From membership to structure: set algebra, ordered pairs and equivalence classes
JEE Main · JEE Advanced75 min study guide10 indexed sections
BY THE END, YOU CAN
Use membership, inclusion and power sets without confusing their levels.
Prove set identities and count overlapping groups with explicit intersection conventions.
Classify relations through their quantified definitions and counterexamples.
Connect equivalence relations to partitions and count constrained finite relations.
BEFORE YOU START
Basic logical words: and, or, not, if and only if.
Finite counting and elementary divisibility.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
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No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. Membership and inclusion answer different questions
A set records which objects belong to it. Repetition and listing order do not create new elements: {2,2,5}={5,2}. The statement a∈A says that a is one element of A. The statement B⊆A says that every element of B belongs to A. A set can itself be an element of another set, so these two statements must not be interchanged.
The empty set ∅ has no elements. It is a subset of every set because there is no element that can violate inclusion. But ∅ is an element only of sets that actually contain it. The set {∅} is not empty: it has one element, namely ∅. Nested braces matter just as parentheses matter in an algebraic expression.
Set equality means identical membership: A=B exactly when A⊆B and B⊆A. The notation A⊊B means proper inclusion, so at least one element of B is missing from A. Some texts use ⊂ ambiguously; using ⊆ and ⊊ avoids that ambiguity.
In set-builder notation, write the ambient set and the condition clearly. The descriptions {x∈ℤ:x²<10} and {x∈ℝ:x²<10} define very different sets. The first is a finite list of integers from −3 to 3, while the second is an open real interval. A condition without its input universe can leave a problem under-specified.
ORIGINAL WORKED EXAMPLE
Read the braces before counting
Let A={1,{1},∅}. Determine |A| and the truth of 1∈A, {1}∈A, {1}⊆A and {∅}⊆A.
The three displayed elements are the number 1, the singleton set {1}, and the empty set. They are distinct, so |A|=3.
The first two membership claims are true because 1 and {1} appear separately as elements.
The only element of {1} is 1, which belongs to A, so {1}⊆A is true.
The only element of {∅} is ∅, which also belongs to A. Thus {∅}⊆A is true, even though {∅} is not itself one of the listed elements of A.
Answer |A|=3; all four stated claims are true.
02 / UNDERSTAND & APPLY
2. Set operations are membership rules
The union A∪B contains elements in A or B, including those in both. The intersection A∩B contains elements in both. The difference A∖B contains elements in A but not B, so its order matters. The symmetric difference A△B contains elements in exactly one of the two sets and equals (A∖B)∪(B∖A).
A complement requires a fixed universal set U: Aᶜ=U∖A. Changing U changes the complement. When multiple complements appear in one identity, keep the same universe throughout. In particular, A∩Aᶜ=∅ and A∪Aᶜ=U depend on A being a subset of that universe.
Translate membership into logical conditions to derive identities. For example, x∈A∩(B∪C) means x∈A and at least one of x∈B or x∈C. This is equivalent to belonging to A∩B or A∩C, proving distributivity. A diagram can suggest the answer, but the membership argument proves it for arbitrary sets.
De Morgan’s laws say that the complement of a union is the intersection of the complements, and the complement of an intersection is the union of the complements. In words, being outside every group differs from being outside at least one group. This distinction frequently controls whether a problem is asking for neither, not both, or exactly one.
(A∪B)ᶜ=Aᶜ∩Bᶜ; (A∩B)ᶜ=Aᶜ∪Bᶜ
All complements use the same universal set.
A△B=(A∪B)∖(A∩B)
Membership in exactly one set excludes their overlap.
ORIGINAL WORKED EXAMPLE
Simplify by following one arbitrary element
Prove A∖(B∖C)=(A∖B)∪(A∩C).
An element x belongs to the left side exactly when x∈A and it is not true that x∈B and x∉C.
Negating that conjunction gives x∉B or x∈C. Thus x∈A and at least one of those two conditions holds.
Distribute the membership condition: either x∈A and x∉B, or x∈A and x∈C.
These alternatives describe (A∖B)∪(A∩C). The equivalence works in both directions, proving equality rather than only one inclusion.
Answer A∖(B∖C)=(A∖B)∪(A∩C).
03 / UNDERSTAND & APPLY
3. Count disjoint regions, not overlapping labels
For finite sets, |A∪B|=|A|+|B|−|A∩B| because the sum on the right initially counts each shared element twice. This formula counts the union, not the number in exactly one set. Exactly one has size |A|+|B|−2|A∩B|, while neither has size |U|−|A∪B|.
For three sets, add their sizes, subtract the three pairwise intersections and add the triple intersection once. An element in all three is counted 3−3+1=1 time. The pairwise intersections in this formula include elements in the triple intersection unless the problem explicitly says “only.”
If t elements are in all three, then the number in exactly two is |A∩B|+|B∩C|+|C∩A|−3t. Each triple member appears in all three pairwise totals and must be removed from each. The number in exactly one is |A|+|B|+|C|−2(|A∩B|+|B∩C|+|C∩A|)+3t.
Always check that every resulting disjoint region has a nonnegative integer size and that the regions sum to the universe size. Impossible survey data can satisfy one headline formula while forcing a negative “only A” region. These feasibility checks are part of the solution rather than optional arithmetic decoration.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Counts label mutually disjoint regions; circle areas are schematic, not proportional.
ORIGINAL WORKED EXAMPLE
A three-group survey with every region checked
In a group of 80, sets A,B,C have sizes 35,30,28. Their pairwise intersections have sizes 12,10,9 for AB,AC,BC, and the triple has size 4. Find exactly one, exactly two, and none.
The pair-only regions are AB only=12−4=8, AC only=10−4=6 and BC only=9−4=5. Exactly two totals 19.
The single-only regions are A only=35−12−10+4=17, B only=30−12−9+4=13 and C only=28−10−9+4=13. Exactly one totals 43.
The union is 35+30+28−12−10−9+4=66, so none is 80−66=14.
Check the disjoint regions: 43+19+4+14=80, all nonnegative integers. The pairwise labels were inclusive of the triple.
The power set P(A) is the set of all subsets of A. If |A|=n, each element independently has two choices, included or excluded, giving |P(A)|=2ⁿ. The empty subset and A itself are included. There are 2ⁿ−1 proper subsets and, for n≥1, 2ⁿ−2 nonempty proper subsets.
If fixed sets L⊆H are given, the subsets X satisfying L⊆X⊆H number 2^{|H∖L|} for finite H. Elements of L are compulsory and elements outside H are forbidden; only H∖L supplies independent decisions. If L is not a subset of H, there are no such X.
The Cartesian product A×B consists of ordered pairs (a,b) with a∈A and b∈B. The coordinates have roles, so (a,b) generally differs from (b,a). For finite sets, |A×B|=|A||B|. If either factor is empty, the product is empty; otherwise exchanging factors usually changes the set of pairs.
Power sets and products operate at different levels. P(A×B) contains sets of ordered pairs, while P(A)×P(B) contains ordered pairs of subsets. Their sizes are 2^{|A||B|} and 2^{|A|+|B|}, respectively, and they are not interchangeable even when the numerical sizes accidentally coincide.
ORIGINAL WORKED EXAMPLE
Separate forced and optional members
Let H={1,2,3,4,5,6}, L={1,4}. How many X satisfy L⊆X⊆H, and how many additionally have |X|=4?
The elements 1 and 4 must be included. The other four elements may independently be included or excluded.
There are therefore 2⁴=16 admissible intermediate subsets in total.
To obtain size 4, choose exactly two of the four optional elements, giving C(4,2)=6.
The endpoints X=L and X=H are included in the first count because both inequalities are non-strict. Neither endpoint has size 4.
Answer 16 in total; 6 have size 4.
05 / UNDERSTAND & APPLY
5. A relation selects pairs from a Cartesian product
A relation R from A to B is any subset of A×B. Writing aRb means (a,b)∈R. A relation on A is a subset of A×A. It need not give each input an output, and it may relate one input to many outputs; a function is a more restricted kind of relation.
The domain of a relation consists of first coordinates that occur, and its range consists of second coordinates that occur. These can be smaller than the sets A and B used to specify the relation. The inverse relation R⁻¹ reverses every pair, so it is a relation from B to A whether or not R represents a function.
For finite sets with |A|=m and |B|=n, there are mn available ordered pairs. Choosing any subset gives 2^{mn} relations from A to B. If A is nonempty and B has n elements, a function A→B instead requires exactly one choice in each input row, giving nᵐ functions when A is finite.
A relation on a finite set can be represented by a directed graph: draw an arrow a→b for each pair (a,b), including a loop when a=b. Alternatively use a zero-one matrix whose row and column labels are the elements. Keep the labels and order visible, since a matrix without them can obscure which pair an entry represents.
06 / UNDERSTAND & APPLY
6. Test the quantifiers, and use decisive counterexamples
Reflexive means aRa for every a in the underlying set. Irreflexive means no aRa occurs. A relation can be neither: having a loop at one element but not another fails both conditions. Reflexivity is relative to the specified set, not merely the elements that happen to occur in the relation.
Symmetric means aRb implies bRa for every pair. Antisymmetric means that if both aRb and bRa hold, then a=b. Antisymmetric does not mean “not symmetric.” The identity relation is both symmetric and antisymmetric; a relation with a one-way arrow and a different two-way arrow can be neither.
Transitive means aRb and bRc together imply aRc for every triple a,b,c. It is not enough to check distinct elements: a two-way connection aRb and bRa forces loops aRa and bRb in a transitive relation. Conversely, if no composable two-arrow path exists, the implication is true without any conclusion needing to be added.
To disprove a universal property, give one explicit witness: a missing loop for reflexivity, a missing reverse pair for symmetry, a two-way pair of distinct elements for antisymmetry, or a two-step path without its direct pair for transitivity. To prove a property, use the definition for arbitrary elements or inspect all finite cases systematically.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Exactly seven relation pairs are drawn. No end-to-end edge is present.
ORIGINAL WORKED EXAMPLE
Symmetric does not automatically mean transitive
On A={1,2,3}, let R={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}. Classify its four main properties.
Every diagonal pair occurs, so R is reflexive.
Each off-diagonal pair appears with its reverse: 1↔2 and 2↔3. Thus R is symmetric.
The distinct elements 1 and 2 are related both ways, so R is not antisymmetric.
The pairs (1,2) and (2,3) occur but (1,3) does not. This is a transitivity counterexample. No further search is needed to disprove it.
Answer Reflexive and symmetric; neither antisymmetric nor transitive.
07 / UNDERSTAND & APPLY
7. Equivalence relations organize a set into blocks
An equivalence relation is reflexive, symmetric and transitive. For a∈A, its equivalence class [a] is the set of elements related to a. Reflexivity puts a in its own class, so every class is nonempty and every element belongs to some class.
If two equivalence classes overlap, they are equal. To see the mechanism, take an element c related to both representatives a and b. Symmetry and transitivity show a is related to b, and every element related to one representative is then related to the other. Thus distinct classes are disjoint.
The distinct classes therefore form a partition: nonempty disjoint blocks whose union is A. Conversely, any partition defines an equivalence relation by declaring two elements related when they lie in the same block. Being in the same block is reflexive, symmetric and transitive. This provides a practical way to construct or count equivalence relations.
Equality of a computed label often generates an equivalence relation. For any function h on A, define aRb by h(a)=h(b). Equality immediately supplies the three properties. Examples include equal remainders, equal absolute values or equal squared values. The class description still depends on the original underlying set.
ORIGINAL WORKED EXAMPLE
A finite remainder partition and pair count
On A={0,1,2,3,4,5,6,7}, define aRb if 3 divides a−b. Find the classes and |R|.
Reflexivity follows because a−a=0 is divisible by 3. Symmetry follows by negating a−b, and transitivity by adding two divisible differences.
The classes are {0,3,6}, {1,4,7} and {2,5}, corresponding to the three remainders.
Within a block of size k, all k² ordered pairs occur, including diagonal pairs. The classes have sizes 3,3,2.
Therefore |R|=3²+3²+2²=22. Products between different blocks are excluded.
Answer Three classes: {0,3,6}, {1,4,7}, {2,5}; |R|=22.
08 / UNDERSTAND & APPLY
8. Order relations and independent pair choices
A partial order is reflexive, antisymmetric and transitive. The usual ≤ on real numbers is one example; divisibility on positive integers is another. A partial order need not compare every pair. Under divisibility, neither 2 divides 3 nor 3 divides 2, although both divide 6. A total order adds comparability for every pair.
On a labelled n-element set, a reflexive relation must contain all n diagonal pairs and may freely choose the other n²−n pairs. Thus there are 2^{n²−n} reflexive relations. This count imposes no symmetry or transitivity unless explicitly added.
For a symmetric relation, each diagonal pair is independently present or absent, and each unordered pair of distinct elements contributes one all-or-neither choice for its two directions. Hence there are 2^{n+C(n,2)} symmetric relations, and 2^{C(n,2)} relations that are both reflexive and symmetric.
For antisymmetry, each unordered distinct pair has three choices: neither direction, just the first, or just the second. Both directions are forbidden, but diagonal pairs remain optional. The count is 2ⁿ3^{C(n,2)}. Transitivity couples different pair choices, so these independent-choice formulas cannot simply be reused to count partial orders or equivalence relations.
ORIGINAL WORKED EXAMPLE
Count properties without accidentally imposing another
On a three-element labelled set, count reflexive relations, symmetric relations, reflexive symmetric relations and equivalence relations.
Reflexive relations: fix three diagonal pairs and choose the other six independently, giving 2⁶=64.
Symmetric relations: three diagonal choices and three unordered off-diagonal pair choices give 2⁶=64. The equal answer is a coincidence of these exponents.
Reflexive symmetric relations: diagonals are compulsory, leaving three choices, giving 2³=8.
