Relations and Functions

The chapter runs in two connected halves. The first half, types of relations, opens with the two trivial extremes (empty and universal relations), then builds the three defining properties — reflexive, symmetric, transitive — and assembles them into the equivalence relation, together with the equivalence classes it generates and the way those classes partition the underlying set. The second half, types of functions, revisits domain, codomain and range from Class XI just long enough to define one-one and onto precisely, combines them into bijective, and then builds the composition of two functions before asking the central question of the chapter: which functions can be inverted? The answer — a function is invertible if and only if it is bijective — is the theorem the whole chapter has been building toward, and the closing remark about finite versus infinite sets (one-one and onto coincide only on a finite self-map) is the chapter's most tested subtlety.

Reflexive, symmetric and transitive relations

A relation $R$ in a set $A$ is, by definition, nothing more than a subset of $A \times A$ — membership in $R$ is the only rule there is. The two extreme cases fix the vocabulary before anything more refined is possible.

where: the empty relation relates nothing to anything (no pair satisfies its defining condition), while the universal relation relates every element of $A$ to every element of $A$, with no exceptions. Both are called trivial relations precisely because neither one discriminates between elements the way an interesting relation should. Between these two extremes sit the three properties that matter most:

where: $a, a_1, a_2, a_3$ range over $A$ in each clause, and every implication must hold for every choice of the elements involved — a single counterexample anywhere destroys the property. Reflexivity is checked one element at a time and must never be skipped for an "obvious" or isolated element; symmetry is checked pair by pair, in both directions; transitivity is checked over chains of two pairs at a time. For the set $\{1,2,3\}$, $R = \{(1,1),(2,2),(3,3),(1,2),(2,3)\}$ is reflexive (all three diagonal pairs are present) but neither symmetric ($(1,2) \in R$ while $(2,1) \notin R$) nor transitive ($(1,2), (2,3) \in R$ but $(1,3) \notin R$). By contrast, $R = \{(a,b) : 2 \text{ divides } a-b\}$ on the integers $\mathbb{Z}$ is reflexive ($a - a = 0$ is divisible by $2$), symmetric (if $2 \mid a-b$ then $2 \mid b-a$), and transitive (if $2 \mid a-b$ and $2 \mid b-c$, then $a - c = (a-b)+(b-c)$ is a sum of two even numbers, hence even) — all three at once, which is exactly what earns it the name equivalence relation.

Definition. A relation $R$ in a set $A$ is an equivalence relation if $R$ is reflexive, symmetric and transitive. Equivalence relations are the single most important type of relation in mathematics because they formalise the everyday idea of "essentially the same," whether that means congruent triangles, parallel lines, or integers with the same remainder on division by some fixed number.

Equivalence classes and partitions of a set

Once a relation is known to be an equivalence relation, it does something remarkable: it sorts the whole set into non-overlapping groups automatically. For the relation $R = \{(a,b) : 2 \text{ divides } a-b\}$ on $\mathbb{Z}$, every even integer turns out to be related to $0$, and every odd integer to $1$, and never the two mixed — the set of even integers $E$ and the set of odd integers $O$ are disjoint, their union is all of $\mathbb{Z}$, and all elements inside each one are mutually related. This subset is called an equivalence class.

where: $[a]$, the equivalence class containing $a$, collects every element of $X$ related to $a$; the second pair of statements records that the full collection of distinct classes $A_i$ (also written $[a]$ for various choices of $a$) covers $X$ completely and never overlaps. In the example above, $[0] = E$ and $[1] = O$, and note $[0] = [2r]$ for every even $2r$, since any two even numbers generate the same class. A richer illustration: on $X = \{1,2,\ldots,9\}$, the relation $R_1 = \{(x,y) : x-y \text{ is divisible by } 3\}$ produces exactly three classes, $\{1,4,7\}$, $\{2,5,8\}$ and $\{3,6,9\}$ — every element of $X$ falls into exactly one of the three, and the three classes are disjoint and exhaust $X$. This reverse direction works too: starting from any partition of a set into disjoint, exhaustive subsets, defining "related" as "lying in the same subset" always produces a genuine equivalence relation — partitions and equivalence relations are two descriptions of the same underlying structure.

One-one, onto and bijective functions

Turn now from arbitrary relations to the special, well-behaved subset of relations called functions, and ask two independent questions about a function $f: X \to Y$: does it ever send two different inputs to the same output, and does its output ever miss part of $Y$?

where: a function that is not one-one is called many-one — some element of the codomain has more than one preimage. A function that is not onto has a range strictly smaller than its codomain — some element of $Y$ is never hit. Definition. A function that is both one-one and onto is bijective, and only bijective functions can be paired with a true two-sided inverse. Three worked contrasts fix the ideas: $f: \mathbb{N} \to \mathbb{N}$, $f(x) = 2x$ is one-one (doubling never collides) but not onto (odd numbers are never hit); $f: \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$ is neither one-one ($f(-1) = f(1) = 1$) nor onto (negative numbers are never hit, since a square is never negative); $f: \mathbb{R} \to \mathbb{R}$, $f(x) = 2x$ is both one-one and onto, since $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$ and every real $y$ has preimage $y/2$.

