📌 Snapshot
- Reflexive, symmetric and transitive relations, and one one, onto and invertible functions, explained for CUET.
- How to test each relation property, and why an equivalence relation always splits a set into equal groups.
- One one, onto and bijective functions, and why the domain and co domain decide the answer as much as the formula.
- Composition gof, the inverse of a function, and the exam traps on order, co domains and finite sets.
📖 Detailed Notes
Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XII, Chapter 1.
Tickets and seats
A cinema sells tickets, and each ticket is assigned a seat. If every ticket gets exactly one seat, that is a function from tickets to seats. If no two tickets share a seat, it is one one. If no seat is left empty, it is onto. If both hold, every ticket is paired with its own seat and every seat with its own ticket, which is a bijection, and the pairing can be run backwards from seat to ticket. That backward pairing is the inverse. Before functions, there is the looser idea of a relation, where an element may be linked to many, one or no others. This chapter takes both ideas and classifies them carefully.
Relations and their types
A relation $R$ in a set $A$ is any subset of $A \times A$, the set of all ordered pairs $(a, b)$ with $a, b \in A$. We write $a\,R\,b$ when $(a, b) \in R$. It simply records which pairs are related.
Two extreme relations are called trivial. The empty relation has no pairs at all, for example $R = \{(a, b) : a - b = 10\}$ in $A = \{1, 2, 3, 4\}$. The universal relation has every pair, $R = A \times A$.
Reflexive, symmetric, transitive
| Property | Meaning in words | In symbols | To disprove it |
|---|---|---|---|
| Reflexive | Everything is related to itself | $(a, a) \in R$ for every $a \in A$ | Find one $a$ with $(a, a) \notin R$ |
| Symmetric | Relations run both ways | $(a, b) \in R \Rightarrow (b, a) \in R$ | Find one pair present without its reverse |
| Transitive | Relations chain | $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$ | Find a chain whose end pair is missing |
A relation that is reflexive, symmetric and transitive all together is an equivalence relation. It captures "alike in some chosen way". Here are the standard examples, which are worth knowing as a set.
| Relation | Set | Reflexive | Symmetric | Transitive |
|---|---|---|---|---|
| Triangles are congruent | All triangles | Yes | Yes | Yes (equivalence) |
| $a = b$ | Any set | Yes | Yes | Yes (equivalence) |
| $2$ divides $a - b$ | $\mathbb{Z}$ | Yes | Yes | Yes (equivalence) |
| $a \le b$ | $\mathbb{R}$ | Yes | No | Yes |
| $a$ divides $b$ | $\mathbb{N}$ | Yes | No | Yes |
| $a < b$ | $\mathbb{R}$ | No | No | Yes |
| $L_1$ is perpendicular to $L_2$ | Lines in a plane | No | Yes | No |
| $b = a + 1$ | $\mathbb{N}$ | No | No | No |
Why perpendicular lines are not transitive. If $L_1 \perp L_2$ and $L_2 \perp L_3$, then $L_1$ and $L_3$ are both perpendicular to the same line, so they are parallel, not perpendicular. And no line is perpendicular to itself, which kills reflexivity.
Pause and predict: take $A = \{1, 2, 3\}$ and the empty relation $R = \varnothing$. Is it reflexive, symmetric, transitive?
Check your answer: it is symmetric and transitive, but not reflexive. Symmetry and transitivity are "if... then" rules, and with no pairs present nothing can break them. Reflexivity demands that $(1,1)$, $(2,2)$ and $(3,3)$ be present, and they are not. So the empty relation on a non-empty set is not an equivalence relation. The universal relation is.
Worked example: a rule based relation. Show that $R = \{(a, b) : 3 \text{ divides } (a - b)\}$ on $\mathbb{Z}$ is an equivalence relation.
| Step | What we do | Result |
|---|---|---|
| 1 | Reflexive: check $(a, a)$ | $a - a = 0 = 3 \times 0$, so $(a, a) \in R$ |
| 2 | Symmetric: let $a - b = 3k$ | $b - a = 3(-k)$, so $(b, a) \in R$ |
| 3 | Transitive: let $a - b = 3k$ and $b - c = 3m$ | Add: $a - c = 3(k + m)$, so $(a, c) \in R$ |
| 4 | Conclude | All three hold, so $R$ is an equivalence relation |
Common wrong step. Testing a few numbers such as $1, 4, 7$ and calling it proved. For a rule on an infinite set, the argument must be written for general $a$, $b$, $c$. A few examples can only disprove a property, never prove it.
