Types of relations

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: For ((1,1)), the maximum is (1), so the relation is not reflexive. Step 2: The maximum value does not change on reversing order, so it is symmetric. Step 3: ((1,5)) and ((5,2)) exist but ((1,2)) does not, so it is not transitive.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: For ((1,1)), the minimum is (1), so not all self-pairs occur. Step 2: The minimum value remains the same when order is reversed, so it is symmetric. Step 3: ((3,2)) and ((2,4)) exist but ((3,4)) does not, so transitivity fails.

Correct answer: A. Reflexive and symmetric but not transitive

Step 1: (a\cdot a=a^2\ge0), so the relation is reflexive. Step 2: Since (ab=ba), it is symmetric. Step 3: (1R0) and (0R(-1)) hold but (1R(-1)) does not, so transitivity fails.

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 it is symmetric. Step 3: The relation stays within positive and negative groups separately, so it is transitive.

Correct answer: A. Transitive but neither reflexive nor symmetric

Step 1: No number is less than itself, so reflexivity fails. Step 2: If (a<b), then (b<a) cannot hold, so it is not symmetric. Step 3: From (a<b) and (b<c), we get (a<c), so it is transitive.

Correct answer: A. Reflexive and transitive but not symmetric

Step 1: For every (a), (a\le a), so the relation is reflexive. Step 2: (1\le2) is true but (2\le1) is not, so symmetry fails. Step 3: (a\le b) and (b\le c) imply (a\le c), so it is transitive.

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 it is symmetric. Step 3: ((1,2)) and ((2,1)) exist but ((1,1)) does not, so transitivity fails.

Correct answer: A. Equivalence relation

Step 1: All self-pairs are present, so it is reflexive. Step 2: Reverse pairs of ((1,3)) and ((2,4)) are present, so it is symmetric. Step 3: ({1,3}) and ({2,4}) are closed blocks, so transitivity holds.

Correct answer: A. Equivalence relation

Step 1: For every (a), (a^3=a^3), so it is reflexive. Step 2: Equality of cubes works in both directions, so it is symmetric. Step 3: If (a^3=b^3) and (b^3=c^3), then (a^3=c^3), so it is transitive.

Correct answer: A. (aRd)

Step 1: An equivalence relation is transitive. Step 2: From (aRb) and (bRc), we get (aRc); then from (aRc) and (cRd), we get (aRd). Step 3: In a long chain, remove middle elements step by step.

Correct answer: A. (bRa)

Step 1: Symmetry is compulsory in an equivalence relation. Step 2: If (aRb), symmetry gives (bRa). Step 3: Relation with any third element is not guaranteed without more information.

Correct answer: A. (1R4)

Step 1: An equivalence relation is transitive. Step 2: From (1R2) and (2R4), we get (1R4). Step 3: No conclusion involving (3) can be drawn without more information.

Correct answer: A. It is an equivalence relation

Step 1: The universal relation 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 all possible pairs are present, transitivity also holds.

Correct answer: A. Equivalence relation

Step 1: The identity relation has all self-pairs, so it is reflexive. Step 2: The reverse of ((a,a)) is itself, so it is symmetric. Step 3: A chain of self-pairs again gives a self-pair, so it is transitive.

Correct answer: A. (5)

Step 1: A reflexive relation must contain the self-pair of every element. Step 2: There are (5) elements, so (5) self-pairs are compulsory. Step 3: In the minimum case, only these (5) pairs are present.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: No self-pair is present, so the relation is not reflexive. Step 2: Every listed pair has its reverse, 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. Reflexive and symmetric but not transitive

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

Correct answer: C. Reflexive but neither symmetric nor transitive

Step 1: For every (a), (a\le a) and (a-a=0\le2), so it is reflexive. Step 2: ((1,2)) exists but ((2,1)) does not, so it is not symmetric. Step 3: ((1,3)) and ((3,5)) exist but ((1,5)) does not, so it is not transitive either.

Correct answer: C. ({5k+2:k\in Z})

Step 1: (a\equiv2 \pmod{5}) means (a) leaves remainder (2) on division by (5). Step 2: Such integers are written as (5k+2). Step 3: While writing an equivalence class, include all elements with the same remainder.

Correct answer: C. It is an equivalence relation

Step 1: Every number divides itself, so the relation is reflexive. Step 2: The condition already includes divisibility in both directions, so it is symmetric. Step 3: For positive numbers, mutual divisibility means equality, so transitivity also holds.

Correct answer: D. Equivalence relation

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

Correct answer: B. Symmetric but neither reflexive nor transitive

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

Correct answer: B. It is an equivalence relation

Step 1: All self-pairs are present, so it is reflexive. Step 2: Every extra pair has its reverse pair, so it is symmetric. Step 3: ({1,2,4}) is a closed block and (3) is alone, so transitivity also holds.

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

Step 1: Reflexivity forces the (6) self-pairs. Step 2: The (\frac{6\cdot5}{2}=15) unordered pairs of distinct elements give independent choices. Step 3: Hence the number of such relations is (2^{15}).

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

Step 1: There are (n) independent choices for self-pairs. Step 2: There are (\frac{n(n-1)}{2}) unordered pairs of distinct elements. Step 3: Total independent choices are (\frac{n(n+1)}{2}), so the number is (2^{\frac{n(n+1)}{2}}).