Types of relations

Correct answer: B. Reflexive and transitive but not symmetric

Step 1: (a^2\le a^2), so it is reflexive. Step 2: (1^2\le2^2) is true but (2^2\le1^2) is false, so it is not symmetric. Step 3: The order relation (\le) remains transitive on squares, so it is transitive.

Correct answer: C. Symmetric and transitive but not reflexive

Step 1: Over real numbers, (a^2+b^2=0) only when (a=0,b=0). Step 2: So 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: B. (3)

Step 1: On division by (3), possible remainders are (0,1,2). Step 2: Elements with the same remainder form one class. Step 3: Therefore (3) equivalence classes are formed.

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

Step 1: The condition (a+b=6) relates (1) with (5), and (2) with (4). Step 2: (3) remains with itself because (3+3=6) and self-pairs are included. Step 3: Hence the classes are ({1,5},{2,4},{3}).

Correct answer: B. 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 required result of a forward chain is present, so the relation is transitive.

Correct answer: C. Transitive but neither reflexive nor symmetric

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

Correct answer: A. It is always an equivalence relation

Step 1: Both relations are reflexive, so their intersection remains reflexive. Step 2: Both are symmetric, so reverse pairs of common pairs remain common. Step 3: Both are transitive, so the result of a chain in the intersection also lies in the intersection.

Correct answer: B. It is always reflexive and symmetric but need not be transitive

Step 1: Both relations contain all self-pairs, so the union is reflexive. Step 2: Both are symmetric, so the union is symmetric. Step 3: Transitivity may fail because one pair of a chain may come from (R) and the other from (S).

Correct answer: C. ((1,1),(2,2),(3,3))

Step 1: ((1,2)) and ((2,1)) force ((1,1)) and ((2,2)). Step 2: ((2,3)) and ((3,2)) force ((2,2)) and ((3,3)). Step 3: Hence all three self-pairs appear in the transitive closure.

Correct answer: D. (6)

Step 1: The original (3) pairs are already present. Step 2: Transitivity forces ((1,3)), ((2,4)), and ((1,4)). Step 3: Therefore the total number of pairs is (3+3=6).

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

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

Correct answer: B. Symmetric but neither reflexive nor transitive

Step 1: For ((2,2)), the greatest common divisor is (2), so the relation is not reflexive. Step 2: The greatest common divisor does not change on reversing order, so it is symmetric. Step 3: (2R3) and (3R4) hold but (2R4) does not, so it is not transitive.

Correct answer: B. Symmetric but neither reflexive nor transitive

Step 1: (a+a=0) is not true for every real number, so it is not reflexive. Step 2: If (a+b=0), then (b+a=0), so it is symmetric. Step 3: (1R(-1)) and ((-1)R1) hold but (1R1) does not, so transitivity fails.

Correct answer: C. It is an equivalence relation

Step 1: Since (a=a), every element is related to itself. Step 2: The condition (a=b) or (a=-b) remains true when reversed. Step 3: This is equality of absolute values, which is also transitive, so it is an equivalence relation.

Correct answer: C. ({2,4,6})

Step 1: This relation divides elements into even and odd classes. Step 2: (4) is even, so its class contains all even elements. Step 3: The even elements in (A) are (2,4,6), so the class is ({2,4,6}).

Correct answer: C. Equivalence relation

Step 1: The condition (a=b) gives all self-pairs, so it is reflexive. Step 2: ({a,b}={1,3}) is independent of order, so it is symmetric. Step 3: The classes are ({1,3},{2},{4}), so transitivity holds too.

Correct answer: C. (8)

Step 1: (a-b) being even means the two numbers have the same parity. Step 2: The odd class ({1,3}) gives (2^2=4) pairs, and the even class ({2,4}) gives (2^2=4) pairs. Step 3: Total pairs are (4+4=8).

Correct answer: B. (2^6-1)

Step 1: The number of reflexive relations is (2^{9-3}=2^6). Step 2: The universal relation is one of these reflexive relations. Step 3: Removing it gives (2^6-1).

Correct answer: A. When (A) has exactly one element

Step 1: The identity relation contains only self-pairs. Step 2: The universal relation contains every possible ordered pair. Step 3: If the set has only one element, both contain the same single self-pair, so they are equal.

Correct answer: C. It is the empty relation itself

Step 1: To form the inverse relation, existing pairs are reversed. Step 2: The empty relation has no pair, so there is nothing to reverse. Step 3: Therefore its inverse relation is also the empty relation.

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

Step 1: ((1,2)) and ((2,1)) are already both present. Step 2: The reverse of ((3,4)), which is ((4,3)), is missing. Step 3: Adding only ((4,3)) makes the relation symmetric.

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

Step 1: ((1,2)) and ((2,3)) are in the relation. Step 2: Transitivity therefore forces ((1,3)). Step 3: No given pair involves (4), so pairs with (4) are not forced.

Correct answer: B. (S) may or may not be symmetric

Step 1: A symmetric larger relation does not force every subrelation to contain reverse pairs. Step 2: If (S) contains ((a,b)) but not ((b,a)), symmetry fails. Step 3: Hence (S) may or may not be symmetric.

Correct answer: A. (S) will be reflexive

Step 1: Since (R) is reflexive, it contains all self-pairs. Step 2: Because (R\subseteq S), all those self-pairs are also in (S). Step 3: Therefore (S) must be reflexive.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: ((2,2)) is missing, so the relation is not reflexive. Step 2: The removed pair is its own reverse, so symmetry remains. Step 3: ((2,1)) and ((1,2)) are present but ((2,2)) is missing, so transitivity fails.