Composition of functions and invertibility

Given a function $f$ from $A$ to $B$ and a second function $g$ from $B$ to $C$, feeding the output of $f$ straight into $g$ builds a single new function from $A$ to $C$, called the composition of $f$ and $g$.

where: the notation $g \circ f$ is read "$g$ of $f$," and the order matters critically — $f$ acts first, on the original input, and $g$ acts second, on $f$'s output. For $f(x) = \cos x$ and $g(x) = 3x^2$, both mapping $\mathbb{R}$ to $\mathbb{R}$, $(g \circ f)(x) = g(\cos x) = 3\cos^2 x$, while $(f \circ g)(x) = f(3x^2) = \cos(3x^2)$ — genuinely different functions, since $3\cos^2 x \neq \cos(3x^2)$ at $x=0$ already ($3 \neq 1$). This asymmetry is the source of the chapter's most reliable trap: never assume $g \circ f = f \circ g$ without checking. With composition available, invertibility can be defined precisely: a function $f: X \to Y$ is invertible if there is a function $g: Y \to X$ that undoes it in both directions at once.

where: $I_X$ and $I_Y$ denote the identity functions on $X$ and $Y$ ($I_X(x) = x$ for every $x \in X$), and the function $g$, when it exists, is called the inverse of $f$ and written $f^{-1}$. The double condition matters: $g \circ f = I_X$ alone (a "left inverse") is not the same guarantee as $f \circ g = I_Y$ (a "right inverse") — invertibility demands both simultaneously. NCERT states, as a working remark, that this happens exactly when $f$ is one-one and onto; the full two-directional proof is worked out below.

Board level - testing reflexivity, symmetry and transitivity (Exercise 1.1, Q3 pattern)

Check whether the relation R defined on the set A = {1, 2, 3, 4, 5, 6} by R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Step 1. List the pairs directly from the rule. For $a = 1,2,3,4,5$, $b = a+1$ gives $2,3,4,5,6$, all inside $A$; for $a=6$, $b=7$ falls outside $A$, so it is excluded. Hence $R = \{(1,2),(2,3),(3,4),(4,5),(5,6)\}$.

Step 2. Reflexive test: we would need $(a,a) \in R$ for every $a \in A$, i.e. $a = a+1$, which is false for every real number. In particular $(1,1) \notin R$, so $R$ fails reflexivity at once — no need to check further elements once a single failure is found.

Step 3. Symmetric test: $(1,2) \in R$; is $(2,1) \in R$? That would require $1 = 2+1 = 3$, false. So $(2,1) \notin R$, and $R$ is not symmetric.

Step 4. Transitive test: $(1,2) \in R$ and $(2,3) \in R$; transitivity would demand $(1,3) \in R$, i.e. $3 = 1+1 = 2$, false. So $(1,3) \notin R$, and $R$ is not transitive. Answer: R is neither reflexive, nor symmetric, nor transitive.

Board level - proving a linear function is bijective and finding its inverse (Exercise 1.2, Q7(i) and Example 17 pattern)

Let f : R to R be defined by f(x) = 3 - 4x. Show that f is one-one and onto, and hence find its inverse.

Step 1. One-one: assume $f(x_1) = f(x_2)$. Then $3 - 4x_1 = 3 - 4x_2 \Rightarrow -4x_1 = -4x_2 \Rightarrow x_1 = x_2$. By the definition in equation $(4)$, $f$ is one-one.

Step 2. Onto: take an arbitrary $y \in \mathbb{R}$ (the codomain) and try to solve $y = 3 - 4x$ for $x$: $x = \dfrac{3-y}{4}$. Since $y$ is any real number, $\dfrac{3-y}{4}$ is also a well-defined real number, so a preimage $x \in \mathbb{R}$ exists for every $y$. Hence $f$ is onto.

Step 3. Since $f$ is both one-one and onto, it is bijective, and by the invertibility theorem proved above, $f$ is invertible. Its inverse is exactly the rule found while proving onto-ness: $g(y) = \dfrac{3-y}{4}$, so $f^{-1}(x) = \dfrac{3-x}{4}$.

Step 4. Verify both compositions: $(g \circ f)(x) = g(3-4x) = \dfrac{3-(3-4x)}{4} = \dfrac{4x}{4} = x$, and $(f \circ g)(y) = f\left(\dfrac{3-y}{4}\right) = 3 - 4 \cdot \dfrac{3-y}{4} = 3-(3-y) = y$. Both check out. Answer: f is one-one and onto, hence invertible, with inverse f⁻¹(x) = (3 - x)/4.

Competitive level (JEE / NEET) - counting equivalence relations containing two given pairs (Example 24 pattern)

Let A = {1, 2, 3}. Find the number of equivalence relations on A that contain the pairs (1, 2) and (2, 1).

