Types of relations

Correct answer: A. Equivalence relation

Step 1: All self-pairs are present, so the relation is reflexive. Step 2: Reverse pairs among (1,2,3) are complete, and (4) is related only to itself. Step 3: Transitivity holds inside each connected group, so it is an equivalence relation.

Correct answer: A. ({1,2,3}) and ({4})

Step 1: (1,2,3) are mutually connected, so they form one class. Step 2: (4) is connected only to itself through ((4,4)), so it forms a separate class. Step 3: While finding equivalence classes, group mutually related elements together.

Correct answer: A. It is a partial order relation

Step 1: All self-pairs are present, so reflexivity holds. Step 2: No reverse pair for unequal elements occurs, so antisymmetry holds. Step 3: ((1,2)) and ((2,3)) require ((1,3)), so it is a partial order relation.

Correct answer: A. Because ((1,3)) is missing

Step 1: A partial order must be transitive. Step 2: Since ((1,2)) and ((2,3)) are present, ((1,3)) is required, but it is missing. Step 3: Even if reflexivity and antisymmetry hold, failure of transitivity prevents partial order.

Correct answer: A. It is a partial order relation

Step 1: Every number divides itself, so it is reflexive. Step 2: If (a\mid b) and (b\mid a), then (a=b) here, so antisymmetry holds. Step 3: Divisibility is also transitive, so it is a partial order relation.

Correct answer: A. ((2,6))

Step 1: ((a,b)) occurs when (a) divides (b). Step 2: (2) divides (6), so ((2,6)) belongs to the relation. Step 3: In divisibility questions, check whether the first element divides the second exactly.

Correct answer: A. Transitive but not reflexive

Step 1: No set is a proper subset of itself, so it is not reflexive. Step 2: If (A\subset B) and (B\subset C), then (A\subset C), so it is transitive. Step 3: Carefully distinguish proper subset from subset.

Correct answer: A. Because it is reflexive, antisymmetric and transitive

Step 1: Every set is a subset of itself, so reflexivity holds. Step 2: If two sets are subsets of each other, they are equal, so antisymmetry holds. Step 3: The subset relation passes through another set, so transitivity holds.

Correct answer: A. Multiples of (4)

Step 1: Numbers related to (0) must satisfy (a-0) divisible by (4). Step 2: This means (a) is a multiple of (4). Step 3: In same-remainder relations, a class contains numbers with the same remainder.

Correct answer: A. Numbers leaving remainder (1) on division by (4)

Step 1: (a) is related to (1) when (a-1) is divisible by (4). Step 2: This means (a) leaves remainder (1) when divided by (4). Step 3: Equivalence classes are easy to identify by remainders.

Correct answer: B. (3)

Step 1: On division by (3), possible remainders are (0,1,2). Step 2: The given set contains elements of all three remainder types. Step 3: Each same-remainder group forms one equivalence class.

Correct answer: A. ({2,5})

Step 1: (2) leaves remainder (2) on division by (3). Step 2: (5) also leaves remainder (2). Step 3: Hence the equivalence class of (2) is ({2,5}).

Correct answer: A. ((b,a)\in R)

Step 1: A symmetric relation is identified by reverse pairs. Step 2: If ((a,b)) is present, ((b,a)) must be present. Step 3: Symmetry does not guarantee self-pairs or transitivity.

Correct answer: A. Both ((a,b)\in R) and ((b,a)\in R)

Step 1: Antisymmetry does not allow reverse pairs together for unequal elements. Step 2: If (a\ne b), both ((a,b)) and ((b,a)) cannot occur. Step 3: Self-pairs do not violate antisymmetry.

Correct answer: A. ({(1,1),(2,2)}) on ({1,2})

Step 1: A relation with only self-pairs is symmetric because each reverse is the same pair. Step 2: It has no reverse pairs between unequal elements, so it is also antisymmetric. Step 3: Antisymmetry fails when reverse pairs of unequal elements occur together.

Correct answer: A. Reflexive

Step 1: Both relations contain every self-pair ((a,a)). Step 2: The union contains all pairs from both relations, so self-pairs remain. Step 3: The union of reflexive relations is reflexive.

Correct answer: A. It will be symmetric

Step 1: (R\cap S) contains only pairs common to both relations. Step 2: If ((a,b)) is in both, symmetry gives ((b,a)) in both. Step 3: Therefore the intersection is also symmetric.

Correct answer: A. Always transitive

Step 1: Linked pairs in the intersection are present in both relations. Step 2: Since both relations are transitive, the required third pair is present in both. Step 3: Hence that third pair is also in the intersection, making it transitive.

Correct answer: A. Because linked pairs may come from different relations and the third pair may be missing

Step 1: Transitivity requires ((a,c)) from ((a,b)) and ((b,c)). Step 2: In a union, the first two pairs may come from different relations, so the third pair may not be present. Step 3: Always check transitivity of a union separately.

Correct answer: A. Yes

Step 1: The main linked pairs are ((1,2)) and ((2,3)), requiring ((1,3)). Step 2: ((1,3)) is present, and ((3,3)) does not create a missing requirement. Step 3: Transitivity does not require all self-pairs, only required linked conclusions.

Correct answer: A. Because ((2,2)) is missing

Step 1: ((2,1)) and ((1,2)) are both in the relation. Step 2: Transitivity requires ((2,2)) from them. Step 3: Since ((2,2)) is missing, transitivity fails.

Correct answer: A. (2^{25})

Step 1: (A\times A) has (5^2=25) ordered pairs. Step 2: Every relation is a subset of (A\times A). Step 3: The number of subsets of (25) pairs is (2^{25}).

Correct answer: A. (2^{20})

Step 1: Total pairs are (5^2=25). Step 2: Reflexivity fixes (5) self-pairs. Step 3: The remaining (25-5=20) pairs are free, so the count is (2^{20}).

Correct answer: A. (2^{15})

Step 1: The number of symmetric relations is (2^{\frac{n(n+1)}{2}}). Step 2: For (n=5), (\frac{5\cdot6}{2}=15) independent choices. Step 3: Therefore the number is (2^{15}).

Correct answer: A. (2^{10})

Step 1: Reflexivity fixes the (5) self-pairs. Step 2: The non-self reverse-pair groups are (\frac{5\cdot4}{2}=10). Step 3: These (10) groups are independent, so the count is (2^{10}).