Worked example: a relation given as a list. Classify $R = \{(1,1), (2,2), (3,3), (4,4), (1,2), (1,3), (3,2)\}$ on $A = \{1, 2, 3, 4\}$.
| Step | What we do | Result |
|---|---|---|
| 1 | Reflexive: is every $(a, a)$ there for $a = 1, 2, 3, 4$? | Yes |
| 2 | Symmetric: $(1, 2)$ is present, is $(2, 1)$? | No, so not symmetric |
| 3 | Transitive: go through every chain | $(1,3), (3,2)$ gives $(1,2)$, which is present. Every other chain passes as well |
| 4 | Conclude | Reflexive and transitive, but not symmetric |
Common wrong step. Checking only a couple of chains. For a list, transitivity means testing every pair of pairs where the second element of the first equals the first element of the second. A single missing end pair is enough to fail.
Equivalence classes
For an equivalence relation on $X$, the equivalence class of $a$ is $[a] = \{x \in X : x\,R\,a\}$, all the elements related to $a$. The classes are mutually disjoint, no element lies in two classes, and together they cover $X$. In other words, an equivalence relation partitions the set into groups of alike elements, and every partition arises from some equivalence relation.
For "2 divides $a - b$" on $\mathbb{Z}$ there are two classes, the even integers $[0]$ and the odd integers $[1]$. For "3 divides $a - b$" there are three:
| Class | Members | Form |
|---|---|---|
| $[0]$ | $\dots, -3, 0, 3, 6, \dots$ | $3r$ |
| $[1]$ | $\dots, -2, 1, 4, 7, \dots$ | $3r + 1$ |
| $[2]$ | $\dots, -1, 2, 5, 8, \dots$ | $3r + 2$ |
In general "$n$ divides $a - b$" gives $n$ classes, one for each possible remainder.
Beyond the textbook. For the exam: an equivalence relation splits a set into disjoint classes, as in the remainders above. In reality: this is the idea behind clock arithmetic. Days of the week repeat every 7 days, so any two dates 7 days apart are "alike", and the seven weekdays are the seven classes. Checksums on ISBNs and bank cards use the same remainders to catch typing errors. Why it matters: when many objects can be treated as one, a huge problem collapses to a few classes.
Pause and predict: how many equivalence relations on $\{1, 2, 3\}$ contain both $(1, 2)$ and $(2, 1)$?
Check your answer: $2$. The pair forces $1$ and $2$ into the same class. The element $3$ is then either alone (classes $\{1,2\}$ and $\{3\}$) or joins them (one class $\{1,2,3\}$). There is no third choice.
Functions: one one, onto, bijective
A function $f : X \to Y$ assigns to every element of $X$ exactly one element of $Y$. $X$ is the domain, $Y$ is the co domain and the set of values actually taken is the range. The range lies inside the co domain, and the two need not be equal.
| Type | Meaning | Test |
|---|---|---|
| One one (injective) | Different inputs give different outputs | $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$ |
| Many one | Two or more inputs share an output | Find $x_1 \ne x_2$ with $f(x_1) = f(x_2)$ |
| Onto (surjective) | Every element of $Y$ is hit | For arbitrary $y \in Y$, solve $y = f(x)$ and check $x \in X$ |
| Into | Some element of $Y$ is missed | Range is smaller than the co domain |
| Bijective | One one and onto | Both tests pass |
On a graph, a function is one one when every horizontal line cuts it at most once, and onto when every horizontal line through the co domain cuts it at least once.
The formula alone does not decide the type. The same formula changes behaviour when the sets change.
