes are not.

What makes a relation a function?

A function f from a set A to a set B assigns every element of A exactly one element of B. The notation f : A → B specifies this direction of assignment. A function is also called a mapping.

If f maps x to y, write y = f(x). Here x is an input, y is its output, and f names the function. The output y is the image of x, and x is a preimage of y.

Definition: For f : A → B, A is the domain, B is the codomain, and the range is the set of actual outputs f(x) as x runs through A.

The domain supplies all permitted inputs. The codomain specifies where outputs belong. The range records which outputs are actually attained. Thus the range is a subset of the codomain, but need not equal it.

Why are both existence and uniqueness necessary?

Every element of A must receive an image, and that image must be unique. Different inputs may share an image. That does not violate the function condition, which restricts the number of outputs for each input.

For A = {1, 2, 3, 4, 5, 6}, the relation R = {(x, y) ∈ A × A : y = x + 1} contains (1, 2), (2, 3), (3, 4), (4, 5) and (5, 6). It is not a function from A to A because 6 has no image.

A real-valued function has real numbers as its outputs. We write ℝ for the set of real numbers. If its inputs also form ℝ or a subset of ℝ, it is a real function. The specified domain remains part of its definition.

How are one-one and many-one functions distinguished?

A function is one-one, or injective, when different inputs have different images. Equivalently, if x₁ and x₂ denote any two elements of its domain, equality f(x₁) = f(x₂) must force x₁ = x₂.

A function is many-one when it is not one-one: at least two distinct inputs share an image. The distinction concerns inputs producing the same output, not whether all elements of the codomain are reached.

How do equal outputs test injectivity?

Worked example 5. Let ℕ = {1, 2, 3, ...} denote the natural numbers, with the dots indicating continuation. Show that f : ℕ → ℕ, defined by f(x) = 2x, is one-one.

Answer: Choose arbitrary x₁ and x₂ in ℕ and suppose f(x₁) = f(x₂). Then 2x₁ = 2x₂. Dividing by 2 gives x₁ = x₂. Thus equal outputs require equal inputs, proving that f is one-one.

To disprove injectivity, one pair of distinct inputs with equal images is sufficient. For f : ℝ → ℝ defined by f(x) = x², the superscript 2 means the square of x. Since f(−1) = f(1) = 1, this function is many-one.

An arrow diagram joins each input to its image with an arrow. An onto function reaches every element of its codomain; this requirement is examined separately from injectivity.

Four arrow diagrams are labelled (i) to (iv). Diagram (i) gives distinct images but leaves codomain elements unused. Diagram (ii) sends 1 and 2 to b. Diagram (iii) reaches all three codomain elements. Diagram (iv) pairs four inputs with four distinct outputs.

See Fig. 1.2 in your NCERT textbook

Shared arrow endpoints reveal many-one behaviour. Unused codomain elements answer a separate question, which is whether the function is onto.

How do into and onto functions depend on the codomain?

A function f : A → B is onto, or surjective, when every element of B has at least one preimage in A. Equivalently, its range equals B. A function is into when some element of its codomain is not reached.

To prove that a function is onto, begin with an arbitrary y in the codomain. Find an input x in the stated domain satisfying f(x) = y. An algebraic solution outside the domain does not establish surjectivity.

Worked example 6. Is f : ℕ → ℕ, defined by f(x) = 2x, onto?

Answer: The codomain contains 1, but 2x = 1 requires x = 1/2, which is not a natural number. Therefore 1 has no permitted preimage. The function is into, with range equal to the even natural numbers.

What changes when the sets change?

Worked example 7. Show that f : ℝ → ℝ, defined by f(x) = 2x, is onto and one-one.

Answer: For any real output y, choose x = y/2. This is a real input, and f(y/2) = y, so f is onto. Also, f(x₁) = f(x₂) gives 2x₁ = 2x₂ and hence x₁ = x₂, so f is one-one.

The formula is unchanged between these two examples, but the permitted inputs and the codomain have changed. On natural numbers, 1 is missed. On real numbers, its required input is permitted, and the same reasoning works for every real output.

For f : ℝ → ℝ given by f(x) = x², the real number −2 has no preimage because a real square cannot be negative. Thus this function is both many-one and into. These classifications describe two separate features.

What is a bijection, and what changes for finite sets?

A bijection is a function that is both one-one and onto. Every codomain element is reached, and no two distinct inputs share an image. Consequently, each codomain element has exactly one preimage.

The function f : ℝ → ℝ, f(x) = 2x, is bijective because both properties have been proved. Showing only injectivity or only surjectivity would leave half of that conclusion unproved for these infinite sets, which do not have finitely many elements.

