Sets, Relations and Functions

Modern mathematics begins with a single quiet idea: collect objects together and reason about the collection as a thing in itself. That collection is called a set, and Class 11 — the first major JEE Main topic — is where it formally enters the syllabus. Master sets cleanly and the whole of Relations, Functions and even Probability becomes far easier later.

Definition: A SET is a well-defined collection of distinct objects. The phrase "well-defined" carries weight — given any object, you must be able to say with certainty yes, it belongs or no, it doesn't. "Distinct" means no element is counted twice.

Notation: a set can be described two ways.

• Roster (tabular) form — list the elements inside curly braces, e.g. A = {1, 2, 3}.
• Set-builder form — describe the rule that elements satisfy, e.g. B = {x : x is an even natural number less than 10}, read as "the set of all x such that x is an even natural number less than 10".

The colon : (or vertical bar |) is read as "such that". Order does not matter (so {1, 2, 3} = {3, 1, 2}), and repetition is ignored (so {1, 1, 2} = {1, 2}).

Key types of sets

A handful of named sets recur throughout every chapter. Learn them by their defining property, not just by their name.

• Empty set ∅ (also called null or void set). It has no elements. By convention it is written as ∅ or { }. Crucially, ∅ is a subset of every set — including itself.
• Singleton set. Exactly one element, like {0} or {π}.
• Finite set. A definite, countable number of elements; you can list them all and stop.
• Infinite set. Cannot be listed exhaustively, e.g. the set of natural numbers ℕ.
• Equal sets. Two sets A and B are equal if every element of A is in B and vice versa. Same elements, regardless of order or repetition.
• Equivalent sets. Same number of elements, not necessarily the same elements. {1, 2, 3} and {a, b, c} are equivalent but not equal.
• Subset. A ⊆ B means every element of A is also in B. We say "A is a subset of B".
• Proper subset. A ⊂ B means A ⊆ B but A ≠ B — A is "strictly smaller" than B.
• Universal set U. The fixed master set that contains all the elements under discussion in a given problem. Choosing U is part of setting up the question; different problems choose different universal sets.
• Power set P(A). The set of all subsets of A — including ∅ and A itself.

Two relationships are always true and worth memorising as reflexes: ∅ ⊆ A and A ⊆ A for every set A.

Counting subsets — the 2ⁿ rule

This is the single highest-yield numerical skill in the chapter.

Result: If a set A has n elements, then it has exactly 2ⁿ subsets, so |P(A)| = 2ⁿ.

Why? For each element of A, there are exactly two independent choices when forming a subset: include it, or leave it out. With n elements, that gives 2 × 2 × … × 2 = 2ⁿ possibilities. Every distinct sequence of in/out choices produces a distinct subset; nothing is double-counted; ∅ corresponds to "leave them all out" and A itself corresponds to "include them all".

From this, two derived counts:

• Number of proper subsets = 2ⁿ − 1 (we remove the case where every element is included, which is A itself).
• Number of non-empty subsets = 2ⁿ − 1 (we remove the empty subset).

Notice the trap: both expressions are 2ⁿ − 1, but they correspond to different deletions. If the question asks for proper non-empty subsets, the answer is 2ⁿ − 2.

Worked example — power set count

Question: If A = {a, b, c, d}, find (i) the number of subsets, (ii) the number of proper subsets, (iii) the number of proper non-empty subsets.

Solution:
Step 1: n = 4.
Step 2: Total subsets = 2⁴ = 16.
Step 3: Proper subsets = 2⁴ − 1 = 15 (drop A itself).
Step 4: Proper non-empty subsets = 2⁴ − 2 = 14 (drop A and ∅).
Conclusion: 16, 15 and 14 respectively.

Worked example — set-builder to roster

Question: Write A = {x : x ∈ ℕ, x² ≤ 20} in roster form.

Solution:
Step 1: x must be a natural number with x² ≤ 20.
Step 2: Test x = 1 → 1 ≤ 20 ✓, x = 2 → 4 ✓, x = 3 → 9 ✓, x = 4 → 16 ✓, x = 5 → 25 ✗.
Conclusion: A = {1, 2, 3, 4}.

This conversion skill is tested both ways in JEE Main; expect at least one question that hides a tricky condition (like x ∈ ℤ vs x ∈ ℕ, or strict vs non-strict inequality).

