Math notesAlgebraRelations
Independent topic page
Relations full marks come from independent property checks: reflexive, symmetric, transitive (and equivalence ⇒ partition) — never ‘two out of three feels like equivalence’.
01. Why Relations is independent: the exam asks property T/F and equivalence recognition far more than set-algebra. Bundling with P&C made students rush property checks.
02. Common process: write the relation rule → test R, S, T on separate lines → conclude type → if equivalence, think partitions / number of relations of that type when asked.
03. Quick formulas: counts of relations on a finite set, equivalence ↔ partition link, matrix/digraph views when helpful.
04. Full marks mindset: each property is a mini true/false. One counterexample kills a property — keep counterexample hunting automatic.
05. Advanced lane: equivalence claims are three independent T/F lines — reflexive, symmetric, transitive. Fail any one and the ‘equivalence’ option is dead.
06. Counting relations on a finite set: think matrix/grid of pairs; reflexive forces the diagonal, symmetric forces upper/lower ties — formulae become pictures.
Official · 2026
Official syllabus points for Relations only — not the full Algebra portion map. Portion hub: Algebra · Algebra analysis.
No topic-specific exclusions flagged — finish the unit points below for Relations.
Source: https://jeeadv.ac.in/documents/jee-advanced-2026-syllabus.pdf
Exam focusEquivalence relations and invertible / one-one / onto function checks.
Syllabus points
Do / check
2020–2026 paper weight
Topic match inside parent portion Algebra (37.2% of Math tags). Stats below are portion backdrop; matched subtopics show where Relations sits.
Portion-wide type/difficulty mix lives on the Algebra analysis — not repeated here so this page stays on Relations.
01. Dominant subtopics: Probability (18), Matrices (15), Sets/Relations (12), Complex (12).
Cross-topic stems
Advanced rarely asks a chapter in isolation. These are tagged multi-idea Math stems that touch Relations — practise the knowledge bridge, not only the single-chapter drill.
Step 1: solve x y' = y − x³ (homogeneous/linear in y/x) to get the family y = Cx − x³/2. Step 2: impose the given condition to fix C, then use y' = 0 to locate the local max. Step 3: compare monotonicity on (1,2) and count positive roots of f = g.
Multi-correct options test DE, calculus and graph reading together; one sign error in the DE kills several options.
After every DE, immediately ask: extrema? monotonicity? intersections with a given g?
Step 1: recover points P, Q, R from the slope/tangent conditions on the parabola. Step 2: find the circumcircle of triangle PQR (perpendicular bisectors). Step 3: read the required radius / relation that also ties to an ellipse option.
Geometry hops across three curve types in one SC item.
Practise ‘points on parabola → circle through them → conic check’ as a single drill.
Step 1: place the ellipse in standard form and introduce eccentric angle or slope form. Step 2: impose the geometric condition (tangent/chord/focus). Step 3: match the simplified relation to the correct option.
Parametric slips create attractive wrong options.
Keep both slope and parametric forms of ellipse tangents fluent.
Step 1: for the ellipse–parabola common-tangent MC, write both tangent conditions and compare. Step 2: later area NUM — find intersection limits first. Step 3: integrate difference of functions; keep trig-equation NUM technique nearby for neighbouring items.
Classic Advanced join: geometry conditions feeding calculus numerics in the same paper.
Any week you revise conics, end with one area-between-curves NUM.
Why methods matter
Relations is three T/F lines. These methods train you never to say ‘equivalence’ early — that single discipline protects multi-correct and matching marks.
Makes equivalence claims honest. This habit alone protects matching and multi-correct.
Do thisAlways write R / S / T as separate verdicts with one-line reason.
Stops false equivalence claims and protects MC partial logic.
R={(1,1),(2,2),(1,2)} on {1,2}: reflexive yes, symmetric no, transitive yes → not equivalence.
Faster than full proofs when the relation is oddly defined.
Do thisTo kill a property, hunt a minimal counterexample before proving.