Worked example: same formula, different sets. Classify $f(x) = x^2$ for (a) $f : \mathbb{R} \to \mathbb{R}$, (b) $f : \mathbb{R} \to [0, \infty)$, (c) $f : [0, \infty) \to [0, \infty)$.
| Step | What we do | Result |
|---|---|---|
| 1 | (a) One one? $f(-1) = f(1) = 1$ with $-1 \ne 1$ | Not one one |
| 2 | (a) Onto? Is $x^2 = -2$ solvable in $\mathbb{R}$? | No, so not onto. (a) is neither |
| 3 | (b) Now every $y \ge 0$ has $x = \sqrt{y}$ | Onto, but still not one one |
| 4 | (c) Restrict the domain to $x \ge 0$ as well | One one and onto, so bijective |
Common wrong step. Saying "$x^2$ is not one one" and stopping, without reading the domain. In (c) the negative inputs are gone, so the two roots $\pm\sqrt{y}$ no longer both appear.
More standard cases to remember:
| Function | Type |
|---|---|
| $f : \mathbb{N} \to \mathbb{N}$, $f(x) = 2x$ | One one, not onto (nothing maps to 1) |
| $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 2x$ | Bijective |
| $f : \mathbb{N} \to \mathbb{N}$, $f(x) = x + 1$ | One one, not onto (1 is missed) |
| $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^3$ | Bijective |
| Constant function on a set with 2 or more elements | Many one, not onto |
| Identity $I_X(x) = x$ | Bijective |
Worked example: a piecewise bijection. Show that $f : \mathbb{N} \to \mathbb{N}$ given by $f(x) = x + 1$ when $x$ is odd and $f(x) = x - 1$ when $x$ is even is a bijection.
| Step | What we do | Result |
|---|---|---|
| 1 | See the pattern | $f(1) = 2$, $f(2) = 1$, $f(3) = 4$, $f(4) = 3$: it swaps neighbours |
| 2 | One one: odd inputs give even outputs, even inputs give odd outputs | If $f(x_1) = f(x_2)$, the inputs have the same parity, so $x_1 \pm 1 = x_2 \pm 1$ with the same sign and $x_1 = x_2$ |
| 3 | Onto: take any $y \in \mathbb{N}$ | If $y$ is even, then $y - 1$ is odd and $f(y-1) = y$. If $y$ is odd, then $y + 1$ is even and $f(y+1) = y$ |
| 4 | Conclude | Both tests pass for all inputs, so $f$ is a bijection |
Common wrong step. Checking a few swaps like $f(1) = 2$ and $f(2) = 1$ and stopping. A proof must handle an arbitrary input and an arbitrary target.
Finite sets are special. For a finite set $X$, a function $f : X \to X$ that is one one is automatically onto, and one that is onto is automatically one one. The reason is the pigeonhole idea: $n$ different inputs give $n$ different outputs, and there are only $n$ elements available. This is why the number of one one functions from $\{1, 2, 3\}$ to itself is $3! = 6$, all of them rearrangements. The shortcut fails for infinite sets, as $f(x) = 2x$ on $\mathbb{N}$ shows.
Pause and predict: $f : \{1, 2, 3\} \to \{1, 2, 3\}$ is one one. Must it be onto?
Check your answer: Yes. Three different inputs give three different outputs, and there are only three elements to use, so each one is hit. The same question for $f(x) = 2x$ on $\mathbb{N}$ gives "no", because the infinite set always has room to spare.
Beyond the textbook. For the exam: for finite sets one one means onto, and for infinite sets it need not. In reality: this is how mathematicians recognise infinity. A set is infinite when it can be matched one one with part of itself: the map $x \to 2x$ pairs all of $\mathbb{N}$ with just the even numbers. Why it matters: it shows that "same size" for infinite sets is defined by the existence of a bijection, which is the idea this chapter formalises.
Composition of functions
If $f : A \to B$ and $g : B \to C$, the composition $g \circ f : A \to C$ is defined by
$$(g \circ f)(x) = g(f(x))$$
It is also written $gof$. Read it as "$g$ after $f$": the function written nearest to $x$ acts first. The output of $f$ must be a valid input for $g$. Composition is associative, $h \circ (g \circ f) = (h \circ g) \circ f$, but it is not commutative.
