Types of relations

Correct answer: A. ((1,3))

Step 1: Transitivity requires ((a,c)) whenever ((a,b)) and ((b,c)) exist. Step 2: Here ((1,2)) and ((2,3)) are present. Step 3: Therefore ((1,3)) is necessary and is the minimum addition.

Correct answer: A. ((1,1)) and ((2,2))

Step 1: From ((1,2)) and ((2,1)), ((1,1)) is required. Step 2: From ((2,1)) and ((1,2)), ((2,2)) is required. Step 3: No chain involves the third element, so ((1,1)) and ((2,2)) are the minimum additions.

Correct answer: A. Symmetric and transitive but not reflexive

Step 1: (1) and (4) are not prime, so their self-pairs will not occur and reflexivity fails. Step 2: The condition that both are prime does not change with order, so symmetry holds. Step 3: The relation is complete within ({2,3}), so transitivity holds.

Correct answer: A. Reflexive and transitive but not symmetric

Step 1: Every number is a multiple of itself, so the relation is reflexive. Step 2: (4) is a multiple of (2), but (2) is not a multiple of (4), so it is not symmetric. Step 3: If (a) is a multiple of (b) and (b) is a multiple of (c), then (a) is a multiple of (c), so it is transitive.

Correct answer: A. Equivalence relation

Step 1: Every person has the same blood group as themselves, so reflexivity holds. Step 2: Equality is unchanged by reversing order, so symmetry holds. Step 3: If two pairs have the same blood group through a common person, the first and third also have the same group, so transitivity holds.

Correct answer: A. Equivalence relation

Step 1: (a+a=2a) is even, so every self-pair is present. Step 2: If (a+b) is even, then (b+a) is even, so symmetry holds. Step 3: This is the same-parity relation, so it is transitive too.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: (a+a=2a) is even, so no element is related to itself. Step 2: If (a+b) is odd, then (b+a) is also odd, so symmetry holds. Step 3: ((1,2)) and ((2,1)) exist but ((1,1)) does not, so transitivity fails.

Correct answer: A. Reflexive and symmetric but not transitive

Step 1: All self-pairs are present, so the relation is reflexive. Step 2: Reverse pairs for ((1,2)) and ((1,3)) are also present, so symmetry holds. Step 3: From ((2,1)) and ((1,3)), ((2,3)) is required but missing, so transitivity fails.

Correct answer: A. (2^3:1)

Step 1: The total number of relations is (2^{3^2}=2^9). Step 2: In reflexive relations, three self-pairs are compulsory, so the number is (2^{9-3}=2^6). Step 3: The ratio is (2^9:2^6=2^3:1).

Correct answer: A. (2^{\frac{n(n-1)}{2}})

Step 1: Reflexivity forces the (n) self-pairs. Step 2: The remaining unordered pairs of distinct elements are (\frac{n(n-1)}{2}). Step 3: For symmetry, each such pair is chosen or omitted together, so the number is (2^{\frac{n(n-1)}{2}}).

Correct answer: A. Equivalence relation

Step 1: The condition (a=b) gives all self-pairs, so it is reflexive. Step 2: ({a,b}={1,2}) does not depend on order, so both ((1,2)) and ((2,1)) occur. Step 3: It forms closed classes ({1,2}), ({3}), and ({4}), so transitivity also holds.

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

Step 1: The condition (a+b=5) relates (1) with (4), and (2) with (3). Step 2: The condition (a=b) keeps each element in its own class. Step 3: Hence the classes are ({1,4}) and ({2,3}).

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

Step 1: Option A contains all self-pairs, so it is reflexive. Step 2: ((1,2)) is present but ((2,1)) is missing, so it is not symmetric. Step 3: ((1,2)) and ((2,3)) are present but ((1,3)) is missing, so it is not transitive.

Correct answer: A. Symmetric and transitive but not reflexive

Step 1: (3) and (4) are not related to themselves, so the relation is not reflexive. Step 2: The condition that both elements are less than (3) remains true after reversing order, so it is symmetric. Step 3: The relation is complete within ({1,2}), so it is transitive.

Correct answer: A. Symmetric and transitive but not reflexive

Step 1: (0\cdot0=0), so (0) is not related to itself and reflexivity fails. Step 2: If (ab>0), then (ba>0), so symmetry holds. Step 3: Same-sign non-zero numbers keep the sign class through a chain, so transitivity holds.

Correct answer: A. Reflexive and symmetric but not transitive

Step 1: (a\cdot a=a^2\ge0), so it is reflexive. Step 2: Since (ab=ba), the relation is symmetric. Step 3: (1R0) and (0R(-1)) are true, but (1R(-1)) is false, so transitivity fails.

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

Step 1: Divisibility of (a-b) by (2) means same parity. Step 2: (3) is odd, so its class contains all odd elements. Step 3: In (A), the odd elements are (1,3,5), so the class is ({1,3,5}).

Correct answer: A. ((2,1))

Step 1: Symmetry is compulsory in an equivalence relation. Step 2: If ((1,2)) is present, its reverse ((2,1)) must also be present. Step 3: ((2,3)) or ((1,3)) would be forced only with additional information.

Correct answer: A. (1R4)

Step 1: An equivalence relation is transitive. Step 2: From (1R2) and (2R4), transitivity gives (1R4). Step 3: No conclusion involving the separate element (3) can be drawn without more information.

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

Step 1: Since (1R2), (1) and (2) must lie in the same class. Step 2: Since (3) is not related to any distinct element, its class is ({3}). Step 3: Option A satisfies both conditions, with (4) as a singleton class.

Correct answer: A. Reflexive and symmetric but not transitive

Step 1: (|a-a|=0<1), so it is reflexive. Step 2: (|a-b|=|b-a|), so it is symmetric. Step 3: (0R0.6) and (0.6R1.2) are true, but (0R1.2) is false, so it is not transitive.

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

Step 1: (a-a=0), so it is not reflexive. Step 2: If (a-b=1), then (b-a=-1), so it is not symmetric. Step 3: (2R1) and (1R0) are true, but (2R0) has difference (2), so transitivity also fails.

Correct answer: A. It is an equivalence relation with classes ({2,4,6},{1,3,5})

Step 1: Every number has the same divisibility type as itself, so reflexivity holds. Step 2: Same divisibility type is unchanged by reversing order, so symmetry holds. Step 3: A chain of the same type remains in that type, so transitivity holds and the classes are even and odd numbers.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: For ((2,2)), the sum (4) is a perfect square, but for ((1,1)), the sum (2) is not, so it is not reflexive. Step 2: (a+b) does not change when order is reversed, so symmetry holds. Step 3: ((1,3)) and ((3,1)) exist but ((1,1)) does not, so transitivity fails.

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

Step 1: Divisibility of (a-b) by (3) means having the same remainder modulo (3). Step 2: (2) and (5) both leave remainder (2) on division by (3). Step 3: Put only those elements in the class that have the same remainder as the chosen element.