Types of relations

Correct answer: A. Reflexive and transitive but not symmetric

Step 1: Every number divides itself, so the relation is reflexive. Step 2: If (a) divides (b) and (b) divides (c), then (a) divides (c), so it is transitive. Step 3: For symmetry, always check the reverse pair; (1) divides (2) but (2) does not divide (1).

Correct answer: B. It is an equivalence relation

Step 1: ((1,1),(2,2),(3,3)) are present, so the relation is reflexive. Step 2: ((1,2)) and ((2,1)) both exist, so symmetry holds. Step 3: Checking all needed chains shows transitivity is not broken, so it is an equivalence relation.

Correct answer: A. Transitive but not reflexive

Step 1: ((3,3)) is missing, so the relation is not reflexive. Step 2: From ((1,2)) and ((2,3)), ((1,3)) is required and it is present. Step 3: In hard questions, first look for missing self-pairs to reject reflexivity quickly.

Correct answer: A. It is symmetric and transitive but not reflexive

Step 1: No ((a,a)) pair is present, so it is not reflexive. Step 2: There is no pair that can violate symmetry or transitivity, so those conditions hold vacuously. Step 3: For an empty relation, reflexivity is the main condition to check.

Correct answer: A. It is always reflexive, symmetric and transitive

Step 1: (A\times A) contains every ((a,a)), so it is reflexive. Step 2: For every ((a,b)), ((b,a)) is also present, so it is symmetric. Step 3: Since every possible pair exists, transitivity also holds automatically.

Correct answer: A. Equivalence relation

Step 1: (a-a=0) is even, so the relation is reflexive. Step 2: If (a-b) is even, then (b-a) is also even, so it is symmetric. Step 3: The sum of even differences is even, so transitivity holds and the relation is an equivalence relation.

Correct answer: A. Equivalence relation

Step 1: For every integer, (a \equiv a \pmod{3}), so it is reflexive. Step 2: If (a \equiv b \pmod{3}), then (b \equiv a \pmod{3}), so it is symmetric. Step 3: Having the same remainder is transitive, so it is an equivalence relation.

Correct answer: A. It is reflexive and transitive but not symmetric

Step 1: For every (a), (a\le a), so the relation is reflexive. Step 2: If (a\le b) and (b\le c), then (a\le c), so it is transitive. Step 3: (2\le 3) is true but (3\le 2) is false, so it is not symmetric.

Correct answer: A. Transitive but neither reflexive nor symmetric

Step 1: (a<a) is never true, so the relation is not reflexive. Step 2: If (a<b), then (b<a) cannot be true, so it is not symmetric. Step 3: From (a<b) and (b<c), we get (a<c), so it is transitive.

Correct answer: A. It is symmetric but neither reflexive nor transitive

Step 1: (|a-a|=0), so no self-pair occurs and the relation is not reflexive. Step 2: If (|a-b|=1), then (|b-a|=1), so it is symmetric. Step 3: ((1,2)) and ((2,3)) exist but ((1,3)) does not, so it is not transitive.

Correct answer: A. Symmetric but not reflexive and not transitive

Step 1: The self-pairs ((1,1),(2,2),(3,3)) are not all present, so it is not reflexive. Step 2: Every listed pair has its reverse pair, so it is symmetric. Step 3: From ((1,2)) and ((2,1)), ((1,1)) is required but missing, so it is not transitive.

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 also even, so symmetry holds. Step 3: The same parity condition preserves transitivity, so the relation is an equivalence relation.

Correct answer: A. Symmetric but not reflexive and not transitive

Step 1: A self-pair like ((1,1)) gives sum (2), so it is not reflexive. Step 2: If (a+b=5), then (b+a=5), so the relation is symmetric. Step 3: From ((1,4)) and ((4,1)), ((1,1)) would be needed, but it is missing, so transitivity fails.

Correct answer: A. Reflexive but neither symmetric nor transitive

Step 1: All self-pairs are present, so the relation is reflexive. Step 2: ((1,2)) exists but ((2,1)) does not, so it is not symmetric. Step 3: ((1,2)) and ((2,3)) exist but ((1,3)) is missing, so it is not transitive.