Worked example: order matters. Let $f(x) = \cos x$ and $g(x) = 3x^2$. Find $g \circ f$ and $f \circ g$, and compare them at $x = 0$.
| Step | What we do | Result |
|---|---|---|
| 1 | $g \circ f$: put $\cos x$ into $g$ | $g(\cos x) = 3\cos^2 x$ |
| 2 | $f \circ g$: put $3x^2$ into $f$ | $f(3x^2) = \cos(3x^2)$ |
| 3 | At $x = 0$ | $(g \circ f)(0) = 3$ but $(f \circ g)(0) = 1$ |
So $g \circ f \ne f \circ g$.
Common wrong step. Writing $(g \circ f)(x) = 3x^2\cos x$, multiplying the two functions, or doing $g$ first and then $f$. Composition feeds one function's output into the other.
Invertible functions
A function $f : X \to Y$ is invertible if there is a function $g : Y \to X$ with $g \circ f = I_X$ and $f \circ g = I_Y$, where $I$ is the identity. This $g$ is unique and is written $f^{-1}$. The key theorem:
$$f \text{ is invertible} \iff f \text{ is one one and onto (bijective)}$$
Why both conditions. To undo $f$, every $y$ in $Y$ must come from some $x$, so $f$ must be onto. And that $x$ must be the only one, otherwise the inverse would have to return two values for one $y$, so $f$ must be one one. Then $f^{-1}(y)$ is simply the one $x$ with $f(x) = y$.
To find an inverse, solve $y = f(x)$ for $x$ in terms of $y$, and then check both compositions.
Worked example: finding an inverse. Let $Y = \{y \in \mathbb{N} : y = 4x + 3 \text{ for some } x \in \mathbb{N}\}$, and $f : \mathbb{N} \to Y$ with $f(x) = 4x + 3$. Show that $f$ is invertible and find $f^{-1}$.
| Step | What we do | Result |
|---|---|---|
| 1 | The co domain is exactly the range, so $f$ is onto. One one: $4x_1 + 3 = 4x_2 + 3$ gives $x_1 = x_2$ | $f$ is bijective, so invertible |
| 2 | Solve $y = 4x + 3$ for $x$ | $x = \dfrac{y - 3}{4}$, so $g(y) = \dfrac{y - 3}{4}$ |
| 3 | Check $g \circ f$ | $g(4x + 3) = \dfrac{4x + 3 - 3}{4} = x$ |
| 4 | Check $f \circ g$ | $f\!\left(\dfrac{y-3}{4}\right) = (y - 3) + 3 = y$ |
For instance, $f^{-1}(19) = \dfrac{16}{4} = 4$.
Common wrong step. Taking the co domain to be $\mathbb{N}$. Then $f$ is not onto (the number $1$ is never reached), so no inverse can exist. That is exactly why $Y$ is defined as the range.
Beyond the textbook. For the exam: a function is invertible exactly when it is a bijection. In reality: this is the rule behind every code that can be decoded. A cipher that shifts letters by 3 is a bijection on the alphabet, and shifting by $-3$ undoes it. A scheme that sent two letters to the same letter could never be decoded. The same idea explains why $\sqrt{\ }$ undoes $x^2$ only after the domain is restricted to $x \ge 0$, as in the earlier worked example. Why it matters: inverse trigonometric functions in the next chapter all begin by restricting a domain until the function becomes a bijection.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| "A reflexive relation must be the universal relation." | Everything is related to itself | Reflexive only needs the pairs $(a, a)$. Other pairs may be absent |
| "$a \le b$ is symmetric." | $a = b$ is | $a \le b$ does not give $b \le a$. It is reflexive and transitive only |
| "The empty relation is an equivalence relation." | It breaks no rule | On a non-empty set it is not reflexive |
| "Perpendicular lines form a transitive relation." | The relation feels symmetric and tidy | Perpendicular to the same line means parallel, so it is not transitive |
| "$f(x) = 2x$ is onto." | True on $\mathbb{R}$ | On $\mathbb{N} \to \mathbb{N}$ the number $1$ is missed. The co domain matters |
| $(g \circ f)(x) = f(g(x))$ | Reading left to right | The function nearest to $x$ acts first: $g(f(x))$ |
| "One one means onto" for any set | True for finite self maps | Fails for infinite sets, as in $x \to 2x$ on $\mathbb{N}$ |