Standard number sets in JEE

The Indian Class 11 NCERT fixes a notation you should treat as universal:

• ℕ — natural numbers {1, 2, 3, …}.
• W — whole numbers {0, 1, 2, 3, …}.
• ℤ — integers, both positive and negative including 0.
• ℚ — rationals, p/q form with q ≠ 0.
• ℝ — real numbers.

The inclusion chain is ℕ ⊂ W ⊂ ℤ ⊂ ℚ ⊂ ℝ, with each step strict.

Why it matters: every later topic — Relations, Functions, Probability, Limits — is phrased in the language of sets. Domain and range are sets. Events in probability are subsets of a sample space. Even definitions in topology and analysis trace back to subset language. Getting the basics into your muscle memory in Class 11 pays interest for years.

Real-world example: every WhatsApp group is a set. Members are elements; "WhatsApp groups inside a college" form a power-set-like collection of subsets of the larger student set. The phrase "you must be in the college to join the cricket group" is the subset relation in everyday language: cricket group ⊆ college.

Common misconception: students often think ∅ is "nothing" and so cannot be a subset of any set. In fact ∅ is a perfectly legal set (the set with no elements), and the statement "every element of ∅ is in A" is vacuously true — there are no elements in ∅ that could possibly violate the condition. Hence ∅ ⊆ A for every set A.

:::compare

:::keypoints

• A set is a well-defined collection of distinct objects; order and repetition don't change it.
• Empty set ∅ is unique, has no elements, and is a subset of every set.
• A ⊆ B allows equality; A ⊂ B forbids it (proper subset).
• A set with n elements has 2ⁿ subsets; this is |P(A)|.
• Number of proper subsets = 2ⁿ − 1; number of proper non-empty subsets = 2ⁿ − 2.
• The choice of universal set U depends on the problem context.
• Standard inclusion: ℕ ⊂ W ⊂ ℤ ⊂ ℚ ⊂ ℝ.
  :::

:::memory
"IN or OUT" rule: each element either goes IN or stays OUT of a subset → 2 choices per element → 2ⁿ subsets total. The same coin-flip image gives the power-set count.
:::

:::recap

• A set is well-defined and contains distinct elements; ∅ and A itself are always subsets of A.
• Equal sets share all elements; equivalent sets only share count.
• Subset means contained; proper subset means contained and not equal.
• The power set of an n-element set has 2ⁿ elements — the single most asked formula in this chapter.
  :::

Core operations:

• Union A ∪ B: in A OR B.
• Intersection A ∩ B: in A AND B.
• Difference A − B: in A but NOT B.
• Complement A' = U − A.
• Symmetric difference A Δ B = (A−B) ∪ (B−A) = (A∪B) − (A∩B).

Key identities:

• Commutative, Associative for ∪ and ∩.
• Distributive: A ∩ (B ∪ C) = (A∩B) ∪ (A∩C); A ∪ (B ∩ C) = (A∪B) ∩ (A∪C).
• DE MORGAN'S LAWS (very high-yield): (A ∪ B)' = A' ∩ B'; (A ∩ B)' = A' ∪ B'.
  Memory: 'Break the bar, flip the operation.'
• A − B = A ∩ B'.

Inclusion–Exclusion (counting):
|A ∪ B| = |A| + |B| − |A ∩ B|
|A ∪ B ∪ C| = |A|+|B|+|C| − |A∩B| − |B∩C| − |C∩A| + |A∩B∩C|.

Q: In a class of 100, 60 study Maths, 45 study Physics, 30 study both. How many study (a) at least one, (b) neither, (c) exactly one?

Let |M|=60, |P|=45, |M∩P|=30, |U|=100.

(a) At least one = |M ∪ P| = |M| + |P| − |M∩P| = 60 + 45 − 30 = 75.

(b) Neither = |U| − |M ∪ P| = 100 − 75 = 25.
Using De Morgan: |M' ∩ P'| = |(M ∪ P)'| = 100 − 75 = 25. ✓

(c) Exactly one = |M ∪ P| − |M ∩ P| = 75 − 30 = 45.
Or: (only M) + (only P) = (60−30) + (45−30) = 30 + 15 = 45. ✓

Exam tip: Draw a Venn diagram and fill the INTERSECTION first, then subtract to get 'only' regions. Most JEE set problems are inclusion–exclusion with 2 or 3 sets — memorise the 3-set formula's sign pattern (+ singles, − pairs, + triple).

