Types of relations

Correct answer: B. It is an equivalence relation

Step 1: No self-pair is removed, so reflexivity holds. Step 2: Both ((1,2)) and ((2,1)) are removed, so symmetry holds. Step 3: But ((1,3)) and ((3,2)) are present while ((1,2)) is missing, so it is not transitive.

Correct answer: B. Reflexive and symmetric but not transitive

Step 1: All self-pairs are still present, so the relation is reflexive. Step 2: The two removed pairs are reverses of each other, so symmetry remains. Step 3: ((1,3)) and ((3,2)) are present but ((1,2)) is missing, so transitivity fails.

Correct answer: C. (10)

Step 1: In an equivalence relation, all ordered pairs inside each class are included. Step 2: ({1,2,3}) gives (3^2=9) pairs and ({4}) gives (1^2=1) pair. Step 3: Total pairs are (9+1=10).

Correct answer: C. Symmetric but neither reflexive nor transitive

Step 1: A number cannot have a different remainder from itself, so it is not reflexive. Step 2: The condition of different remainders remains true when order is reversed, so it is symmetric. Step 3: (1R2) and (2R3) hold but (1R3) does not, so transitivity fails.

Correct answer: B. Symmetric but neither reflexive nor transitive

Step 1: For ((3,3)), the sum is (6), so reflexivity fails. Step 2: In (a+b\le5), reversing order does not change the sum, so it is symmetric. Step 3: ((4,1)) and ((1,4)) exist but ((4,4)) does not, so transitivity fails.

Correct answer: C. Reflexive and symmetric but not transitive

Step 1: (|a-a|=0<2), so it is reflexive. Step 2: (|a-b|=|b-a|), so it is symmetric. Step 3: (0R1.5) and (1.5R3) hold but (0R3) does not, so transitivity fails.

Correct answer: B. Equivalence relation

Step 1: The condition (a=b) gives all self-pairs, so it is reflexive. Step 2: The condition that both are odd is unchanged by reversal, so it is symmetric. Step 3: Odd elements ({1,3,5}) form one closed class and even elements stay single, so it is transitive.

Correct answer: A. (aRb)

Step 1: Equal equivalence classes mean both elements are in the same group. Step 2: Two elements in the same equivalence class are related to each other. Step 3: Therefore ([a]=[b]) implies (aRb).

Correct answer: B. (aRb) must be false

Step 1: Equivalence classes divide the set into disjoint parts. Step 2: Elements from different classes are not related to each other. Step 3: Hence if (a) and (b) are in different equivalence classes, (aRb) is false.

Correct answer: C. (14)

Step 1: Every equivalence class contributes all ordered pairs within it. Step 2: The classes contribute (2^2=4), (3^2=9), and (1^2=1) pairs. Step 3: Total pairs are (4+9+1=14).

Correct answer: A. (2^{16}-2^{12})

Step 1: The total number of relations is (2^{16}). Step 2: The number of relations containing all self-pairs is (2^{12}). Step 3: Hence the number not containing all self-pairs is (2^{16}-2^{12}).

Correct answer: A. Symmetric and transitive but not reflexive

Step 1: (4) is not related to itself, so the relation is not reflexive. Step 2: The condition that both elements are less than (4) remains true after reversal. Step 3: The relation is complete inside ({1,2,3}), so it is transitive.

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

Step 1: (1) and (2) are both less than (3), so they lie in one class. Step 2: (3) and (4) are related only to themselves. Step 3: Hence the classes are ({1,2},{3},{4}).

Correct answer: A. (R^{-1}) will also be transitive

Step 1: If ((a,b)) and ((b,c)) are in (R^{-1}), then ((b,a)) and ((c,b)) are in (R). Step 2: By transitivity of (R), ((c,a)) is in (R). Step 3: Therefore ((a,c)) is in (R^{-1}), so (R^{-1}) is transitive.

Correct answer: B. It is an equivalence relation

Step 1: All self-pairs are present, so the relation is reflexive. Step 2: Within ({1,2,3}), all reverse pairs are present and (4) is related to itself, so it is symmetric. Step 3: ({1,2,3}) is a complete closed class and (4) is a singleton class, so transitivity holds too.

Correct answer: B. Reflexive and transitive but not symmetric

Step 1: All self-pairs are present, so the relation is reflexive. Step 2: ((1,2)) is present but ((2,1)) is not, so it is not symmetric. Step 3: Every needed forward-chain result such as ((1,3),(2,4),(1,4)) is present, so transitivity holds.

Correct answer: A. Reflexive and symmetric but not transitive

Step 1: The greatest common divisor of each element with itself is the element, so reflexivity holds. Step 2: The greatest common divisor does not change when order is reversed, so symmetry holds. Step 3: (2R6) and (6R3) hold, but (2R3) does not, so transitivity fails.

Correct answer: B. It is an equivalence relation

Step 1: (\frac{a}{a}=1) is rational, so reflexivity holds. Step 2: If (\frac{a}{b}) is rational, then (\frac{b}{a}) is rational, so symmetry holds. Step 3: If (\frac{a}{b}) and (\frac{b}{c}) are rational, then (\frac{a}{c}) is rational, so transitivity holds.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: For ((1,1)), the sum (2) is prime, but for ((2,2)), the sum (4) is not, so the relation is not reflexive. Step 2: (a+b) is unchanged by reversing order, so it is symmetric. Step 3: ((2,3)) and ((3,4)) hold, but ((2,4)) does not, so transitivity fails.

Correct answer: D. ({4k+3:k\in Z})

Step 1: Divisibility of (a-b) by (4) means the same remainder modulo (4). Step 2: The remainder of (3) is (3), so its class contains all integers of the form (4k+3). Step 3: Include only elements with the same remainder in the class.

Correct answer: B. Symmetric but neither reflexive nor transitive

Step 1: (1) is divisible by neither (2) nor (3), so ((1,1)) is not in the relation and reflexivity fails. Step 2: Both conditions are unchanged by reversing order, so symmetry holds. Step 3: (2R6) and (6R3) hold, but (2R3) does not, so transitivity fails.

Correct answer: B. Reflexive and symmetric but not transitive

Step 1: No self-pair has been removed, so reflexivity holds. Step 2: The two removed pairs are reverses of each other, so symmetry remains. Step 3: ((1,2)) and ((2,4)) are present, but ((1,4)) is missing, so transitivity fails.

Correct answer: B. (2^4:1)

Step 1: The number of symmetric relations on (4) elements is (2^{10}). Step 2: The number of reflexive and symmetric relations is (2^6). Step 3: The ratio is (2^{10}:2^6=2^4:1).

Correct answer: C. (9)

Step 1: Self-pairs give (5) pairs. Step 2: The condition (a+b=6) adds ((1,5),(5,1),(2,4),(4,2)); ((3,3)) is already counted. Step 3: Total pairs are (5+4=9).

Correct answer: A. Reflexive and transitive but not symmetric

Step 1: (a^2-a^2=0), so the relation is reflexive. Step 2: (2^2-1^2\ge0) is true, but (1^2-2^2\ge0) is not, so it is not symmetric. Step 3: (a^2\ge b^2) and (b^2\ge c^2) imply (a^2\ge c^2), so it is transitive.