Types of relations

Correct answer: A. Both symmetry and transitivity

Step 1: All three self-pairs are present, so reflexivity holds. Step 2: ((1,3)) is present but ((3,1)) is not, so symmetry fails. Step 3: From ((2,1)) and ((1,3)), ((2,3)) is required but missing, so transitivity also fails.

Correct answer: A. Symmetric and transitive but not reflexive

Step 1: Over real numbers, (a^2+b^2=0) only when (a=0) and (b=0). Step 2: Hence the relation contains only ((0,0)), which is symmetric and transitive. Step 3: It is not reflexive because self-pairs for all real numbers are not present.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: For ((1,1)), the sum is (2), not divisible by (3), so it is not reflexive. Step 2: The sum remains the same when order changes, so symmetry holds. Step 3: ((1,2)) and ((2,1)) exist but ((1,1)) does not, so transitivity fails.

Correct answer: A. (2^{30})

Step 1: (A\times A) has (6^2=36) ordered pairs. Step 2: A reflexive relation must contain (6) self-pairs. Step 3: The remaining (30) pairs are optional, so the number is (2^{30}).

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

Step 1: Reflexivity forces all five self-pairs. Step 2: The (10) unordered pairs of distinct elements remain. Step 3: For symmetry, each pair is chosen or omitted together, so the total number is (2^{10}).

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

Step 1: The elements (1,2,3) are mutually related in all directions. Step 2: The element (4) is related only to itself. Step 3: Therefore the partition is ({1,2,3}) and ({4}).

Correct answer: A. Transitive but neither reflexive nor symmetric

Step 1: There is no self-pair, so it is not reflexive. Step 2: ((1,2)) is present but ((2,1)) is not, so it is not symmetric. Step 3: Every needed forward result such as ((1,3),(2,4),(1,4)) is present, so transitivity holds.

Correct answer: A. Universal relation and equivalence relation

Step 1: Every element of the set is not greater than (4). Step 2: Hence every ((a,b)) with (a,b\in A) belongs to the relation. Step 3: When all pairs are present, the relation is universal and also an equivalence relation.

Correct answer: A. Equivalence relation

Step 1: Since (a=a), every number is related to itself, so the relation is reflexive. Step 2: If (a=b) or (a=-b), the reverse relation satisfies the same condition, so it is symmetric. Step 3: This is the relation of equal absolute values, so transitivity also holds.

Correct answer: A. Reflexive and transitive but not symmetric

Step 1: (a-a=0), so every number is related to itself. Step 2: (a-b\ge0) means (a\ge b), which is transitive. Step 3: (3R2) is true but (2R3) is false, so symmetry does not hold.

Correct answer: A. Equivalence relation

Step 1: Every element belongs to the same divisibility-by-(3) type as itself, so reflexivity holds. Step 2: The same type condition is unchanged by reversing order, so symmetry holds. Step 3: A chain within the same type remains in that type, so transitivity holds.

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

Step 1: Multiples of (3) form one class. Step 2: (3) and (6) are multiples, while (1,2,4,5) are not. Step 3: Hence the two classes are ({3,6}) and ({1,2,4,5}).

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 ((2,4)) are also present, so symmetry holds. Step 3: From ((1,2)) and ((2,4)), ((1,4)) is required but missing, so it is not transitive.

Correct answer: A. Symmetric and transitive but not reflexive

Step 1: ((3,3)) is missing, so the relation is not reflexive. Step 2: Both ((1,2)) and ((2,1)) are present, so symmetry holds. Step 3: All required pairs inside ({1,2}) are present and no chain involves (3), so transitivity holds.

Correct answer: A. Transitive but neither reflexive nor symmetric

Step 1: No number can be at least (2) less than itself, so it is not reflexive. Step 2: If (a) is less than (b), the reverse cannot hold, so it is not symmetric. Step 3: If (a\le b-2) and (b\le c-2), then (a\le c-4), so (a) is at least (2) less than (c).

Correct answer: A. Reflexive and symmetric but not transitive

Step 1: The condition (a=b) puts every self-pair in the relation, so it is reflexive. Step 2: Since (|a-b|=|b-a|), symmetry holds. Step 3: ((1,2)) and ((2,3)) exist, but ((1,3)) does not, so transitivity fails.

Correct answer: A. (bRa)

Step 1: Every equivalence relation is symmetric. Step 2: If (aRb), symmetry necessarily gives (bRa). Step 3: No statement involving an arbitrary third element (c) follows without more information.

Correct answer: A. Symmetric and transitive but not reflexive

Step 1: (5) will not be related to itself, so the relation is not reflexive. Step 2: The condition that both elements are less than (5) remains true after reversal, so symmetry holds. Step 3: The relation is complete inside ({1,2,3,4}), so transitivity also holds.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: (a+a=0) only for (a=0), so reflexivity does not hold for all real numbers. Step 2: If (a+b=0), then (b+a=0), so symmetry holds. Step 3: (1R(-1)) and ((-1)R1) are true, but (1R1) is false, so transitivity fails.

Correct answer: A. Reflexive and transitive but not symmetric

Step 1: ((1,1),(2,2),(3,3)) are present, so reflexivity holds. Step 2: ((1,2)) exists but ((2,1)) does not, so it is not symmetric. Step 3: ((1,2)) and ((2,3)) require ((1,3)), which is present, and other required chains also do not fail.

Correct answer: A. (R^{-1}=R)

Step 1: In a symmetric relation, whenever ((a,b)) is present, ((b,a)) is also present. Step 2: The inverse relation reverses these pairs, so it does not create a different set. Step 3: Therefore, for a symmetric relation, (R^{-1}=R).

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

Step 1: Transitivity requires ((1,3)) from ((1,2)) and ((2,3)). Step 2: No other chain forces any additional pair. Step 3: Hence the smallest transitive extension is obtained by adding only ((1,3)).

Correct answer: A. (R) itself

Step 1: Symmetry requires the reverse of every pair. Step 2: ((1,2)) has ((2,1)), and ((2,3)) has ((3,2)) already. Step 3: Therefore nothing needs to be added to make it symmetric.

Correct answer: A. ((4,3))

Step 1: ((1,2)) and ((2,1)) are already balanced. Step 2: The reverse of ((3,4)), which is ((4,3)), is missing. Step 3: Adding only ((4,3)) completes symmetry.

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

Step 1: From ((1,2)) and ((2,3)), ((1,3)) is required. Step 2: From ((2,3)) and ((3,4)), ((2,4)) is required. Step 3: Then ((1,3)) and ((3,4)) also require ((1,4)).