Types of relations

Correct answer: A. It is an equivalence relation with singleton classes

Step 1: Both same-remainder conditions are reflexive, symmetric and transitive. Step 2: Having the same remainders modulo (2) and modulo (3) means being the same number in this set. Step 3: Hence every element forms a singleton class and the relation is an equivalence relation.

Correct answer: B. (2)

Step 1: (1,2,3) are related to each other in all directions, so they form one class. Step 2: (4) is related only to itself, so it forms another class. Step 3: The classes are ({1,2,3}) and ({4}), so there are (2) classes.

Correct answer: A. It is always symmetric

Step 1: If ((a,b)) is in (R\cap S), it is in both (R) and (S). Step 2: Since both are symmetric, ((b,a)) is in both. Step 3: Hence ((b,a)) is in (R\cap S), so the intersection is symmetric.

Correct answer: A. It is always reflexive

Step 1: Since (R) is reflexive, it contains every self-pair. Step 2: (S) also contains every self-pair. Step 3: The union contains all pairs from both relations, so all self-pairs remain and the union is reflexive.

Correct answer: A. (R\cup{(1,3),(2,4),(1,4)})

Step 1: ((1,3)) is required from ((1,2)) and ((2,3)). Step 2: ((2,4)) is required from ((2,3)) and ((3,4)). Step 3: Further, ((1,4)) is required from ((1,3)) and ((3,4)).

Correct answer: A. (R\cup{(3,2)})

Step 1: ((1,2)) and ((2,1)) are already present in both directions. Step 2: The reverse of ((2,3)), namely ((3,2)), is missing. Step 3: Adding only ((3,2)) makes the relation symmetric.

Correct answer: D. All of the above

Step 1: ((1,3)) is forced by ((1,2)) and ((2,3)). Step 2: New chains and given pairs also force ((1,4)) and, in the reverse direction, ((4,1)). Step 3: While forming transitive closure, chains created by newly added pairs must also be checked.

Correct answer: A. (R^{-1}) will be reflexive and transitive

Step 1: Self-pairs remain the same after reversal, so (R^{-1}) is reflexive. Step 2: Transitive chains in (R) reverse into transitive chains in the inverse. Step 3: Therefore reflexivity and transitivity are both preserved in the inverse.

Correct answer: B. No, for example (R={(1,1)}) on (A={1,2})

Step 1: Symmetry and transitivity do not automatically provide all self-pairs. Step 2: In the given example, ((2,2)) is missing, so reflexivity fails. Step 3: For such theory questions, a small counterexample is the strongest method.

Correct answer: C. Equivalence relation

Step 1: (a-a=0) is divisible by (2), so reflexivity holds. Step 2: If (a-b) is divisible by (2), then (b-a) is also divisible by (2), so symmetry holds. Step 3: The sum of divisible differences is again divisible by (2), so transitivity holds.

Correct answer: A. ({1,2,3},{4},{5},{6})

Step 1: (1,2,3) are all less than (4), so they are mutually related. Step 2: (4,5,6) are related only to themselves through (a=b). Step 3: Hence the classes are ({1,2,3},{4},{5},{6}).

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: (|a-a|=0), so no self-pair occurs and reflexivity fails. Step 2: (|a-b|=|b-a|), so symmetry holds. Step 3: ((1,3)) and ((3,1)) exist but ((1,1)) does not, so transitivity fails.

Correct answer: A. Neither reflexive, symmetric, nor transitive

Step 1: (a-a=0), so the relation is not reflexive. Step 2: If (a-b=2), then (b-a=-2), so it is not symmetric. Step 3: (4R2) and (2R0) hold, but (4R0) has difference (4), so transitivity fails.

Correct answer: A. Transitive but neither reflexive nor symmetric

Step 1: (a-a=0), which is not at least (2), so the relation is not reflexive. Step 2: If (a) is at least (2) greater than (b), the reverse cannot hold. Step 3: From (a-b\ge2) and (b-c\ge2), we get (a-c\ge4), so it is transitive.

Correct answer: A. ((2,3)) and ((3,2))

Step 1: The relation is already reflexive and symmetric. Step 2: ((2,1)) and ((1,3)) force ((2,3)), while ((3,1)) and ((1,2)) force ((3,2)). Step 3: Adding these two pairs makes ({1,2,3}) a closed class and leaves (4) alone.

Correct answer: C. (9)

Step 1: Each equivalence class contributes all ordered pairs within itself. Step 2: For sizes (2,2,1), the count is (2^2+2^2+1^2). Step 3: Hence the total is (4+4+1=9).

Correct answer: C. (5)

Step 1: The number of equivalence relations equals the number of partitions of the set. Step 2: For three elements, partitions are all singletons, all together, and one pair with one singleton. Step 3: The total is (1+1+3=5).

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

Step 1: Equivalence classes form a complete partition of the set. Step 2: ({1,2}) and ({3,4}) are two closed classes covering the whole set. Step 3: The empty relation is not reflexive, and the universal relation has only one class.

Correct answer: A. It is an equivalence relation

Step 1: Every element has the same parity as itself, so reflexivity holds. Step 2: Same parity is unchanged by reversing order, so symmetry holds. Step 3: A chain of the same parity remains in the same parity, so transitivity also holds.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: No element can have a different remainder from itself, so reflexivity fails. Step 2: The different-remainder condition remains true after reversal, so symmetry holds. Step 3: (1R2) and (2R3) hold, but (1R3) does not, so transitivity fails.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: For (aRa), we need (2a^2=1), which is not true for every real (a). Step 2: (a^2+b^2) is unchanged when order is reversed, so symmetry holds. Step 3: (0R1) and (1R0) hold, but (0R0) does not, so transitivity fails.

Correct answer: A. ([a]=[b])

Step 1: Equivalence classes are either disjoint or exactly the same. Step 2: If two classes share even one element, they cannot remain distinct. Step 3: Therefore ([a]\cap[b]\neq\varnothing) implies ([a]=[b]).

Correct answer: B. They will be disjoint

Step 1: In an equivalence relation, two elements lie in the same class exactly when they are related. Step 2: Since (aRb) is false, (a) and (b) are not in the same class. Step 3: Distinct equivalence classes are disjoint.

Correct answer: A. It is an equivalence relation

Step 1: The condition (a=b) gives all self-pairs. Step 2: (a+b) being even means same parity, which is unchanged by reversal. Step 3: A chain of same parity is transitive, so the relation is an equivalence relation.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: For ((4,4)), the sum is (8), so reflexivity fails. Step 2: (a+b\le6) remains the same after reversing order, so symmetry holds. Step 3: ((5,1)) and ((1,5)) exist but ((5,5)) does not, so transitivity fails.