Before functions, before graphs, before calculus — there is the humble idea of a relation. Every time you say "x is the father of y" or "x is greater than y", you are silently picking out some ordered pairs and ignoring the rest. JEE Main expects you to formalise that intuition and count the possibilities precisely.

Definition: A relation R from a set A to a set B is any subset of the Cartesian product A x B. If the ordered pair (a, b) belongs to R, we write aRb and read it as "a is related to b".

Definition: The Cartesian product A x B is the set of all ordered pairs (a, b) where a comes from A and b comes from B.

What "any subset" really means

The power of the definition lies in the word "any". A relation does not need a formula. It does not need a pattern. It can be the empty subset (no element of A is related to anything — the void relation), or the full A x B (every a relates to every b — the universal relation), or anything in between. The only requirement is that R consists of ordered pairs whose first entry is from A and second entry is from B.

So if A = {1, 2} and B = {p, q}, then A x B = {(1,p), (1,q), (2,p), (2,q)}. Each of these four pairs is either "in" R or "out" of R — like a four-bit switch. That gives 2^4 = 16 possible relations from A to B, from the empty relation up to all of A x B.

Why it matters: This single combinatorial fact — "each pair is in or out" — is the engine behind almost every counting question on relations in JEE. Internalise it once, and the formulas below stop feeling like memorisation.

Domain, range and codomain

Definition: The domain of a relation R is the set of all first coordinates of the pairs that appear in R.

Definition: The range of R is the set of all second coordinates that appear in R.

Definition: The codomain of R is the entire target set B, whether or not every element of B actually shows up.

The distinction between range and codomain trips up many aspirants. Range is what is achieved; codomain is what is available. Range is always a subset of codomain, but the two are equal only when R is "onto" in the function sense. For example, if A = {1, 2, 3}, B = {a, b, c, d}, and R = {(1,a), (2,a), (3,b)}, then the domain is {1, 2, 3}, the range is {a, b}, and the codomain is {a, b, c, d}.

The counting formulas you must own

For a relation from A to B with |A| = m and |B| = n:

• |A x B| = mn
• Total number of relations from A to B = 2^(mn)
• Total number of relations on a single set A with |A| = n = 2^(n^2) (because we count subsets of A x A, and |A x A| = n^2)

These follow from the same "each pair: in or out" idea applied to mn or n^2 pairs respectively.

:::compare

A worked counting example

Question: Let A = {1, 2, 3} and B = {x, y}. How many relations from A to B contain the pair (1, x) but do not contain (2, y)?

Solution:
Step 1: |A x B| = 3 x 2 = 6. The six pairs are (1,x), (1,y), (2,x), (2,y), (3,x), (3,y).
Step 2: We force (1,x) IN and (2,y) OUT — two pairs are fixed. The remaining 6 - 2 = 4 pairs are free.
Step 3: Each of the 4 free pairs is independently in or out, giving 2^4 = 16 relations.
Conclusion: There are 16 such relations.

The same "freeze some pairs, count the rest" trick handles almost every JEE counting question on relations.

Relations versus functions

Every function from A to B is a relation, but not every relation is a function. A relation is a function when each element of the domain A appears as the first coordinate of exactly one ordered pair. Many JEE questions sneak in this filter: "how many functions" instead of "how many relations". The counting answer changes from 2^(mn) to n^m, because each of the m elements of A must independently choose one of the n elements of B.

Real-world example: In an Indian census database, "x lives in district y" is a relation between people and districts. Each person lives in exactly one district, so this particular relation is also a function. But "x has visited district y" is only a relation — one person can map to many districts — so the function-counting formula no longer applies.

Common misconception: Students often confuse domain with the entire set A. If A = {1, 2, 3} but R = {(1, a), (1, b)}, the domain of R is just {1}, not {1, 2, 3}. The set A is sometimes called the co-domain of the first coordinate or simply "the set from which R is drawn", but the domain is the actual collection of first coordinates that appear in R.

