On (\mathbb{R}), (aRb) if and only if (a-b>0). Which property of (R) is correct?
Correct answer: A
Since (a-a=0), (aRa) is never true, so it is irreflexive. If (a>b) and (b>c), then (a>c), so it is transitive.
Mathematics
Relations
संबंध
In this Class 11 Mathematics topic, students learn how a relation connects elements of one set with elements of another, building the foundation for the chapter Relations and Functions. The topic introduces ordered pairs, Cartesian products, and the representation of relations as subsets of a Cartesian product. Students also examine the domain, codomain, and range of a relation and learn to interpret relations through rosters, diagrams, and set notation, preparing them to distinguish relations from functions.
TOPIC PRACTICE
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Expert · Level 6
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- Irreflexive and transitive
- Reflexive and symmetric
- Symmetric and transitive
- Equivalence relation
Expert · Level 6
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- (a\leq b)
- (a\geq b)
- (a=b^3)
- (|a|\leq|b|)
Expert · Level 6
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- Symmetric but not reflexive
- Reflexive and transitive
- Equivalence relation
- Transitive but not symmetric
Expert · Level 6
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- Symmetric but not transitive
- Reflexive and transitive
- Equivalence relation
- Antisymmetric
Expert · Level 6
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- (a\leq b)
- (a=b)
- (a\neq b)
- (a+b) is even
Expert · Level 6
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- ((1,4))
- ((4,1))
- ((2,1))
- ((4,2))
Expert · Level 6
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- (3)
- (2)
- (4)
- (6)
Expert · Level 6
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- ((1,1),(2,2),(3,3))
- ((1,3),(3,1))
- ((2,3),(3,2))
- ((1,2),(2,1))
Expert · Level 6
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- ((2,1),(3,2))
- ((1,1),(2,2),(3,3))
- ((1,3),(3,1))
- ((2,2),(3,3))
Expert · Level 6
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- Yes, because no distinct reverse pair appears together
- No, because diagonal pairs are present
- No, because it is transitive
- Yes, because it is symmetric
Expert · Level 6
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- Equivalence relation
- Only reflexive
- Only symmetric
- Antisymmetric relation
Expert · Level 6
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- ([a]={a+q:q\in\mathbb{Q}})
- ([a]={aq:q\in\mathbb{Q}})
- ([a]={q-a:q\in\mathbb{Q}})
- ([a]=\mathbb{Q})
Expert · Level 6
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- Symmetric but not reflexive
- Reflexive and transitive
- Equivalence relation
- Antisymmetric
Expert · Level 6
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- (R\cap S) is also an equivalence relation
- (R\cap S) is never reflexive
- (R\cap S) is always empty
- (R\cap S) is always the universal relation
Expert · Level 6
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- ({1,3}) and ({2,4})
- ({1,2}) and ({3,4})
- ({1,4}) and ({2,3})
- ({1},{2},{3},{4})
On (\mathbb{R}), (aRb) if and only if (a^3\leq b^3). This relation is equivalent to which relation?
Correct answer: A
The function (x^3) is strictly increasing, so (a^3\leq b^3) is exactly equivalent to (a\leq b). Use monotonicity to identify relation properties quickly.
On (\mathbb{N}), (aRb) if and only if (a+b) is odd. What is the correct statement about (R)?
Correct answer: A
Addition is commutative, so the relation is symmetric. But (a+a=2a) is always even, so (aRa) never holds.
On (A={1,2,3,4}), (aRb) if and only if (a\neq b). Which statement about (R) is correct?
Correct answer: A
If (a\neq b), then (b\neq a), so it is symmetric. But (1R2) and (2R1) hold while (1R1) does not, so it is not transitive.
On (A={1,2,3}), relation (R={(1,1),(1,2),(2,2),(2,3),(1,3),(3,3)}) is given. Which familiar order does this relation resemble?
Correct answer: A
The given pairs are exactly those where the first element is less than or equal to the second. Thus it is the usual order (\leq) on (A).
Which pair is compulsory in the transitive closure of a relation (R), if (R) contains ((1,2)) and ((2,4))?
Correct answer: A
By transitivity, ((1,2)) and ((2,4)) force ((1,4)). A closure adds the minimum necessary pairs.
On (A={1,2,3,4}), (R={(1,2),(2,3),(3,4)}). What is the minimum number of new pairs needed in the transitive closure of (R)?
Correct answer: A
The pairs to add are ((1,3),(2,4),(1,4)). Note that newly added ((1,3)) with ((3,4)) also requires ((1,4)).
On (A={1,2,3}), (R={(1,2),(2,1)}). Which pairs will be added in the reflexive closure?
Correct answer: A
In a reflexive closure, ((a,a)) must be present for every (a\in A). Therefore all three diagonal pairs are added.
On (A={1,2,3}), (R={(1,2),(2,3)}). Which pairs will be added in the symmetric closure?
Correct answer: A
In a symmetric closure, ((b,a)) is added for every ((a,b)). Thus ((2,1)) is needed for ((1,2)), and ((3,2)) for ((2,3)).
On (A={1,2,3}), (R={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}). Is (R) antisymmetric?
Correct answer: A
Antisymmetry is not affected by diagonal pairs. It fails only when ((a,b)) and ((b,a)) both occur for (a\neq b).
On (A={1,2,3,4,5,6}), (aRb) if and only if (a) and (b) have the same greatest prime factor. What kind of relation is it?
Correct answer: A
Having the same greatest prime factor behaves like equality, so it is reflexive, symmetric, and transitive. Such same invariant conditions often give equivalence relations.
On (\mathbb{R}), (aRb) if and only if (a-b) is rational. How are the equivalence classes of (R) formed?
Correct answer: A
(xRa) means (x-a\in\mathbb{Q}), so (x=a+q) where (q\in\mathbb{Q}). When writing a class, solve the relation condition for the variable.
On (A={1,2,3,4,5}), (R={(a,b):a+b\leq6}). Choose the correct option about (R).
Correct answer: A
Since (a+b=b+a), the relation is symmetric. But ((4,4)\notin R) because (4+4\leq6) is false, so it is not reflexive.
If (R) and (S) are both equivalence relations on (A), which statement about (R\cap S) is always true?
Correct answer: A
The intersection retains common diagonal pairs, and symmetry and transitivity are also preserved. Hence the intersection of equivalence relations is again an equivalence relation.
On the set (A={1,2,3,4}), relation (R={(a,b)\in A\times A:a+b) is even(}) is given. What are the equivalence classes of (R)?
Correct answer: A
(a+b) is even exactly when (a) and (b) have the same parity. Hence the equivalence classes are the odd class ({1,3}) and the even class ({2,4}).
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