Equivalence relations correspond to partitions. There is one partition into three singletons, three partitions into a pair and a singleton, and one partition with a single three-element block. Total 5.
Answer 64; 64; 8; 5, respectively.
09 / UNDERSTAND & APPLY
9. Original practice with full reasoning
For a claimed identity, justify membership in both directions. For a failed relation property, identify the exact offending element, pair or triple.
ORIGINAL PRACTICE · 1
For U={1,2,3,4,5}, A={1,3,5}, B={3,4}, find A△B and (A∪B)ᶜ.
Reveal answer & reasoning
A△B={1,4,5}; complement={2}.
The shared element 3 is removed from the symmetric difference. The union is {1,3,4,5}, leaving only 2 in the fixed universe.
ORIGINAL PRACTICE · 2
If |A|=18, |B|=15 and |A∩B|=7, how many elements belong to exactly one set?
Reveal answer & reasoning
19.
Exactly one is (18−7)+(15−7)=11+8=19. The union would be 26, which includes the overlap.
ORIGINAL PRACTICE · 3
A set has four elements. How many nonempty proper subsets does it have?
Reveal answer & reasoning
14.
There are 2⁴=16 subsets. Remove the empty set and the whole set, which are distinct because the original set is nonempty.
ORIGINAL PRACTICE · 4
How many relations are there from a two-element set to a three-element set?
Reveal answer & reasoning
64.
The Cartesian product has six ordered pairs. Any subset is a relation, giving 2⁶=64.
ORIGINAL PRACTICE · 5
On A={1,2}, classify the empty relation as reflexive, symmetric, antisymmetric and transitive.
Reveal answer & reasoning
Not reflexive; symmetric, antisymmetric and transitive.
The loops are missing, so reflexivity fails. The other three are implications whose pair premises never hold, so no counterexample exists.
ORIGINAL PRACTICE · 6
On ℝ, is aRb defined by |a−b|<1 an equivalence relation?
Reveal answer & reasoning
No: it is reflexive and symmetric but not transitive.
The inputs 0,0.6,1.2 give two successive differences below 1, while the endpoint difference is 1.2. Thus 0R0.6 and 0.6R1.2 but not 0R1.2.
ORIGINAL PRACTICE · 7
On ℝ, define aRb by a²=b². Describe [2] and [0].
Reveal answer & reasoning
[2]={−2,2}; [0]={0}.
Equal-square solutions are b=±a, with the two possibilities coinciding when a=0. Equality of squared labels also proves this is an equivalence relation.
ORIGINAL PRACTICE · 8
How many antisymmetric relations exist on a two-element labelled set?
Reveal answer & reasoning
12.
The two diagonal pairs have 2² choices. The single unordered distinct pair has three permitted directional choices, giving 4·3=12.
ORIGINAL PRACTICE · 9
A partition has block sizes 2,3,4. How many ordered pairs belong to its equivalence relation?
Reveal answer & reasoning
29.
Each block contributes every internal ordered pair, so the total is 2²+3²+4²=4+9+16=29.
ORIGINAL PRACTICE · 10
Is divisibility a total order on {1,2,3,6}?
Reveal answer & reasoning
It is a partial order but not a total order.
Positive-integer divisibility is reflexive, antisymmetric and transitive. However, 2 and 3 are incomparable, so total comparability fails.
10 / UNDERSTAND & APPLY
10. Final checks: universe, quantifier, counterexample
Before manipulating sets, identify the universe and distinguish elements from subsets. Before applying inclusion-exclusion, decide whether pairwise counts include the triple overlap. Before counting subsets or relations, identify which choices are forced, forbidden or genuinely independent.
For relations, write the underlying set beside the pair condition. Symmetry reverses arrows, antisymmetry forbids two-way arrows between distinct points, and transitivity completes two-step paths. These are independent requirements; neither familiar names nor a nearly complete diagram substitutes for the definitions.
Equivalence relations group elements into disjoint classes, whereas partial orders compare elements consistently. Both use reflexivity and transitivity, but their middle conditions differ: symmetry for equivalence, antisymmetry for partial order. The identity relation satisfies both and reminds us that these families can overlap.
Check an answer by another representation where possible: list small subsets, use a zero-one relation matrix, enumerate partitions, or verify disjoint region totals. These checks often reveal a missing loop, a reversed pair or an overlap counted twice before it becomes a larger algebraic error.
KEEP THE BIG IDEAS
Quick revision
Membership is not inclusion, and the empty set is not its singleton.
Set identities can be proved by following an arbitrary element.
Inclusive pairwise intersections contain the triple intersection.
Relation properties are quantified conditions, best disproved by explicit witnesses.
Equivalence classes partition a set; partial orders need not compare every pair.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.