Why JEE loves this topic

Relations are the foundation on which functions, equivalence classes, and partial orders rest. By forcing you to count subsets of A x A, JEE quietly trains you in combinatorics, binary thinking, and careful reading. A typical paper asks: "If A has 3 elements, how many reflexive relations are there on A?" The answer uses the same in-or-out idea: 3 reflexive pairs (1,1), (2,2), (3,3) are forced IN, and the other 9 - 3 = 6 pairs are free, giving 2^6 = 64.

:::keypoints

• A relation from A to B is any subset of A x B — no formula needed.
• Domain = first coordinates appearing in R; range = second coordinates appearing; codomain = entire B.
• |A x B| = mn; number of relations from A to B = 2^(mn).
• Number of relations on a single set A with |A| = n is 2^(n^2).
• Range is always a subset of codomain; they need not be equal.
• A function is a special relation where each element of A appears as a first coordinate exactly once.
• Use the "freeze the forced pairs, count free pairs as 2^k" technique for JEE counting questions.
  :::

:::memory
"R-D-R-C: Relation, Domain, Range, Codomain."

• Relation: any subset of A x B.
• Domain: first coordinates used.
• Range: second coordinates used.
• Codomain: the whole B (available, not necessarily used).
  Each ordered pair is a light switch — on (in R) or off (not in R) — so 2^(number of pairs) relations are possible.
  :::

:::recap

• A relation is just a subset of A x B; nothing fancier.
• Domain and range are what appear; codomain is what is available.
• Counting relations is counting subsets: 2^(mn) from A to B, 2^(n^2) on A itself.
• Functions are the well-behaved relations where every input has exactly one output.
  :::

Every NEET aspirant who has ever opened NCERT Class 11 Chapter 10 has seen the same diagram — two homologous chromosomes lying side by side, with a cross-shaped link between non-sister chromatids. That little X is called a chiasma, and the event that produces it — crossing over — is arguably the single most important reason you do not look like a clone of one of your parents. Master this stage and you have unlocked half of all NEET questions on meiosis.

Definition: Meiosis is a special type of cell division in diploid (2n) reproductive cells that produces four haploid (n) daughter cells, each genetically distinct from the parent, and is essential for sexual reproduction.
Definition: Crossing over is the reciprocal exchange of genetic material between non-sister chromatids of homologous chromosomes during Prophase I of meiosis.
Definition: A chiasma (plural: chiasmata) is the visible X-shaped point at which two non-sister chromatids remain attached after crossing over has taken place.

The five sub-stages of Prophase I — memorise the order

Prophase I of meiosis I is long and elaborate. NCERT splits it into five sub-stages, in this fixed order: Leptotene → Zygotene → Pachytene → Diplotene → Diakinesis. NEET examiners love to test "which event happens in which sub-stage", so the mapping matters.

• Leptotene: chromosomes start condensing and become visible as thin threads.
• Zygotene: synapsis begins — homologous chromosomes pair up, forming the synaptonemal complex. Each pair is now called a bivalent or tetrad (four chromatids in close apposition).
• Pachytene: crossing over occurs. This is the headline event.
• Diplotene: the synaptonemal complex dissolves, homologues begin to separate, and the chiasmata become visible as the X-shaped points of crossover.
• Diakinesis: chromosomes condense fully, chiasmata terminalise (slide to the ends), and the cell is ready for metaphase I.

The classic NEET trap: "When does crossing over occur?" Answer — Pachytene. "When are chiasmata first seen?" Answer — Diplotene. The events are sequential, not simultaneous. Mix them up and you lose a one-mark question every time.

The molecular machinery — what NCERT actually expects

Crossing over is not a random tangle; it is a precisely orchestrated, enzyme-catalysed exchange. The key players NCERT names are:

• Recombination nodules — protein assemblies that sit at specific points along the paired homologues and mark the sites where crossover will happen.
• Recombinase enzyme — the enzyme that cuts and rejoins the DNA strands so that a stretch from one chromatid is welded onto its non-sister partner, and vice versa.
• The synaptonemal complex — a ladder-like protein structure that holds the homologues in perfect register so that the cut-and-paste is reciprocal (each chromatid loses what the other gains, no DNA is lost overall).

Note carefully: crossing over happens between non-sister chromatids of homologous chromosomes, not between sister chromatids of the same chromosome (which would change nothing — sisters are identical) and not between non-homologous chromosomes (which would be a translocation, a chromosomal mutation).

The significance of meiosis — the two big functions

NCERT puts these in bold and so does NEET.