Faster than full proofs when the relation is defined oddly.
‘aRb iff |a−b|=1’ on integers: not transitive because 1R2,2R3 but not 1R3.
Makes counting formulae obvious instead of memorised and swapped.
Do thisVisualise n×n yes/no grid; impose reflexive/symmetric constraints on cells.
Makes 2^(n²−n) and symmetric counts obvious.
Reflexive: n diagonal cells forced 1; remaining n²−n cells free → 2^(n²−n).
Turns abstract R into concrete classes you can count and describe.
Do thisIf equivalence, describe classes; counting equivalences ↔ counting partitions when asked that way.
Connects abstract R to concrete groupings.
‘Same remainder mod m’ on ℤ → m classes. On finite A, build classes explicitly.
Processed examples
Read the impact line first. Then notice which family/model was chosen. The solution is written the way your rough sheet should look in the hall.
On ℤ, aRb iff |a−b|=1. Is R transitive?
Transitive failures need a chain counterexample — one is enough.
Model chosenHunt (a,b),(b,c) in R but (a,c) not.
AnswerNot transitive
Takeaway: One counterexample kills the property line.
Number of reflexive relations on a set with 3 elements.
Grid free-cell model prevents wrong exponents under stress.
Model chosenDiagonal forced; 9−3=6 free cells → 2⁶.
Answer64
Takeaway: Free cells → 2^(free), not 2^(n²).
How many relations on a 4-element set are reflexive, symmetric, and contain exactly 8 ordered pairs?
Symmetry forces off-diagonal pairs in blocks of 2 — wrong exponents are the usual leak.
Model chosenForce diagonal; choose unordered off-diagonal blocks for the rest.
Answer15
Takeaway: Count symmetric off-diagonal entries as unordered pair-blocks.
Reusable models
Advanced stems cluster. Name the family in 10–15s, write the fast model, then run the same path. Do not invent a new plan each time.
Looks likeWhich properties hold? Is R equivalence?
Fast modelThree separate T/F lines with counterexample-to-kill — never ‘two out of three’.
Independent lines protect MC and stop false equivalence.
Do not call equivalence after only two properties.
Looks likeNumber of relations of a type on an n-set
Fast modeln×n yes/no grid → count free cells → 2^(free).
Grid model replaces memorising wrong exponents.
Do not use 2^(n²) for reflexive counts.
Looks likeClasses / ‘same type’ / number of equivalences via partitions
Fast modelIf equivalence, describe classes; counting ↔ partitions when asked that way.
Partition picture is faster than listing all pairs.
Do not list all ordered pairs when classes suffice.
Looks likeone-one / onto / invertible / number of functions of a type
Fast modelDecide injection/surjection separately; invertible ⇔ bijective on finite equal sets.
Separate T/F lines stop ‘bijective’ guesses from two properties.
Do not call invertible after only checking one-one.
Hall timing
Clarify domain. For ‘on A’, relation is among elements of A only. Circle whether ask is properties or counting.
Do not merge lines. Equivalence requires all three yes. Partial orders need reflexive+antisymmetric+transitive — only if syllabus ask appears.
Reflexive fail: find a with (a,a)∉R. Symmetric fail: (a,b) in but (b,a) out. Transitive fail: (a,b),(b,c) in but (a,c) out.
Never upgrade ‘symmetric+reflexive’ to equivalence. If counting, sanity-check against 2^n² total relations.
After the model is locked
2^(n²) for |A|=n
Each of n² pairs yes/no
2^(n²−n)
Diagonal pairs fixed in; off-diagonal free
2^(n(n+1)/2)
Diagonal free; off-diagonal pairs tied (a,b)↔(b,a)
Equivalence relation ⇔ partition into classes
Class counting / ‘related means same type’ stems
∀a∈A, (a,a)∈R
First property line
(a,b)∈R ⇒ (b,a)∈R
Second property line
(a,b)∈R and (b,c)∈R ⇒ (a,c)∈R
Third property line — needs careful counterexamples
Make it automatic