Result: a self-map of a finite set is one-one exactly when it is onto

A finite set has finitely many elements. A self-map is a function whose domain and codomain are the same set. For a finite set X, a function f : X → X is one-one if and only if it is onto.

“If and only if” asserts both directions. If distinct inputs have distinct images, the images fill the finite codomain. If two inputs share an image, the range cannot have as many elements as that codomain, so the function cannot be onto.

Worked example 8. Explain why an onto function f : {1, 2, 3} → {1, 2, 3} must be one-one.

Answer: Suppose two domain elements had the same image. The remaining domain element could supply at most one additional image. The range would then contain at most two elements, contradicting the requirement that all three codomain elements are reached. Therefore f is one-one.

The finite-set condition is essential. The self-map f : ℕ → ℕ, f(x) = 2x, is one-one but not onto. An infinite set therefore need not satisfy the implication established for finite self-maps.

Do not apply the result merely because a formula looks simple. Check that the domain and codomain are the same finite set before using one property to infer the other.

How is a composite function formed?

Let f : A → B and g : B → C be functions, where C is the final codomain and g names the second function. Their composite function, written g ∘ f, applies f first and then applies g to the resulting output.

Definition: The symbol ∘ denotes composition. The composite g ∘ f : A → C is defined by (g ∘ f)(x) = g(f(x)) for each x ∈ A.

The intermediate value f(x) must be an allowed input for g. The displayed sets guarantee this: outputs of f lie in B, the domain of g. The composite's output is obtained only after both assignments have been performed.

How do you calculate a composition from listed values?

Worked example 9. Let f : {2, 3, 4, 5} → {3, 4, 5, 9} satisfy f(2) = 3, f(3) = 4 and f(4) = f(5) = 5. Let g : {3, 4, 5, 9} → {7, 11, 15} satisfy g(3) = g(4) = 7 and g(5) = g(9) = 11. Find g ∘ f.

Answer: Follow f and then g for each input. The results are (g ∘ f)(2) = g(3) = 7, (g ∘ f)(3) = g(4) = 7, (g ∘ f)(4) = g(5) = 11 and (g ∘ f)(5) = g(5) = 11.

For algebraic formulas, replace the input variable in g by the whole expression f(x). Composition is not multiplication. Reversing the order asks for f(g(x)), which need not have the same values and may require different domain checks.

Three ovals are labelled A, B and C. An arrow labelled f takes x to f(x), and an arrow labelled g takes f(x) to g(f(x)). A longer arrow labelled gof runs from x to the final image.

See Fig. 1.5 in your NCERT textbook

When does a function have an inverse?

The identity function on a set X, denoted Iₓ, sends each element to itself: Iₓ(x) = x. Unlike a constant function, which assigns the same fixed output to every input, the identity function returns whichever input it receives.

A function f : X → Y is invertible when there is a function g : Y → X such that g ∘ f = Iₓ and f ∘ g = Iᵧ. Here Y is the codomain of f, and Iᵧ is its identity function. The function g reverses the assignment made by f.

Theorem: a function is invertible if and only if it is bijective

The inverse function is denoted f⁻¹. The superscript −1 here names an inverse function; it does not mean the reciprocal 1/f(x). The inverse takes each output of the original function back to its unique input.

Injectivity ensures that there is no choice between different preimages. Surjectivity ensures that every element of the codomain has a preimage. Together, these conditions allow the reverse assignment to be a function on the entire original codomain.

How do you find and verify the inverse algebraically?

Worked example 10. Let f : ℕ → Y be defined by f(x) = 4x + 3, where Y = {4x + 3 : x ∈ ℕ}. Find and verify its inverse.

Answer: For y ∈ Y, solve y = 4x + 3 to obtain x = (y − 3)/4. Thus g : Y → ℕ is given by g(y) = (y − 3)/4. Its values belong to ℕ by the definition of Y.

For x ∈ ℕ, g(f(x)) = (4x + 3 − 3)/4 = x. For y ∈ Y, f(g(y)) = 4((y − 3)/4) + 3 = y. Both identity conditions hold, so f⁻¹ = g.

The inverse's domain is Y, not all real numbers. Solving an equation supplies a candidate formula, but its permitted inputs and outputs must also be stated. The two composition checks verify the reversal in both directions.

How are a function and its inverse represented on a graph?

The graph of a real function y = f(x) is the set of points with coordinates (x, f(x)) for permitted inputs x. The first coordinate records the input; the second records the output. Both coordinates are measured on the corresponding axes.

For a bijection, if (a, b) lies on its graph, then f(a) = b and f⁻¹(b) = a. Thus (b, a) lies on the inverse's graph. The input and output exchange places, just as the domain and range exchange roles.