1. Maintenance of chromosome number across generations

Without meiosis, gametes would carry the same chromosome number as somatic cells (2n), and after fertilisation the zygote would be 4n. The next generation would be 8n, then 16n, and the genome would balloon out of control within a few generations. Meiosis halves the chromosome number in gametes (2n → n), and fertilisation restores it (n + n = 2n). This is why humans always stay at 46 chromosomes generation after generation.

2. Generation of genetic variation

This is the engine of evolution. Meiosis creates variation in two independent ways:

• Crossing over (Pachytene of Prophase I) — physically shuffles maternal and paternal alleles between homologous chromosomes. After crossing over, no chromatid is purely "from mom" or "from dad" any more; each is a mosaic.
• Independent assortment (Metaphase I) — each bivalent orients on the spindle independently, so any combination of maternal and paternal chromosomes can end up in a daughter cell. With 23 human bivalents, that alone gives 2^23 ≈ 8.4 million combinations, before crossing over multiplies the figure further.

The product: every sperm or egg you make is genetically unique. Siblings are not clones precisely because of this combinatorial machinery.

Where and when does meiosis occur?

In animals, meiosis happens in the gamete mother cells (spermatogonia and oogonia) during gametogenesis, producing sperm and ova. In plants, it happens in spore mother cells during sporogenesis, producing microspores (which become pollen) and megaspores (which become the embryo sac). In both cases the starting cell is diploid; the four products are haploid.

Why it matters: Roughly 4–5 NEET questions every year come from meiosis — many target the exact sub-stage of an event, the names of the molecular players, or the calculation of variation. Knowing crossing over = Pachytene and chiasmata = Diplotene alone has fetched students an easy mark in NEET 2020, 2022 and 2023.

Real-world example: The reason a brother and sister born to the same Indian parents can look strikingly different — one tall with curly hair, the other short with straight hair — is meiosis. Each parent's spermatogonia and oogonia underwent crossing over and independent assortment, producing gametes with countless allele combinations. The genetic lottery that meiosis runs is what makes every child (except identical twins) a one-of-a-kind individual.

Common misconception: "Crossing over occurs in mitosis too." No. Mitosis produces genetically identical daughter cells; there is no homologue pairing, no synaptonemal complex, no chiasmata. Crossing over is strictly a meiosis-I, Pachytene event.

A second misconception: "Crossing over creates new genes." It does not. It rearranges existing alleles into new combinations. The creation of brand-new alleles requires mutation, a different process altogether.

Question: In a human spermatogonium undergoing meiosis, name (a) the sub-stage where homologues pair, (b) the sub-stage where crossing over occurs, (c) the sub-stage where chiasmata are first clearly visible, and (d) the structure that holds the homologues together during synapsis.

Solution:
Step 1: Pairing of homologues = synapsis = Zygotene.
Step 2: Crossing over = Pachytene.
Step 3: Chiasmata first clearly visible = Diplotene.
Step 4: The structure of synapsis = synaptonemal complex.
Conclusion: Zygotene → Pachytene → Diplotene is the consecutive order, and the synaptonemal complex is the protein scaffold that makes the precision of crossing over possible.

:::compare

:::keypoints

• Meiosis = two divisions (I and II), one round of DNA replication, four haploid products from one diploid cell.
• Prophase I order: Leptotene → Zygotene → Pachytene → Diplotene → Diakinesis.
• Crossing over happens in Pachytene; chiasmata are first seen in Diplotene.
• Crossing over occurs between non-sister chromatids of homologous chromosomes only.
• Recombinase enzyme catalyses the reciprocal exchange at recombination nodules.
• Meiosis maintains chromosome number across generations.
• Meiosis generates variation via crossing over + independent assortment — the raw material of evolution.
• Site: gamete mother cells (animals) or spore mother cells (plants).
  :::

:::memory
Leptotene-Zygotene-Pachytene-Diplotene-Diakinesis = "Lazy Zebras Play Drums Daily."
Crossing over = Pachytene; Chiasmata first visible = Diplotene — remember "Pachy Cuts, Diplo Shows."
:::

:::recap

• Meiosis halves chromosome number and shuffles alleles — two functions, both essential.
• Crossing over is a Pachytene event; chiasmata are a Diplotene visual.
• Reciprocal exchange between non-sister chromatids of homologues, mediated by recombinase.
• Independent assortment in Metaphase I multiplies the variation crossing over begins.
  :::

Equivalence relations look abstract on the page but they are really just a clean way of saying "split a set into baskets where the things in the same basket are treated as the same". Once you see them through that lens — partitions, not pairs — JEE Main questions on this chapter become much faster.