Correct answer: A. It is an equivalence relation

Step 1: In the identity relation, every element is related to itself, so it is reflexive. Step 2: The reverse of ((a,a)) is again ((a,a)), so it is symmetric. Step 3: Combining self-pairs gives the same self-pair, so transitivity also holds.

Correct answer: A. Reflexive and transitive but not symmetric

Step 1: For every (a), (a\ge a), so it is reflexive. Step 2: (3\ge 2) is true but (2\ge 3) is false, so it is not symmetric. Step 3: If (a\ge b) and (b\ge c), then (a\ge c), so it is transitive.

Correct answer: A. Symmetric and transitive but not reflexive

Step 1: Self-pairs for (2) and (4) will not occur, so it is not reflexive. Step 2: If both numbers are odd, changing order keeps both odd, so it is symmetric. Step 3: Within the odd elements, the relation stays connected, so transitivity holds.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: (a+a=2a) is even, so no self-pair occurs and it is not reflexive. Step 2: If (a+b) is odd, then (b+a) is odd, so it is symmetric. Step 3: ((1,2)) and ((2,1)) exist but ((1,1)) does not, so transitivity fails.

Correct answer: A. Failure of transitivity

Step 1: All self-pairs are present, so reflexivity holds. Step 2: Every pair has its reverse, so symmetry holds. Step 3: From ((1,2)) and ((2,3)), ((1,3)) is required but missing, so transitivity fails.

Correct answer: A. (3)

Step 1: A reflexive relation must contain ((a,a)) for every (a\in A). Step 2: There are three elements, so ((1,1),(2,2),(3,3)) are necessary. Step 3: The minimum number is equal to the number of elements in the set.

Correct answer: A. (2^{12})

Step 1: (A\times A) has (4^2=16) ordered pairs. Step 2: Reflexivity forces (4) self-pairs, so the remaining (12) pairs are optional. Step 3: Each optional pair has two choices, giving (2^{12}) relations.

Correct answer: A. (2^6)

Step 1: The three self-pairs can be chosen independently. Step 2: There are three unordered pairs of distinct elements, and both directions must be selected together or omitted together. Step 3: Total independent choices are (3+3=6), so the number is (2^6).

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

Step 1: The (n) self-pairs can be chosen independently. Step 2: There are (\frac{n(n-1)}{2}) unordered pairs of distinct elements, and both directions are decided together. Step 3: Total independent choices are (n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}).

Correct answer: A. 3^3 = 27

Correct answer: A, 3^3 = 27. Let the set be {a,b,c}. Reflexivity requires (a,a), (b,b), and (c,c) to be included, so the three diagonal pairs are fixed and create no choices. For the distinct elements, consider the unordered pairs {a,b}, {a,c}, and {b,c}; there are three such pairs. For each unordered pair, antisymmetry allows exactly three choices: include neither ordered pair, include only the first direction, or include only the reverse direction. Including both directions, such as (a,b) and (b,a), is forbidden. Therefore the total number is 3 × 3 × 3 = 3^3 = 27. Option A is correct. Option B, 2^6, incorrectly treats all six off-diagonal ordered pairs as independent, which would sometimes include both opposite directions and violate antisymmetry. Option C, 2^9, incorrectly treats all nine pairs as independent and ignores the compulsory diagonal pairs. Option D has the same numerical value but is not a mathematically distinct answer; it is deliberately worded as an incorrect method, so A is the unique valid choice. Memory cue: reflexive fixes the diagonal, while antisymmetry gives three choices for each unordered off-diagonal pair.

Correct answer: A. It is reflexive and symmetric but not transitive

Step 1: All self-pairs are present, so the relation is reflexive. Step 2: ((1,2)) has ((2,1)) and ((1,3)) has ((3,1)), so it is symmetric. Step 3: From ((2,1)) and ((1,3)), ((2,3)) is required but missing, so it is not transitive.