Property: inverse graphs are reflections in the line y = x

Reflection in the line y = x exchanges a point's coordinates. Therefore the graph of an inverse function is the reflection of the original graph in that line. The point correspondence follows from the inverse relation, rather than from the appearance of a sketch.

On common coordinate axes, sketch y = 2x and y = x/2 for real x. Add the line y = x as the mirror line. The two function graphs are straight lines through the origin, the point (0, 0).

For f : ℝ → ℝ, f(x) = 2x, solving y = 2x gives x = y/2. Hence f⁻¹(x) = x/2. Applying either function followed by the other returns the original real input, confirming the relationship shown in the sketch.

Keep domain restrictions when drawing graphs. In the function f : ℕ → Y, f(x) = 4x + 3, inputs are natural numbers, so the graph consists of separate permitted points. Its inverse also has separate permitted inputs, namely the elements of Y.

A sketch does not remove a failure of injectivity. The function f : ℝ → ℝ, f(x) = x², has equal outputs for −1 and 1, so it has no inverse function with these stated sets. Check invertibility before labelling a reflected graph an inverse function.

Glossary

• Relation — A subset of a Cartesian product, specifying which ordered pairs of elements are included.
• Identity relation — A relation containing exactly the pairs that relate each element of the underlying set to itself.
• Reflexive relation — A relation in which every element of the underlying set is related to itself.
• Symmetric relation — A relation that contains the reversed pair whenever it contains an ordered pair.
• Transitive relation — A relation containing the direct pair whenever two included pairs form a chain through a common middle element.
• Equivalence relation — A relation satisfying reflexivity, symmetry and transitivity on its stated underlying set.
• Equivalence class — The subset containing all elements related to a chosen element under an equivalence relation.
• Domain — The set of permitted inputs for which a function supplies exactly one output each.
• Codomain — The specified target set within which all the outputs of a function must lie.
• Range — The set of outputs actually obtained as inputs run through a function's entire domain.
• Injective function — A one-one function that assigns different output values to distinct elements of its domain.
• Surjective function — An onto function whose range equals its codomain, so every target element has a preimage.
• Bijection — A function that is both one-one and onto between its stated domain and codomain.
• Composite function — A function obtained by applying one function and then another to the intermediate output.
• Inverse function — The function that reverses a bijection, taking each output back to its unique original input.

Common errors and misconceptions

• Misconception: A relation containing some self-pairs is reflexive. Correct: It must contain (a, a) for every element a of the underlying set.
• Misconception: Symmetry means that every possible pair is present. Correct: It means that each included pair has its reverse; universal membership is a different requirement.
• Misconception: Reflexivity and symmetry are sufficient for equivalence. Correct: Transitivity must also be established, with every relevant chain satisfying its condition.
• Misconception: A function cannot give two inputs the same output. Correct: A many-one function can do this; each individual input must still have exactly one output.
• Misconception: The formula alone determines whether a function is onto. Correct: Its domain and codomain matter, as f(x) = 2x on ℕ and on ℝ demonstrates.
• Misconception: A one-one function is necessarily onto. Correct: The equivalence holds for self-maps of finite sets, but need not hold for infinite sets.
• Misconception: In g ∘ f, apply g first. Correct: Apply f first, then use its output as the input of g.
• Misconception: An inverse means 1/f(x), or exists whenever an equation can be rearranged. Correct: An inverse reverses a bijection and must satisfy both identity compositions on the stated sets.

Exam-style questions with model answers

Q1. On A = {1, 2, 3, 4}, let R = {(a, b) ∈ A × A : a − b = 10}. Identify R and explain why A being non-empty does not change your answer. [2 marks]

1. No pair of elements of A has difference 10, so R contains no ordered pairs and is the empty relation.
2. The elements of A are not themselves elements of R. R consists of selected pairs, and none satisfies the condition.

Q2. For A = {1, 2, 3}, classify R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} as reflexive, symmetric and transitive, giving reasons. [3 marks]

1. R is reflexive because (1, 1), (2, 2) and (3, 3) all belong to it. These are the required self-pairs for every element of A.
2. R is not symmetric: (1, 2) belongs to R, whereas its reverse (2, 1) does not belong to R.
3. R is not transitive: (1, 2) and (2, 3) belong to R, but the required pair (1, 3) does not.

Q3. On the set ℤ of integers, let R = {(a, b) ∈ ℤ × ℤ : 2 divides a − b}. Prove that R is an equivalence relation and describe its equivalence classes. [5 marks]

1. For every integer a, a − a = 0, which is divisible by 2. Therefore every pair (a, a) belongs to R, so R is reflexive.
2. If (a, b) belongs to R, then a − b is divisible by 2. Its negative b − a is also divisible by 2, proving symmetry.
3. If (a, b) and (b, c) belong to R, then a − c = (a − b) + (b − c) is divisible by 2. This proves transitivity.
4. Since R is reflexive, symmetric and transitive on ℤ, it is an equivalence relation.
5. Its classes are the even integers [0] and the odd integers [1]. They are disjoint, and together contain every integer.

