What this chapter is about
In Class 11, you met the basic ideas of relations and functions. This chapter takes those ideas further by asking deeper questions: when is a relation an equivalence relation, and when is a function invertible? These are foundational concepts that appear throughout higher mathematics, from abstract algebra to real analysis.
You will learn to classify relations as reflexive, symmetric and transitive, and recognise when all three properties hold together. For functions, you will study injectivity (one-one), surjectivity (onto) and bijectivity, which determine whether a function can be reversed. You will also learn how to compose two functions and find the inverse of a bijective function.
After studying this chapter, you should be able to determine the type of any given relation, check whether a function is one-one or onto, find the composition of two functions, and compute the inverse of an invertible function. These skills are essential for later chapters on inverse trigonometric functions and for any future study in mathematics.
Key ideas
- A relation R on a set A is a subset of A × A; we write a R b to mean (a, b) belongs to R.
- A relation is reflexive if every element is related to itself: (a, a) ∈ R for all a in A.
- A relation is symmetric if whenever a R b, we also have b R a.
- A relation is transitive if whenever a R b and b R c, we also have a R c.
- An equivalence relation is one that is reflexive, symmetric and transitive; it partitions the set into disjoint equivalence classes.
- A function f : A → B is one-one (injective) if distinct elements in A map to distinct elements in B: f(a₁) = f(a₂) implies a₁ = a₂.
- A function is onto (surjective) if every element in B is the image of at least one element in A.
- A function is bijective if it is both one-one and onto; only bijective functions have inverses.
Formulas and facts to remember
- Formula or Rule: (g ∘ f)(x) = g(f(x)) · Meaning: Composition: first apply f, then apply g to the result
- Formula or Rule: f⁻¹(f(x)) = x for all x in domain of f · Meaning: The inverse undoes the function
- Formula or Rule: f(f⁻¹(y)) = y for all y in range of f · Meaning: Applying f to its inverse returns the original value
- Formula or Rule: If f : A → B and g : B → C are both one-one, then g ∘ f is one-one · Meaning: One-one property is preserved under composition
- Formula or Rule: If f : A → B and g : B → C are both onto, then g ∘ f is onto · Meaning: Onto property is preserved under composition
- Formula or Rule: (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹ · Meaning: Inverse of a composition reverses the order
- Formula or Rule: Number of reflexive relations on a set with n elements = 2^(n² − n) · Meaning: Diagonal pairs must be present; other pairs are optional
Worked examples
Example 1: Checking whether a relation is an equivalence relation
Let A = {1, 2, 3, 4} and define a relation R by: a R b if and only if a − b is divisible by 2.
Step 1 (Reflexive): For any a in A, a − a = 0, which is divisible by 2. So (a, a) ∈ R for all a. R is reflexive.
Step 2 (Symmetric): Suppose a R b, meaning 2 divides (a − b). Then a − b = 2k for some integer k. So b − a = −2k = 2(−k), which is also divisible by 2. Hence b R a. R is symmetric.
Step 3 (Transitive): Suppose a R b and b R c. Then a − b = 2k and b − c = 2m for integers k and m. Adding: a − c = 2k + 2m = 2(k + m), which is divisible by 2. So a R c. R is transitive.
Since R is reflexive, symmetric and transitive, R is an equivalence relation.
Example 2: Determining if a function is one-one and onto
Let f : R → R be defined by f(x) = 3x + 7.
One-one check: Suppose f(a) = f(b). Then 3a + 7 = 3b + 7, which gives 3a = 3b, so a = b. Since equal outputs imply equal inputs, f is one-one.
Onto check: Let y be any real number. We need x such that f(x) = y, that is, 3x + 7 = y. Solving: x = (y − 7)/3, which is a real number for every real y. So every element in R (codomain) has a pre-image. f is onto.
Since f is both one-one and onto, f is bijective.
Example 3: Finding the inverse of a bijective function
Let f : R → R be defined by f(x) = 5x − 2. Find f⁻¹.
Step 1: Write y = f(x) = 5x − 2.
Step 2: Solve for x in terms of y: 5x = y + 2, so x = (y + 2)/5.
Step 3: The inverse function is obtained by interchanging roles of x and y. Thus f⁻¹(x) = (x + 2)/5.
Verification: f(f⁻¹(x)) = f((x + 2)/5) = 5 × (x + 2)/5 − 2 = (x + 2) − 2 = x. Correct.
Common mistakes
- Forgetting to check reflexivity at every element → always verify (a, a) ∈ R for each a in the set, not just one.
- Confusing one-one with onto → one-one is about no two inputs sharing an output; onto is about every output being hit.
- Assuming every function has an inverse → only bijective functions are invertible; check both conditions first.
- Writing composition in wrong order: thinking (g ∘ f)(x) means f(g(x)) → remember g ∘ f means apply f first, then g.
- Believing a relation that is symmetric and transitive must be reflexive → this is false; you need an independent check.
Quick revision
- Equivalence relation = reflexive + symmetric + transitive; it partitions a set into disjoint classes.
- One-one means different inputs give different outputs; onto means every output is achieved.
- Only bijective functions have inverses; to find f⁻¹, solve y = f(x) for x.
- Composition order: (g ∘ f)(x) = g(f(x)) — f acts first.
- Inverse of a composition: (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹.