Q.Check whether the relation R in R defined by R={(a,b):a≤b3} is reflexive, symmetric or transitive.
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 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 to test each property individually — reflexivity, symmetry, transitivity — using the definition a≤b3.
-
Reflexive: For any a∈R, we need a≤a3. This fails for a=21, since 21≤81 is false. So R is not reflexive.
-
Symmetric: If a≤b3, does b≤a3 follow? Take a=1, b=2: 1≤8 holds, but 2≤1 is false. So R is not symmetric. …
The relation R={(a,b):a≤b3} on R is neither reflexive, nor symmetric, nor transitive.
We test each property directly against the defining rule a≤b3. A single counterexample is enough to disprove a property.
1. Reflexivity
Reflexivity would require a≤a3 for every real number a.
Take a=21:
21≤(21)3=81?
This is false, since 21>81. So (21,21)∈/R.
R is not reflexive.
For 0<a<1 the cube is smaller than the number itself, so a≤a3 fails. Cubes only "grow" for a>1.
2. Symmetry
Symmetry would require: whenever a≤b3, also b≤a3.
Take a=1, b=2:
a≤b3:1≤23=8✓so (1,2)∈R.
But
b≤a3:2≤13=1×so (2,1)∈/R.
R is not symmetric.
3. Transitivity …
Method: Testing an Inequality Relation with a Cube
Use this for relations like aRb⟺a≤b3. The cube on one side is exactly what makes each property fail, so aim your test cases at that asymmetry.
Steps
Step 1: Reflexive — check a≤a3 across the domain
Try values in (0,1) where a cube shrinks the number: a=21 gives 21≤81, false, so it is not reflexive.
Step 2: Symmetric — swap a mismatched pair
Pick a small and b larger: (1,2) has 1≤8 true, but (2,1) needs 2≤1, false.
Step 3: Transitive — chain toward a broken bound …
Common Mistakes
Mistake 1: Testing reflexivity only with numbers >1
Why it's wrong: a≤a3 holds for a≥1 but fails for 0<a<1, where the cube is smaller. Correct approach: include fractional test values.
Mistake 2: Assuming a cube relation behaves symmetrically
Why it's wrong: cubing one side breaks the balance, so 1≤23 does not imply 2≤13. Correct approach: test the reverse pair explicitly. …
- KEAM 2026Set eng-2026-04194 marksMCQQ.Let X={a1,a2,a3,…,an} be a set consisting of n elements. The relation R={(a1,a1),(a2,a2),(a3,a3),…,(an,an)} on the set X is (A) reflexive, symmetric but not transitive (B) reflexive, transitive but not symmetric (C) transitive, symmetric but not reflexive (D) reflexive, symmetric and transitive (E) reflexive, not symmetric and not transitive
›Reveal solutionSolution
The diagonal (identity) relation on a set is reflexive, symmetric and transitive.
R={(ai,ai):i=1,…,n} contains exactly the diagonal pairs.
- Reflexive: every (ai,ai) is present. Yes.
- Symmetric: each pair is its own reverse, so (ai,ai)∈R⇒(ai,ai)∈R. Yes. …
- KEAM 2026Set eng-2026-04204 marksMCQQ.Let X={1,2,3,4,5,6,7}. Let R={(1,1),(1,2),(1,3),(5,6),(6,7),(7,5)} be a relation on X. Then the relation R will become reflexive if we include the pairs (A) (2,7),(3,3),(5,5),(6,6) and (7,7) (B) (2,2),(4,3),(5,5),(6,6) and (7,7) (C) (2,2),(3,3),(5,5),(6,6) and (7,6) (D) (2,2),(3,3),(5,5) and (7,7) (E) (2,2),(3,3),(4,4),(5,5),(6,6) and (7,7)
›Reveal solutionSolution
A relation on X is reflexive iff it contains (i,i) for every i∈X={1,…,7}. R already has (1,1), so exactly the six pairs (2,2),(3,3),(4,4),(5,5),(6,6),(7,7) must be added.
Reflexivity requires (i,i)∈R for all i∈{1,2,3,4,5,6,7}.
Currently R contains the diagonal pair (1,1) only. The missing diagonal pairs are:
(2,2),(3,3),(4,4),(5,5),(6,6),(7,7). …
- KEAM 2024Set eng-2024-06054 marksMCQQ.Let N be the set of all natural numbers. Let R be a relation defined on N given by aRb if and only if a+2b=11. Then the relation R is (A) reflexive but not symmetric (B) not reflexive but symmetric (C) reflexive and symmetric (D) neither reflexive nor symmetric (E) an equivalence relation
›Reveal solutionSolution
The relation a+2b=11 is neither reflexive nor symmetric.
Reflexive? aRa requires a+2a=3a=11, which has no natural-number solution. Not reflexive. …
- KEAM 2024Set eng-2024-06054 marksMCQQ.If R={(x,y):x,y∈Z, x2+3y2≤7} is a relation on the set of integers Z, then the range of the relation R is (A) {0,1} (B) {1,−1} (C) {0,−1} (D) {1} (E) {0,−1,1}
›Reveal solutionSolution
Bounding 3y2≤7 limits y to {−1,0,1}, all achievable.
Since x2≥0, a pair (x,y) can exist only if 3y2≤7, i.e. y2≤37≈2.33, so y∈{−1,0,1}. Each is realizable:
- y=0: e.g. x=0 (0≤7), …
- KEAM 2024Set eng-2024-06064 marksMCQQ.Let S denote the set of all subsets of integers containing more than two numbers. A relation R on S is defined by R={(A,B): the sets A and B have at least two numbers in common }. Then the relation R is (A) reflexive, symmetric and transitive (B) reflexive and symmetric but not transitive (C) not reflexive, not symmetric and not transitive (D) not reflexive but symmetric and transitive (E) reflexive but not symmetric and transitive
›Reveal solutionSolution
S = subsets of integers with more than two elements; ARB iff A,B have at least two common numbers.
Properties:
- Reflexive: every set in S has ≥3 elements, so A∩A has ≥2 elements → ARA. Yes.
- Symmetric: "at least two in common" is symmetric in A,B. Yes. …
- KEAM 2024Set eng-2024-06074 marksMCQQ.The relation R in the set of integers Z is given by R={(a,b):b=2a+3}. Then the relation R is (A) reflexive, symmetric and transitive (B) neither reflexive nor symmetric nor transitive (C) not reflexive but symmetric and transitive (D) reflexive and symmetric but not transitive (E) reflexive but not symmetric and transitive
›Reveal solutionSolution
Test each property against the rule b=2a+3. …
- KEAM 2024Set eng-2024-06094 marksMCQQ.Let a relation R on the set N of natural numbers be defined by (x,y)∈R if and only if x2−4xy+3y2=0 for all x,y∈N. Then the relation is (A) reflexive (B) symmetric (C) transitive (D) reflexive and symmetric but not transitive (E) an equivalence relation
›Reveal solutionSolution
The condition factors to x=y or x=3y; only reflexivity holds.
Factor the defining relation:
x2−4xy+3y2=(x−y)(x−3y)=0,
so (x,y)∈R iff x=y or x=3y.
Reflexive: For any x, x=x satisfies the condition, so (x,x)∈R — reflexive. ✓
Symmetric: Take (3,1): 3=3(1) so (3,1)∈R. But (1,3) needs 1=3 or 1=9, both false, so (1,3)∈/R — not symmetric. ✗ …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.