Q4. Compare injectivity and surjectivity for f : ℕ → ℕ, f(x) = 2x, and g : ℝ → ℝ, g(x) = 2x. Here ℕ = {1, 2, 3, ...} and ℝ is the set of real numbers. [4 marks]

1. For f, equality f(x₁) = f(x₂) gives 2x₁ = 2x₂ and hence x₁ = x₂ for natural inputs. Therefore f is one-one.
2. The natural number 1 has no preimage under f: its required input 1/2 is not natural. Thus f is not onto.
3. The same equal-output argument applies to real inputs for g, so g is also one-one.
4. Given any real y, the real input y/2 satisfies g(y/2) = y. Thus g is onto and therefore bijective.

Q5. Let f : {2, 3, 4, 5} → {3, 4, 5, 9}, with f(2) = 3, f(3) = 4, f(4) = f(5) = 5. Let g : {3, 4, 5, 9} → {7, 11, 15}, with g(3) = g(4) = 7 and g(5) = g(9) = 11. Determine g ∘ f. [4 marks]

1. For input 2, first use f(2) = 3 and then g(3) = 7. Hence (g ∘ f)(2) = 7.
2. For input 3, f(3) = 4 and g(4) = 7, giving (g ∘ f)(3) = 7.
3. For input 4, f(4) = 5 and g(5) = 11, giving (g ∘ f)(4) = 11.
4. For input 5, f(5) = 5 and g(5) = 11, giving (g ∘ f)(5) = 11. These specify the composite on its entire domain.

Q6. Let ℕ = {1, 2, 3, ...}, Y = {4x + 3 : x ∈ ℕ}, and f : ℕ → Y be given by f(x) = 4x + 3. Find its inverse, state the inverse's domain and codomain, and verify both compositions. [5 marks]

1. For an arbitrary y ∈ Y, the equation y = 4x + 3 gives x = (y − 3)/4. This supplies the candidate inverse formula.
2. Define g : Y → ℕ by g(y) = (y − 3)/4. The definition of Y ensures that this expression is a natural number for every permitted y.
3. For every x ∈ ℕ, g(f(x)) = (4x + 3 − 3)/4 = x, so the first composition returns each original input.
4. For every y ∈ Y, f(g(y)) = 4((y − 3)/4) + 3 = y, so the second composition is also the required identity.
5. Both inverse conditions hold. Hence f⁻¹ = g, with domain Y, codomain ℕ, and rule f⁻¹(y) = (y − 3)/4.

Key takeaways

• A relation on A is a subset of A × A, so every included pair has both entries in A.
• Reflexivity checks all self-pairs, symmetry checks reversed pairs, and transitivity checks chains through a shared middle element.
• An equivalence relation satisfies all three properties and separates its underlying set into disjoint equivalence classes.
• A function assigns each domain element exactly one image; its range consists of the outputs actually obtained.
• One-one functions distinguish inputs by their outputs; onto functions reach every element of their stated codomain.
• A self-map of a finite set is one-one exactly when it is onto; the finite-set condition matters.
• For g ∘ f, apply f first and g second, checking that each intermediate output is a permitted input.
• A function is invertible exactly when it is bijective, and both compositions with its inverse give identity functions.

Test yourself

What distinguishes the identity relation on A from the universal relation on A?

The identity relation contains exactly the self-pairs (a, a). The universal relation contains every ordered pair from A × A.

What must be missing to disprove reflexivity of a relation on A?

A self-pair (a, a) must be absent for at least one element a belonging to A.

Can two distinct inputs of a function have the same output?

Yes. This is many-one behaviour; the function condition still requires exactly one output for each individual input.

Why is f : ℝ → ℝ, f(x) = x², neither one-one nor onto?

Distinct inputs −1 and 1 both give 1, so it is many-one. The codomain element −2 has no real preimage, so it is not onto.

In g ∘ f, which function is applied first?

Apply f first, then apply g to its output: (g ∘ f)(x) = g(f(x)).

What conditions make f : A → B invertible?

It must be both one-one and onto, ensuring every element of B has exactly one preimage in A.

If (a, b) lies on the graph of an invertible function, which point lies on its inverse's graph?

The point (b, a) lies on the inverse's graph because the inverse exchanges the original input and output.

For f : ℕ → Y, f(x) = 4x + 3 and Y = {4x + 3 : x ∈ ℕ}, what is the inverse?

The inverse is f⁻¹ : Y → ℕ, defined by f⁻¹(y) = (y − 3)/4 for each y belonging to Y.