| "Every one one function has an inverse." | You can solve for $x$ | It must also be onto, so its co domain equals its range |
Formula sheet
| Concept | Condition | How to test |
|---|---|---|
| Relation | $R \subseteq A \times A$ | List or describe the pairs |
| Reflexive | $(a, a) \in R$ for all $a$ | Check every element |
| Symmetric | $(a, b) \in R \Rightarrow (b, a) \in R$ | Reverse each pair |
| Transitive | $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$ | Check every chain |
| Equivalence relation | All three | Gives disjoint classes covering the set |
| Classes for "$n$ divides $a - b$" | $n$ classes | Remainders $0, 1, \dots, n-1$ |
| One one | $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$ | Assume equal outputs |
| Onto | Range equals co domain | Solve $y = f(x)$ |
| Composition | $(g \circ f)(x) = g(f(x))$ | Inner function first |
| Inverse | $g \circ f = I_X$ and $f \circ g = I_Y$ | Solve $y = f(x)$ for $x$ |
| Invertibility | Bijective | Check one one and onto |
| Finite sets | One one iff onto, for $f : X \to X$ | Pigeonhole |
60-second revision
| Topic | Lock this in |
|---|---|
| Relation | A subset of $A \times A$; empty and universal are trivial |
| Equivalence | Reflexive, symmetric, transitive; splits the set into disjoint classes |
| Quick examples | Congruence and $a = b$ are equivalences; $\le$ is not symmetric; $\perp$ is only symmetric |
| One one | Different inputs, different outputs |
| Onto | Range equals co domain |
| Bijective | One one and onto; the same formula can change type with the sets |
| Finite sets | One one iff onto for $f : X \to X$ |
| Composition | $g \circ f$ means $f$ first; not commutative; associative |
| Inverse | Exists iff bijective; solve $y = f(x)$ for $x$ |
| Counts | Bijections of $\{1,2,3\}$: $3! = 6$. Equivalence relations containing $(1,2),(2,1)$ on $\{1,2,3\}$: $2$ |
Where this chapter connects
- Before: Relations and Functions in Class XI introduced ordered pairs, domain and range, and Sets supplied the language of subsets used throughout.
- Next: Inverse Trigonometric Functions restricts the domains of the trig functions until they are bijective, so that the inverses exist.
- Later: Application of Derivatives shows that a function that is always increasing or always decreasing is one one.
🎯 Practice MCQs
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Q1. The relation $R = \{(1,1), (2,2), (3,3), (4,4), (1,2), (1,3), (3,2)\}$ on $\{1, 2, 3, 4\}$ is
▸ Show answer & explanation
Answer: B
All four pairs $(a, a)$ are present, so it is reflexive. The pair $(1, 2)$ is present but $(2, 1)$ is not, so it is not symmetric. Going through the chains, $(1,3), (3,2)$ forces $(1,2)$, which is present, and every other chain also ends in a pair that is present, so it is transitive.
Q2. The number of distinct equivalence classes of the relation $R = \{(a, b) : 3 \text{ divides } (a - b)\}$ on $\mathbb{Z}$ is
▸ Show answer & explanation
Answer: C
Two integers are related when they leave the same remainder on division by $3$. The possible remainders are $0$, $1$ and $2$, so there are three classes: $3r$, $3r + 1$ and $3r + 2$. Option D confuses the number of classes with the number of elements in each.
Q3. The relation $R = \{(a, b) : a \le b\}$ on $\mathbb{R}$ is
▸ Show answer & explanation
Answer: B
$a \le a$ holds, so it is reflexive. If $a \le b$ and $b \le c$ then $a \le c$, so it is transitive. But $1 \le 2$ does not give $2 \le 1$, so it is not symmetric. Option D would describe $a < b$, which is not reflexive.
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Q4. Which of the following functions is a bijection?
