Q.Let the relation R be defined on the set A={1,2,3,4,5} by R={(a,b):∣a2−b2∣<8}. Then R is given by ______.
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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 |
Concept: Relation Properties — we first list all ordered pairs that satisfy the condition, then check which pairs are present.
Step 1: For each a,b∈A, compute ∣a2−b2∣ and check if it is <8.
Step 2:
- 12=1, 22=4, 32=9, 42=16, 52=25. Differences less than 8 occur only between numbers whose squares differ by at most 7.
Step 3: List all pairs:
- With a=1: ∣1−1∣=0, ∣1−4∣=3, ∣1−9∣=8 (not included) → (1,1),(1,2) and symmetrically (2,1).
- With a=2: ∣4−1∣=3, ∣4−4∣=0, ∣4−9∣=5, ∣4−16∣=12 → (2,1),(2,2),(2,3) and (3,2).
- With a=3: ∣9−4∣=5, ∣9−9∣=0, ∣9−16∣=7, ∣9−25∣=16 → (3,2),(3,3),(3,4) and (4,3). …
Listing every ordered pair of A={1,2,3,4,5} with ∣a2−b2∣<8 gives an 11-element relation.
We need all ordered pairs (a,b) with a,b∈{1,2,3,4,5} satisfying ∣a2−b2∣<8. The squares are 1,4,9,16,25.
- Every pair with a=b gives ∣a2−b2∣=0<8: (1,1),(2,2),(3,3),(4,4),(5,5).
- (1,2) and (2,1): ∣1−4∣=3<8 ✓
- (2,3) and (3,2): ∣4−9∣=5<8 ✓
- (3,4) and (4,3): ∣9−16∣=7<8 ✓ …
Method: Building a Relation in Roster Form From a Condition
Use this when a relation on a small finite set is defined by a condition such as ∣a2−b2∣<8 and you must list its pairs.
Steps
Step 1: Precompute the key quantity for each element.
Here, list the squares a2 for every a in the set so comparisons are quick.
Step 2: Test each ordered pair against the condition. …
Common Mistakes
Mistake 1: Including a pair that meets the boundary exactly.
Why it's wrong: the condition is ∣a2−b2∣<8 (strict), so a pair like (1,3) with ∣1−9∣=8 does not qualify. Correct approach: admit only differences strictly less than 8.
Mistake 2: Listing (a,b) but forgetting (b,a). …
- 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 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-2024-06094 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. ✗ …
- 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 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. …
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