Step 1. Any equivalence relation must contain $(1,1), (2,2), (3,3)$ by reflexivity, regardless of anything else. Together with the two given pairs, the smallest possible candidate is $R_1 = \{(1,1),(2,2),(3,3),(1,2),(2,1)\}$. This is already a valid equivalence relation: reflexive by construction, symmetric since $(1,2)$ and $(2,1)$ are both present, and transitive vacuously — the only chain to check is $(1,2)$ with $(2,1)$, which forces $(1,1)$, already present.

Step 2. The only undecided pairs link $3$ to $\{1,2\}$: namely $(1,3), (3,1), (2,3), (3,2)$. Suppose we try adding just one of these, say $(1,3)$. Symmetry immediately forces $(3,1)$ as well. Then transitivity, applied to $(3,1) \in R$ and $(1,2) \in R$, forces $(3,2) \in R$; symmetry then forces $(2,3) \in R$ too. Adding a single cross-connection between $3$ and the $\{1,2\}$ block forces, through the closure demanded by symmetry and transitivity, the addition of all four remaining pairs at once — there is no way to add "a little bit" of a new relationship to an equivalence relation.

Step 3. So there are exactly two possibilities: either $3$ remains in a class of its own, giving $R_1$ with equivalence classes $\{1,2\}$ and $\{3\}$; or all four remaining pairs get added at once, merging $3$ fully into the $\{1,2\}$ block and producing the universal relation $R_2 = A \times A$, with one single class $\{1,2,3\}$. Step 2 shows there is nothing strictly between these two, because admitting any single cross-pair forces closure to the full universal relation.

Step 4. Counting the possibilities from Steps 1 to 3 gives exactly two valid equivalence relations. Answer: there are exactly 2 equivalence relations on {1, 2, 3} containing (1, 2) and (2, 1). Timing note: budget under two minutes. The fast method is to count set partitions rather than re-deriving symmetry and transitivity closure from scratch: an equivalence relation on a finite set corresponds exactly to a partition of that set into disjoint classes, and forcing $(1,2) \in R$ and $(2,1) \in R$ simply forces $1$ and $2$ into the same class. What remains is only the question of where $3$ goes — its own singleton class, or merged into the $\{1,2\}$ block — exactly $2$ partitions, hence $2$ equivalence relations. This partition-counting shortcut is the standard technique for every "count the equivalence relations containing such-and-such pairs" question on JEE and NEET.

The single most common slip is checking transitivity (or symmetry) only against the pairs explicitly listed, mistaking "no counterexample among the pairs I listed" for a genuine proof — a rule-based relation like $\{(a,b): 2 \text{ divides } a-b\}$ needs an argument that holds for every triple, not a spot check of a few numbers. A closely related mistake is forgetting that reflexivity must hold for literally every element of $A$, including isolated elements with no other relationships at all. Many students also believe symmetric-and-transitive together force reflexivity — and indeed, if $(a,b) \in R$ for some $b$, symmetry gives $(b,a) \in R$ and transitivity then gives $(a,a) \in R$ — but this only works for elements that already appear in some pair; an isolated element is never forced into $(a,a)$, so symmetric-and-transitive can still fail to be reflexive overall. On the function side, students often assume an element of the codomain with two preimages breaks onto-ness — it does not; onto only demands at least one preimage per codomain element, which is exactly what "many-one but onto" describes. Composition order is another reliable trap: $(g \circ f)(x)$ applies $f$ first and $g$ second, and $g \circ f \neq f \circ g$ in general. Finally, students over-apply the finite-set shortcut to infinite sets, or to a function between two finite sets of different sizes — it applies only to a self-map $f: X \to X$ on one and the same finite set.

• Test any given relation for reflexivity, symmetry and transitivity directly against the formal definitions in equation $(2)$, checking every element or pair the definition requires, not just a sample.
• State the definition of an equivalence relation and prove a given relation satisfies all three properties in general, using an arbitrary element or triple rather than a finite check.
• Compute the equivalence classes of a given equivalence relation, and explain why they always partition the underlying set into disjoint, exhaustive groups.
• Distinguish one-one, onto and bijective functions using the formal quantifier definitions in equation $(4)$, and correctly identify which property fails for a given function, with a stated reason.
• Compute the composition $g \circ f$ of two functions, evaluate it at specific points, and recognise that $g \circ f$ generally differs from $f \circ g$.
• Prove a function is invertible by showing it is bijective, and construct its explicit inverse using the method in the first derivation.
• State and correctly apply the finite-set shortcut (one-one is onto for a self-map on a finite set), and explain, with the two standard counterexamples on $\mathbb{N}$, exactly why it fails on infinite sets.
• Count the number of relations or equivalence relations on a small finite set satisfying given constraints, using the partition-counting technique from the competitive worked example.

Ask the tutor: Want to work through proving a trickier relation is an equivalence relation, or counting equivalence relations on a four-element set, step by step together?