Types of relations

Correct answer: A. It is an equivalence relation with classes ({1,2},{3})

Step 1: All self-pairs are present, so reflexivity holds. Step 2: Both ((1,2)) and ((2,1)) are present, so symmetry holds. Step 3: All required pairs inside ({1,2}) are present and (3) is alone, so the relation is an equivalence relation.

Correct answer: A. All multiples of (4)

Step 1: (a\equiv 0 \pmod{4}) means (a) leaves remainder (0) when divided by (4). Step 2: Such integers are all multiples of (4). Step 3: An equivalence class contains all elements with the same remainder.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: For ((1,1)), the minimum is (1), so not all self-pairs occur and reflexivity fails. Step 2: The minimum does not change when order is reversed, so symmetry holds. Step 3: ((3,2)) and ((2,4)) exist, but ((3,4)) has minimum (3), so transitivity fails.

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 stays the same when order is reversed, so it is symmetric. Step 3: ((1,5)) and ((5,2)) exist, but ((1,2)) does not, so transitivity fails.

Correct answer: A. (a\ge b)

Step 1: In the inverse relation, each ((a,b)) is changed to ((b,a)). Step 2: If the original relation has (a\le b), the inverse has (b\le a). Step 3: This can be written as (a\ge b).

Correct answer: A. Equivalence relation

Step 1: Because (a=b), all self-pairs are present, so it is reflexive. Step 2: The condition that both are greater than (2) is unchanged by reversing order, so symmetry holds. Step 3: ({3,4}) forms a closed class while (1,2) are singleton classes, so transitivity holds.

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

Step 1: The condition (a+b=6) relates (1) with (5), and (2) with (4). Step 2: Since (3+3=6), (3) stays in its own class. Step 3: Hence the classes are ({1,5},{2,4},{3}).

Correct answer: A. Equivalence relation

Step 1: (a-a=0) is an integer, so reflexivity holds. Step 2: If (a-b) is an integer, then (b-a) is also an integer, so symmetry holds. Step 3: The sum of two integer differences is an integer, so transitivity holds.

Correct answer: A. Equivalence relation

Step 1: (a^2-a^2=0), so reflexivity holds. Step 2: If (a^2-b^2=0), then (b^2-a^2=0), so symmetry holds. Step 3: Equality of squares passes through a third element, so transitivity holds.

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

Step 1: Option A has no self-pair, so it is not reflexive. Step 2: From ((1,2)) and ((2,3)), ((1,3)) is present. Step 3: No other chain forces a new missing pair, so it is transitive.

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

Step 1: Option A misses ((3,3)), so it 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 (3) is separate, so transitivity holds.

Correct answer: A. (R\cap S) will be reflexive

Step 1: Since (R) is reflexive, every ((a,a)) is in (R). Step 2: Since (S) is also reflexive, every ((a,a)) is in (S). Step 3: Therefore each self-pair is common to both, so (R\cap S) is reflexive.

Correct answer: A. (R\cup S) will be symmetric

Step 1: If a pair is in (R\cup S), it is in (R) or in (S). Step 2: In that relation, symmetry gives the reverse pair. Step 3: Hence the reverse pair is also in (R\cup S), so the union remains symmetric.

Correct answer: A. It is not always transitive

Step 1: Transitivity requires a third pair from two connected pairs. Step 2: In (R\cup S), one pair may come from (R) and the other from (S), so the required third pair may be missing. Step 3: Hence the union of transitive relations is not always transitive.

Correct answer: A. (R\cap S) will be transitive

Step 1: If ((a,b)) and ((b,c)) are both in (R\cap S), they are in both (R) and (S). Step 2: Since (R) is transitive, ((a,c)) is in (R); since (S) is transitive, it is also in (S). Step 3: Therefore ((a,c)) is in (R\cap S).

Correct answer: A. Reflexive and symmetric but not transitive

Step 1: The difference of an element from itself is (0), so it is reflexive. Step 2: Difference remains the same when order is reversed, so symmetry holds. Step 3: (1R3) and (3R4) are true, but (1R4) is false because the difference is (3), so it is not transitive.

Correct answer: A. Equivalence relation

Step 1: The condition (a=b) gives all self-pairs, so reflexivity holds. Step 2: (a+b=5) remains true after reversing order, so symmetry holds. Step 3: The relation forms closed classes ({1,4}) and ({2,3}), so transitivity also holds.

Correct answer: A. It is the universal relation

Step 1: A set with three elements has (3^2=9) ordered pairs. Step 2: The given relation contains all (9) pairs. Step 3: Therefore it is (A\times A), the universal relation.

Correct answer: A. (16)

Step 1: A reflexive relation must contain self-pairs, but other pairs may also be included. Step 2: In the maximum case, the relation is (A\times A). Step 3: For (4) elements, (A\times A) has (4^2=16) pairs.

Correct answer: A. (5)

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

Correct answer: A. (8)

Step 1: In an equivalence relation, all ordered pairs within each class are included. Step 2: ({1,3}) gives (2^2=4) pairs and ({2,4}) gives (2^2=4) pairs. Step 3: Total pairs are (4+4=8).

Correct answer: C. It is symmetric and irreflexive but not transitive.

The correct answer is C. First test symmetry: if a+b is odd, then b+a has exactly the same value, so bRa also holds. Next test irreflexivity: for every a, a+a=2a is even, never odd; hence no element is related to itself. Finally test transitivity. We have 1R2 because 1+2=3 is odd, and 2R3 because 2+3=5 is odd, but 1R3 is false because 1+3=4 is even. Therefore the relation is not transitive. A is wrong because the relation is not transitive and is symmetric. B is wrong because it is symmetric and irreflexive. D is wrong because an equivalence relation must be reflexive, symmetric, and transitive. A useful cue is that adding two numbers of opposite parity gives odd, but two successive such steps return to the same parity.

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

The correct answer is A. The relation in A is the complete relation on the subset {1,2}: it contains (1,1), (2,2), (1,2), and (2,1). Any two-step composition whose middle element is 1 or 2 produces a pair still in this set, so it is transitive. Both directions of the only off-diagonal connection are present, so it is symmetric. However, reflexivity on A requires all four diagonal pairs, including (3,3) and (4,4); these are absent. Therefore it is not reflexive on A. B is neither symmetric nor transitive, since (1,2) and (2,3) lack (1,3). C is reflexive as well as symmetric and transitive because it contains every ordered pair. D is not symmetric because (1,2) appears without (2,1). The key distinction is that transitivity and reflexivity are tested on the whole stated set A.