Definition: A relation R on a set A is an equivalence relation if it is reflexive (aRa for every a), symmetric (aRb implies bRa), and transitive (aRb and bRc imply aRc).

Definition: An equivalence class of a, written [a], is the set of all elements of A that are related to a. Two elements are in the same class precisely when they are R-related.

Definition: A partition of a set A is a collection of non-empty subsets of A such that every element of A belongs to exactly one of these subsets (the subsets are mutually disjoint and their union is A).

The fundamental theorem you must internalise

Every equivalence relation on a set A produces a partition of A into equivalence classes, and conversely every partition of A defines an equivalence relation (declare aRb iff a and b are in the same block). This is one of the most important "moral" results in Class 12 Relations and Functions, and JEE Main repeatedly tests it in disguised forms — most often by asking you to count equivalence relations on a small set.

The reason the partition picture is so useful is that two equivalence classes are either identical or disjoint — they never partially overlap. So counting equivalence relations on a finite set becomes the same as counting partitions of that set, which is much easier than wrestling with ordered pairs.

The classic congruence-modulo-5 example

Define a relation R on the set of integers Z by: a R b if and only if (a − b) is divisible by 5.

Reflexive: for any integer a, a − a = 0, and 0 is divisible by 5. So aRa.

Symmetric: if 5 | (a − b), then 5 | −(a − b) = (b − a). So bRa.

Transitive: if 5 | (a − b) and 5 | (b − c), then 5 | [(a − b) + (b − c)] = (a − c). So aRc.

All three checks pass, so R is an equivalence relation. The equivalence classes are determined by the remainder when an integer is divided by 5:

[0] = {…, −10, −5, 0, 5, 10, …} — integers leaving remainder 0
[1] = {…, −9, −4, 1, 6, 11, …} — integers leaving remainder 1
[2] = {…, −8, −3, 2, 7, 12, …} — integers leaving remainder 2
[3] = {…, −7, −2, 3, 8, 13, …} — integers leaving remainder 3
[4] = {…, −6, −1, 4, 9, 14, …} — integers leaving remainder 4

Every integer belongs to exactly one of these five classes. This is the partition of Z corresponding to "congruence modulo 5", written Z/5Z or simply Z_5. The same idea is the bedrock of modular arithmetic used later in Number Theory and even in cryptography. JEE writes this relation as a R b iff a is equivalent to b (mod 5).

A counting question of the JEE Main flavour

Question: How many equivalence relations on A = {1, 2, 3} contain the pair (1, 2)?

Solution:
Step 1: Reframe the problem. An equivalence relation on A corresponds uniquely to a partition of A. Saying that the relation contains (1, 2) means that 1 and 2 sit in the same block of the partition.
Step 2: List all partitions of {1, 2, 3} where 1 and 2 are together.

• Partition P1: blocks {1, 2} and {3}. Two blocks total.
• Partition P2: block {1, 2, 3}. One block.
  Step 3: No other partition keeps 1 and 2 in the same block — once you separate them, the condition fails.
  Conclusion: There are exactly 2 equivalence relations on A that contain (1, 2).

Notice how the partition view collapsed a complicated counting question into a quick two-line check. If you had instead tried to list ordered pairs by hand and verify reflexivity, symmetry and transitivity for each candidate, you could have spent ten minutes — and still missed cases.

A look-up table you should know cold (Bell numbers)

The total number of equivalence relations on a set with n elements is the Bell number B_n. JEE Main expects you to know at least the first five:

:::compare

If a question asks "how many equivalence relations on {1, 2, 3, 4} contain (1, 2)?", note that you are partitioning {1, 2, 3, 4} with 1 and 2 forced into the same block — equivalent to partitioning a 3-element set {[12], 3, 4}, which gives B_3 = 5.

Why it matters

Equivalence relations sit at the centre of Class 12 Relations and Functions, and at least one JEE Main question almost every year is built on this theme — sometimes pure counting, sometimes verification of the three properties on a relation defined on R, Z, or N. They also reappear at the start of group theory, linear algebra (cosets, quotient spaces) and topology in BTech, so the time invested now pays off well beyond the entrance exam.