▸ Show answer & explanation
Answer: C
For option C, $3x_1 - 5 = 3x_2 - 5$ gives $x_1 = x_2$, and $y = 3x - 5$ is solved by $x = \dfrac{y + 5}{3} \in \mathbb{R}$ for every $y$, so it is one one and onto. Options A and D fail because $f(-1) = f(1)$. Option B misses every odd number.
Q5. The function $f : \mathbb{R} \to [0, \infty)$ defined by $f(x) = x^2$ is
▸ Show answer & explanation
Answer: C
Every $y \ge 0$ equals $x^2$ for $x = \sqrt{y}$, so the function is onto the given co domain. But $f(-2) = f(2) = 4$, so it is not one one. Changing the co domain from $\mathbb{R}$ to $[0, \infty)$ is what makes it onto.
Q6. If $f(x) = 2x + 1$ and $g(x) = x^2$ on $\mathbb{R}$, then $(g \circ f)(x)$ is
▸ Show answer & explanation
Answer: A
$(g \circ f)(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2$. Option B is $(f \circ g)(x) = f(x^2) = 2x^2 + 1$, with the order reversed.
Q7. Let $Y = \{y \in \mathbb{N} : y = 4x + 3 \text{ for some } x \in \mathbb{N}\}$ and $f : \mathbb{N} \to Y$ be $f(x) = 4x + 3$. Then $f^{-1}(19)$ is
▸ Show answer & explanation
Answer: B
The inverse is $f^{-1}(y) = \dfrac{y - 3}{4}$, so $f^{-1}(19) = \dfrac{16}{4} = 4$. Check: $f(4) = 19$. Option D is $4 \times 19$, which treats $f^{-1}$ as multiplication.
Q8. Assertion (A): The function $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$ is not invertible. Reason (R): $f$ is not one one.
▸ Show answer & explanation
Answer: A
A function is invertible only if it is a bijection. Since $f(-1) = f(1)$, $f$ is not one one, so it cannot be a bijection, and that is exactly why it has no inverse. (It also fails to be onto, but not being one one is already enough.)
Q9. The number of one one functions from $\{1, 2, 3\}$ to itself is
▸ Show answer & explanation
Answer: B
The image of $1$ can be chosen in $3$ ways, the image of $2$ in $2$ remaining ways, and the image of $3$ in $1$ way, giving $3! = 6$. Each one one self map of a finite set is also onto. Option D, $3^3$, counts all functions, one one or not.
Q10. Consider the statements. I: An equivalence relation on a set divides it into mutually disjoint equivalence classes. II: For a finite set $X$, every onto function $f : X \to X$ is one one. III: The empty relation on a non empty set is reflexive. IV: $g \circ f = f \circ g$ for all functions $f$ and $g$. Which are correct?
▸ Show answer & explanation
Answer: A
I and II are standard results. III is false: reflexivity needs every $(a, a)$ to be present, and the empty relation has none. IV is false, as $\cos(3x^2)$ and $3\cos^2 x$ differ at $x = 0$.
Q11. Match List I with List II. List I: (i) $f : \mathbb{N} \to \mathbb{N}$, $f(x) = 2x$ (ii) $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 2x$ (iii) $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$ (iv) $f : \mathbb{R} \to [0, \infty)$, $f(x) = x^2$. List II: (p) bijective (q) one one but not onto (r) neither one one nor onto (s) onto but not one one.
▸ Show answer & explanation
Answer: A
For (i), $2x = 1$ has no solution in $\mathbb{N}$, so it is one one but not onto. For (ii), $x = \dfrac{y}{2}$ always works, so it is a bijection. For (iii), it fails on both counts. For (iv), it is onto but $f(-2) = f(2)$.
Q12. The number of equivalence relations on the set $\{1, 2, 3\}$ that contain $(1, 2)$ and $(2, 1)$ is
▸ Show answer & explanation
Answer: B
The pair $(1, 2)$ and $(2, 1)$ puts $1$ and $2$ in the same class. The element $3$ is then either in a class of its own, giving classes $\{1, 2\}$ and $\{3\}$, or joins them, giving one class $\{1, 2, 3\}$. Each partition gives exactly one equivalence relation, so there are two.
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