Q.If a relation R on the set {1,2,3} be defined by R={(1,2)}, then R is
(A) reflexive
(B) transitive
(C) symmetric
(D) none of these
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Relation Properties
Properties of a Relation
A relation R on a set A pairs elements of A with one another. Some relations behave in regular, predictable ways, and we name these behaviours properties. Three matter most for CBSE Class 12 — reflexive, symmetric, transitive (together they build an equivalence relation); a fourth, antisymmetric, is worth knowing for order relations.
Reflexive — everything relates to itself
R is reflexive if aRa for every a∈A. "Has the same age as" is reflexive; "is taller than" is not. If even one element misses its self-pair, reflexivity fails: on {1,2,3}, {(1,1),(2,2)} is not reflexive because (3,3) is absent.
Symmetric — the relation runs both ways
R is symmetric if aRb⟹bRa. "Is married to" is symmetric; "is taller than" is not. Symmetry does not demand that every pair be related — only that any pair which appears also appears reversed. So {(1,2),(2,1),(3,3)} is symmetric, but {(1,2),(2,1),(1,3)} is not, since (3,1) is missing.
Transitive — relations chain
R is transitive if aRb and bRc together force aRc. "Is an ancestor of" is transitive; "is a friend of" is not. A single broken chain breaks the property: {(1,2),(2,3)} is not transitive because (1,3) is missing.
Antisymmetric — two-way ties force equality
R is antisymmetric if aRb and bRa together force a=b. The order relation ≤ is antisymmetric: a≤b and b≤a give a=b. It does not ban self-pairs like (1,1); it only rules out distinct elements related both ways.
Test the properties in order of ease — reflexivity first, then symmetry, transitivity. A single counterexample is enough to disprove any of them.
| Property | Condition |
|---|---|
| Reflexive | ∀a, aRa |
The key idea is that a relation can satisfy some properties without satisfying others — here, R is reflexive only if every element relates to itself, symmetric only if every pair has its reverse, and transitive only if chains close.
- Reflexive? No — (1,1),(2,2),(3,3) are all missing.
- Symmetric? No — (1,2) is present but (2,1) is absent. …
On {1,2,3} with R={(1,2)}: not reflexive, not symmetric, but vacuously transitive — option (B).
Check each property for R={(1,2)} on the set {1,2,3}.
Reflexive? Needs (1,1),(2,2),(3,3)∈R. None are present, so R is not reflexive.
Symmetric? Since (1,2)∈R, symmetry would require (2,1)∈R. It is absent, so R is not symmetric. …
Method: Testing a small listed relation for R, S, T
Use this when the relation is given as an explicit short list of pairs on a small set and you must decide reflexive / symmetric / transitive.
Steps
Step 1: Reflexive — check every diagonal pair is present.
For a set {1,2,3} the relation must contain (1,1),(2,2),(3,3). If even one is missing, it is not reflexive.
Step 2: Symmetric — for each listed (a,b) look for (b,a).
Scan the non-diagonal pairs; if any (a,b) has its reverse (b,a) absent, symmetry fails immediately.
Step 3: Transitive — chase every existing chain, and apply vacuous truth. …
Common Mistakes
Mistake 1: Declaring R={(1,2)} 'not transitive' because it has just one pair.
Why it's wrong: transitivity is violated only when a chain (a,b),(b,c) exists without (a,c); here nothing starts with 2, so no chain exists and transitivity holds vacuously. Correct approach: check whether any chain can even form before deciding.
Mistake 2: Thinking a single pair makes the relation symmetric or reflexive. …
- GUJCET 2026Set x1 markMCQQ.Let R be the relation in the set N given by R={(a,b):a=b−2, b<6}, then ______ (A) (6,8)∈R (B) (8,7)∈R (C) (8,3)∈R (D) (2,4)∈R
›Reveal solutionSolution
Test each pair against both conditions a=b−2 and b<6.