Real-world example: When the Indian Railways categorises every train in the country by gauge (broad gauge, metre gauge, narrow gauge), it is implicitly using an equivalence relation on the set of all railway lines. Two lines are "equivalent" if they have the same gauge. This relation is reflexive (every line has the same gauge as itself), symmetric (if line X has the same gauge as line Y, then Y has the same as X), and transitive. The equivalence classes are exactly the gauge categories. The fact that a broad-gauge coach can be moved onto any broad-gauge track — but not a metre-gauge track — is the partition at work in real engineering.

Common misconception: "If a relation is reflexive and symmetric, it must be transitive too." This is wrong, and JEE Main loves to test it. Consider A = {1, 2, 3} and R = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}. This is reflexive and symmetric — but (1, 2) and (2, 3) are both in R, yet (1, 3) is not, so transitivity fails. You must check all three properties independently. Reflexive + symmetric is not enough.

:::keypoints

• An equivalence relation must be reflexive, symmetric and transitive — check all three.
• Equivalence classes partition the underlying set into disjoint, non-empty blocks.
• Counting equivalence relations on a finite set = counting partitions = Bell numbers.
• B_1 = 1, B_2 = 2, B_3 = 5, B_4 = 15, B_5 = 52.
• "Same remainder mod n" is the canonical equivalence relation on Z, with n classes.
• To count relations that contain a fixed pair (a, b), force a and b into the same block, then count partitions of the smaller "merged" set.
• Two equivalence classes are either identical or completely disjoint — never partial.
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:::memory
"RST = Reflexive, Symmetric, Transitive. Drop any one letter and you no longer have an equivalence."
:::

:::recap

• Equivalence relations partition a set into equivalence classes.
• Counting equivalence relations = counting partitions (Bell numbers).
• Congruence mod n on Z is the prototype example with n classes.
• Always verify all three RST properties; reflexive + symmetric alone is not enough.
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Functions are the language in which all of higher mathematics is written, and three properties — one-one, onto, and invertible — are the gatekeepers that decide what a function can do and whether you can undo it. Mastering these three concepts unlocks inverse trigonometric functions, logarithms, and the entire theory of equations.

Definition: Function — a rule f: A → B that assigns each element of the domain A to exactly one element of the codomain B. The set of outputs actually produced is the range (or image), which is a subset of B (range ⊆ B).

One-one (injective) — no two inputs share an output

A function f is one-one (injective) if distinct inputs always produce distinct outputs.

Formal test: f(x₁) = f(x₂) ⟹ x₁ = x₂.

Graphical test (horizontal-line test): every horizontal line meets the graph at most once.

Example 1 — one-one:
f: ℝ → ℝ, f(x) = 2x.
Suppose f(x₁) = f(x₂) → 2x₁ = 2x₂ → x₁ = x₂. One-one confirmed.

Example 2 — not one-one:
f: ℝ → ℝ, f(x) = x².
f(2) = 4 and f(−2) = 4 — two different inputs map to the same output. Not one-one.

However, if we restrict the domain to [0, ∞), then for any x₁, x₂ ≥ 0, f(x₁) = f(x₂) → x₁² = x₂² → x₁ = x₂. On this restricted domain, x² is one-one.

Onto (surjective) — every codomain element is hit

A function f: A → B is onto (surjective) if every element of B is the image of at least one element of A — equivalently, the range equals the entire codomain.

Formal test: For every b ∈ B, there exists at least one a ∈ A such that f(a) = b.

Crucially, being onto depends on how the codomain is declared, not just the rule.

:::compare Onto vs not onto — the same rule, different codomains

This is why declaring the codomain carefully is mathematically important, not a technicality.

Bijection and the inverse

A function is a bijection (one-one and onto) if it is simultaneously injective and surjective. Only bijections have well-defined inverse functions.

Intuition: a bijection is a perfect pairing — every input matches one output, and every output matches one input. Like seats in an exam hall where exactly one student per seat and every seat is occupied — you can uniquely reverse the assignment.

Finding the inverse:

1. Write y = f(x).
2. Solve for x in terms of y.
3. Replace y with x (rename variables): the result is f⁻¹(x).
4. Verify: f⁻¹(f(x)) = x.

