Q.Let R be the relation in the set {1, 2, 3, 4} given by R = {(1,2), (2,2), (1,1), (4,4), (1,3), (3,3), (3,2)}. Choose the correct answer in the given options.
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Reflexive, Transitive, Not Symmetric — A First Look
What kind of relation is reflexive and transitive but not symmetric? "Same grade as" won't do — it's symmetric. We need a relation that only goes one way. Let's build it.
The Intuition: A One-Way Street
The classic example is "divides" on the positive integers:
- Reflexive: every number divides itself. 5∣5. ✓
- Transitive: if a∣b and b∣c, then a∣c. E.g. 2∣4 and 4∣12 gives 2∣12. ✓
- Not symmetric: 2∣4 is true, but 4∣2 is false. The relation goes only one way. ✓
The key insight: the relation can go from smaller to larger (or equal), but not back.
The Precise Statement
Let R be a relation on a set S. Then:
Reflexive: ∀a∈S,aRa
Transitive: ∀a,b,c∈S,(aRb∧bRc)⟹aRc
Not symmetric: ∃a,b∈S such that aRb but bRa
"Reflexive transitive not symmetric" is just a checklist of three properties — not a standard name like "equivalence relation". A relation with these (plus antisymmetry) is a partial order.
Why This Matters
Exams often ask: "Is this relation reflexive? Symmetric? Transitive?" Test each property independently — a relation can be reflexive and transitive but fail symmetry, and that's perfectly fine. For example, on the reals define xRy if x≤y: reflexive yes, transitive yes, symmetric no (3≤5 but 5≤3).
A common mistake: assuming that reflexive + transitive forces symmetry. False — both "divides" and "≤" disprove it. Always test each property separately.
A Quick Table for Clarity
| Property | Meaning | Example: "divides" on N |
|---|---|---|
| Reflexive | Every element relates to itself | 3∣3 ✓ |
Checking a relation's properties means testing each pair systematically: reflexive requires every (a,a) to be present, symmetric requires every pair's reverse to be present, and transitive requires every implied co …
Check reflexivity (every (a,a) present), symmetry (every pair's reverse present) and transitivity (chain closure) directly against the listed pairs.
R = {(1,2), (2,2), (1,1), (4,4), (1,3), (3,3), (3,2)} on {1, 2, 3, 4}.
Reflexive: We need (1,1), (2,2), (3,3), (4,4) ∈ R. All four are present, so R is reflexive.
Symmetric: (1,2) ∈ R but (2,1) ∉ R, so R is NOT symmetric.
…
- CBSE 2026Set 65/2/11 markMCQQ.A relation R on set A={1,2,3} defined as R={(1,1),(2,2),(1,2)} is (A) Reflexive only (B) Reflexive and Transitive (C) Symmetric and Transitive (D) Transitive only
›Reveal solutionSolution
On A={1,2,3}, R is not reflexive (missing (3,3)), not symmetric (has (1,2) but not (2,1)), and is transitive. So R is transitive only — option (D).
Check each property of R={(1,1),(2,2),(1,2)} on A={1,2,3}.
Reflexive? Requires (a,a)∈R for every a∈A, i.e. (1,1),(2,2),(3,3). Since 3∈A but (3,3)∈/R, R is not reflexive.
Symmetric? Requires (b,a)∈R whenever (a,b)∈R. Here (1,2)∈R but (2,1)∈/R, so R is not symmetric.
Transitive? Requires (a,c)∈R whenever (a,b),(b,c)∈R. The only linking pairs are: …
- CBSE 2026Set ANNUAL1 markMCQQ.Let R = {(4, 4), (6, 6), (7, 7), (4, 6), (6, 4), (4, 7), (6, 7)} be a relation defined on A = {4, 6, 7}, then this relation R is ................. .(a) Reflexive, not symmetric and not transitive(b) Reflexive, symmetric and transitive(c) Reflexive and transitive but not symmetric(d) Neither Reflexive, nor symmetric and nor transitive
›Reveal solutionSolution
Check reflexivity, symmetry and transitivity directly against the listed ordered pairs.
Reflexive: (4,4),(6,6),(7,7) are all present in R, so R is reflexive.