- (6,8): b=8<6. ✗
- (8,7): b=7<6. ✗
- (8,3): a=b−2=1=8. ✗ …
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.A relation R on the set N is defined by R={(a,b)∣a=b−2, b>6}.(a) (2,4)∈R(b) (6,8)∈R(c) (3,8)∈R(d) (8,7)∈R
›Reveal solutionSolution
Test each pair against both conditions: a=b−2 and b>6.
- (2,4): 2=4−2 holds, but 4>6 fails.
- (6,8): 6=8−2 holds, and 8>6 holds -- both conditions satisfied. …
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.The relation R={(a,b),(b,a)} is defined on the set {a,b,c}, then R is ______.(a) Reflexive, but not symmetric and transitive(b) Symmetric, but not reflexive and transitive(c) Transitive, but not reflexive and symmetric(d) An equivalence relation
›Reveal solutionSolution
Check each property of R={(a,b),(b,a)} on {a,b,c} directly against its definition.
Reflexive? Needs (a,a),(b,b),(c,c)∈R -- none are present, so R is NOT reflexive.
Symmetric? (a,b)∈R⇒(b,a)∈R -- both pairs are present, so R IS symmetric.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.Let R be the relation in the set {1,2,3} given by R={(1,1),(2,2),(3,3)}. Choose the correct answer.(a) R is an equivalence relation(b) R is reflexive and symmetric but not transitive(c) R is reflexive and transitive but not symmetric(d) R is symmetric and transitive but not reflexive
›Reveal solutionSolution
Check the three properties on R={(1,1),(2,2),(3,3)}.
Reflexive: (1,1),(2,2),(3,3) are all present ✓.
Symmetric: every pair is of the form (a,a), so (a,a)⇒(a,a) ✓. …
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.Consider a binary operation ∗ on N defined as a∗b=∣a−b∣. Choose the correct answer.(a) ∗ is both associative and commutative(b) ∗ is commutative but not associative(c) ∗ is associative but not commutative(d) ∗ is neither commutative nor associative
›Reveal solutionSolution
Test commutativity and associativity of a∗b=∣a−b∣.
Commutative: a∗b=∣a−b∣=∣b−a∣=b∗a ✓.
Associative? (a∗b)∗c=∣a−b∣−c, while a∗(b∗c)=a−∣b−c∣. …
- GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQQ.Let R be the relation on the set N given by R={(a,b):a=b−2, b>6}. Choose the correct answer.(a) (2,4)∈R(b) (3,8)∈R(c) (6,8)∈R(d) (8,7)∈R
›Reveal solutionSolution
Check each ordered pair against both conditions of the relation: a=b−2 AND b>6.
R={(a,b):a=b−2, b>6} on N.
- (2,4): a=b−2⇒2=4−2 ✓, but b>6⇒4>6 ✗. Rejected.
- (3,8): a=b−2⇒3=8−2=6? ✗. Rejected. …
- GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQQ.The relation S={(1,1),(2,2),(3,3),(4,4),(5,5)} on {1,2,3,4,5} is ______.(a) reflexive only(b) symmetric only(c) transistive only(d) an equivalence relation
›Reveal solutionSolution
S={(a,a):a∈{1,...,5}} is the identity relation, and the identity relation is always an equivalence relation.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.If a∗b=a3+b3 on z, then (1∗2)∗0= ___.(a) 0(b) 729(c) 81(d) 27
›Reveal solutionSolution
Evaluate the operation from the inside out.
Given a∗b=a3+b3 on Z.
First 1∗2=13+23=1+8=9.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.If the relation is defined on R−{0} by (x,y)∈S⇔xy>0, then S is ___.(a) an equivalence relation(b) symmetric only(c) reflexive only(d) transitive only
›Reveal solutionSolution
"Same sign" is reflexive, symmetric and transitive on R−{0}.
- Reflexive: x⋅x=x2>0 for xe0, so (x,x)∈S.
- Symmetric: xy>0⇒yx>0. …
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