Question: Find the inverse of f(x) = 3x + 5, f: ℝ → ℝ.
Solution:
Step 1: Write y = 3x + 5.
Step 2: Solve for x: 3x = y − 5 → x = (y − 5)/3.
Step 3: Rename: f⁻¹(x) = (x − 5)/3.
Conclusion: f⁻¹(x) = (x − 5)/3. Check: f⁻¹(f(x)) = f⁻¹(3x + 5) = (3x + 5 − 5)/3 = x ✓.

Question: Is f: ℝ → ℝ, f(x) = x² invertible? If not, how to make it so?
Solution:
Step 1: f is not one-one on ℝ (f(2) = f(−2)), so it is not a bijection.
Step 2: Restrict domain to [0, ∞) — now f is one-one; also redefine codomain as [0, ∞) — now f is onto.
Step 3: On [0, ∞) → [0, ∞), f(x) = x² is a bijection; its inverse is f⁻¹(x) = √x.
Conclusion: You must restrict the domain (and match the codomain) to recover invertibility.

Composition of functions

Composition chains two functions: (g ∘ f)(x) = g(f(x)) — apply f first, then g to the result.

Composition is associative: h ∘ (g ∘ f) = (h ∘ g) ∘ f.
Composition is generally not commutative: g ∘ f ≠ f ∘ g in most cases.

Example: f(x) = x + 1, g(x) = x².
(g ∘ f)(x) = g(x + 1) = (x + 1)² = x² + 2x + 1.
(f ∘ g)(x) = f(x²) = x² + 1.
These are different, confirming non-commutativity.

Defining property of the inverse:
(f⁻¹ ∘ f)(x) = x for all x in the domain of f.
(f ∘ f⁻¹)(x) = x for all x in the domain of f⁻¹.

This means composing a bijection with its inverse always returns the input — the identity function.

Number of one-one and onto functions (finite sets)

If |A| = m and |B| = n:

• Number of one-one functions (requires m ≤ n): ⁿPₘ = n! / (n − m)!
• Number of onto functions: involves inclusion-exclusion: Σ (−1)^k × C(n,k) × (n−k)^m
• Number of bijections (requires m = n): n!

Why it matters: Invertibility is what lets us "undo" operations — solving equations, decrypting messages, reversing geometric transformations, and computing inverse trigonometric or logarithmic values all depend on finding or creating a bijection. Whenever you write y = sin⁻¹ x, you are using a deliberately restricted domain [−π/2, π/2] that turns the non-injective sine into a bijection.

Real-world example: Think of a function as the PIN-to-account mapping in a banking system. One-one means no two accounts share the same PIN (security!). Onto means every PIN in the system is assigned to some account (no wasted PINs). If both hold, the bank can uniquely reverse the query — given a PIN, find exactly one account — which is the inverse function. If two accounts share a PIN (not one-one), the reverse lookup fails.

Common misconception: "Every function has an inverse — just swap x and y." Swapping always produces a relation, but it is a function only when the original is a bijection. For f(x) = x² on ℝ, swapping gives x = y² → y = ±√x, which returns two values for x > 0 — violating the definition of a function. You must restrict the domain first.

:::keypoints Key points

• A function assigns exactly one output per input; range ⊆ codomain.
• One-one: f(x₁) = f(x₂) ⟹ x₁ = x₂ (horizontal-line test: at most one crossing).
• Onto: range = codomain — depends on the declared codomain, not just the formula.
• Bijection = one-one and onto; the only functions with well-defined inverses.
• To find f⁻¹: write y = f(x), solve for x, then relabel.
• Composition (g ∘ f) is associative but generally non-commutative.
• Restricting the domain (and codomain) can make a non-bijective function invertible.
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:::memory
"One-one = no doubling; Onto = no missing; Both = bijection = undo-able" — remember the logic chain: inject means you never double-map, surject means you never miss, and bijection means you can always reverse.
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:::recap

• One-one guards against repeated outputs; onto guards against missed targets in the codomain.
• Invertibility requires both properties — a bijection.
• f⁻¹ ∘ f = identity and f ∘ f⁻¹ = identity; this is the defining test of an inverse.
• When a function fails to be one-one (like x², sin x), restrict the domain to the largest interval where it is monotone, then match the codomain to the range.
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