Symmetric: (4,6)∈R and (6,4)∈R is fine, but (4,7)∈R while (7,4)∈/R. So R is NOT symmetric.
…
- CBSE 2025Set ANNUAL1 markQ.Show that the relation R in R defined as R={(a,b):a≤b} is transitive.
›Reveal solutionSolution
Use the transitivity of the order relation ≤ on real numbers.
The relation is R={(a,b):a≤b} on R. To prove transitivity, assume
(a,b)∈Rand(b,c)∈R.
By definition of R this means
a≤bandb≤c.
The usual order on real numbers is transitive, so …
- CBSE 2024Set ANNUAL1 markMCQQ.Let R be the relation in the set {1, 2, 3, 4} given by R = {(1,2), (2,2), (1,1), (4,4), (1,3), (3,3), (3,2)}. Choose the correct answer in the given options.(a) R is reflexive and symmetric but not transitive.(b) R is reflexive and transitive but not symmetric.(c) R is symmetric and transitive but not reflexive.(d) R is an equivalence relation.
›Reveal solutionSolution
Check reflexivity (every (a,a) present), symmetry (every pair's reverse present) and transitivity (chain closure) directly against the listed pairs.
R = {(1,2), (2,2), (1,1), (4,4), (1,3), (3,3), (3,2)} on {1, 2, 3, 4}.
Reflexive: We need (1,1), (2,2), (3,3), (4,4) ∈ R. All four are present, so R is reflexive.
Symmetric: (1,2) ∈ R but (2,1) ∉ R, so R is NOT symmetric.
…
- CBSE 2020Set 65/2/11 markMCQQ.The relation R in the set {1,2,3} given by R={(1,2),(2,1),(1,1)} is (A) symmetric and transitive, but not reflexive (B) reflexive and symmetric, but not transitive (C) symmetric, but neither reflexive nor transitive (D) an equivalence relation
›Reveal solutionSolution
The relation R={(1,2),(2,1),(1,1)} on the set {1,2,3} is symmetric, but it is neither reflexive nor transitive. Therefore, option (C) is correct.
To determine the properties of the given relation R on the set A={1,2,3}, we need to check if it satisfies the definitions of reflexivity, symmetry, and transitivity. Understanding these definitions precisely is key to avoiding common errors.
A relation R on a set A is:
- Reflexive if for every element a∈A, the ordered pair (a,a) is in R. This means every element must be related to itself.
- Symmetric if for every pair (a,b)∈R, the pair (b,a) is also in R. This means if a is related to b, then b must also be related to a.
- Transitive if for every a,b,c∈A, whenever (a,b)∈R and (b,c)∈R, it must follow that (a,c)∈R. This means if a is related to b and b is related to c, then a must also be related to c.
Let's examine R={(1,2),(2,1),(1,1)} on the set A={1,2,3} step by step.
-
Check for Reflexivity:
For R to be reflexive, every element in A must be related to itself. That is, (1,1), (2,2), and (3,3) must all be present in R.
- We see that (1,1)∈R.
- However, (2,2)∈/R.
- Also, (3,3)∈/R. Since (2,2) and (3,3) are not in R, the relation R is not reflexive.
-
Check for Symmetry:
For R to be symmetric, for every pair (a,b)∈R, the reverse pair (b,a) must also be in R.
Let's check each pair in R:
- For (1,2)∈R, we need to check if (2,1)∈R. Yes, (2,1)∈R.
- For (2,1)∈R, we need to check if (1,2)∈R. Yes, (1,2)∈R.
- For (1,1)∈R, we need to check if (1,1)∈R. Yes, (1,1)∈R. Since for every ordered pair (a,b) in R, the pair (b,a) is also in R, the relation R is symmetric.
-
Check for Transitivity:
For R to be transitive, if (a,b)∈R and (b,c)∈R, then (a,c) must also be in R. We need to check all such combinations.
- Consider (1,2)∈R and (2,1)∈R. Here, a=1,b=2,c=1. According to the definition of transitivity, (a,c)=(1,1) must be in R. We see that (1,1)∈R. This case holds.
- Consider (2,1)∈R and (1,2)∈R. Here, a=2,b=1,c=2. …
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