Introduction to Relations

Option A is correct because whenever (a,b) belongs to R, the reverse pair (b,a) also belongs to R: both (1,2) and (2,1) are present, while diagonal pairs cause no problem. It is not reflexive because (3,3) is absent, not antisymmetric because both distinct pairs (1,2) and (2,1) occur, and not universal because several pairs involving 3 are missing.

Step 1: From ((1,2)) and ((2,3)), transitivity requires ((1,3)). Step 2: This pair is not given in the relation. Step 3: Quickly identify the third pair formed by two connected pairs.

Step 1: For (A={1,2}), (A\times A) has four pairs. Step 2: The given relation contains all four pairs. Step 3: Hence it equals (A\times A), the universal relation.

Step 1: A relation from (A) to (B) is a subset of (A\times B). Step 2: (A\times B) has (3\times3=9) pairs. Step 3: Hence the number of relations is (2^9).

All three diagonal pairs are present, so R is reflexive. The non-diagonal pair (2,3) appears together with (3,2), so symmetry also holds. However, antisymmetry requires that if both (a,b) and (b,a) occur, then a=b. Here both pairs occur although 2≠3, so R is not antisymmetric. Every listed ordered pair still belongs to A×A.

Step 1: All self-pairs are present, so it is reflexive. Step 2: Both ((1,4)) and ((4,1)) are present, so it is symmetric. Step 3: The formed transitive chains are completed by existing self-pairs, so it is an equivalence relation.

Step 1: Symmetry requires ((2,1)) along with ((1,2)). Step 2: ((2,1)) is not in the relation. Step 3: A cyclic-looking relation is not necessarily symmetric.

Step 1: From ((1,2)) and ((2,3)), transitivity requires ((1,3)). Step 2: ((1,3)) is not in the relation. Step 3: Therefore, the relation is not transitive.

Step 1: It contains all self-pairs, so it is reflexive. Step 2: Every pair has its reverse, so it is symmetric. Step 3: Since all possible pairs are present, transitivity also holds.

Step 1: (a\equiv b \pmod{2}) means same parity. Step 2: The odd elements like (1) are (1) and (3). Step 3: Hence the class of (1) is ({1,3}).

Step 1: Dividing (2) by (3) gives remainder (2). Step 2: Dividing (5) by (3) also gives remainder (2). Step 3: Therefore, the class of (2) is ({2,5}).

Step 1: This condition says that two distinct elements cannot be related in both directions. Step 2: This is the identity of antisymmetry. Step 3: Understand symmetry and antisymmetry separately through reverse pairs.

Step 1: Every number is less than or equal to itself, so it is reflexive. Step 2: If (a\le b) and (b\le a), then (a=b), so it is antisymmetric. Step 3: The (\le) chain is also transitive.

Step 1: Antisymmetry fails when both ((a,b)) and ((b,a)) exist for distinct (a,b). Step 2: Here ((1,2)) is present but ((2,1)) is not. Step 3: So antisymmetry is not violated, and the relation is antisymmetric.

Step 1: In antisymmetry, pairs in both directions should not occur between distinct elements. Step 2: Here (1) and (2) are distinct, yet both ((1,2)) and ((2,1)) are present. Step 3: Therefore, antisymmetry fails.

Step 1: A partial order relation needs three special properties. Step 2: These are reflexivity, antisymmetry and transitivity. Step 3: Equivalence relation uses symmetry, while partial order uses antisymmetry.

Step 1: In divisibility, the least element is the one that divides every element. Step 2: (1) divides (1), (2), and (3). Step 3: In divisibility questions, check (1) carefully.

Step 1: A least element must divide every element in the set. Step 2: (2) does not divide (3), and (3) does not divide (2). Step 3: Hence no element divides all elements of this set.

Step 1: A greatest element is divisible by every element of the set. Step 2: (1), (2), (3), and (6) all divide (6). Step 3: Therefore, (6) is the greatest element.

Step 1: All self-pairs are present, so the relation is reflexive. Step 2: No reverse pair exists for distinct elements, so it is antisymmetric. Step 3: ((1,2)) and ((2,3)) require ((1,3)), which is present, so it is transitive.