Introduction to Relations

Step 1: For (a=1), there are (4) choices; for (a=2), (3); for (a=3), (2); and for (a=4), (1). Step 2: Total pairs are (4+3+2+1=10). Step 3: To count ordered pairs, fix the first element and count possible second elements.

Step 1: The sum is even when both numbers are odd or both are even. Step 2: (A) has (3) odd and (2) even numbers, so pairs are (3^2+2^2=9+4=13). Step 3: Since the pairs are ordered, use square counts for each group.

Step 1: (3) divides (12), so ((3,12)) belongs to the relation. Step 2: (12) does not divide (3), so ((12,3)) does not belong. Step 3: This is why the divisibility relation is generally not symmetric.

Step 1: (aRb) means (a) and (b) belong to the same group. Step 2: In an equivalence relation, elements of the same group have the same equivalence class. Step 3: Equivalence classes are either identical or disjoint.

Step 1: All diagonal pairs are present, so reflexivity holds. Step 2: The reverses of ((1,2)) and ((1,3)) are also present, so symmetry holds. Step 3: From ((2,1)) and ((1,3)), ((2,3)) is required but missing; hence transitivity fails.

Step 1: The identity relation contains only pairs of the form ((a,a)). Step 2: Since (A) has (4) elements, it has (4) diagonal pairs. Step 3: Do not confuse identity relation with the universal relation, which would have (16) pairs.

Step 1: (A\times B) has (3\times 2=6) ordered pairs. Step 2: Any relation from (A) to (B) is a subset of (A\times B). Step 3: Hence the total number of relations is (2^6).

Step 1: (A\times A) contains all possible ordered pairs. Step 2: When a relation contains all these pairs, it is called the universal relation. Step 3: The identity relation contains only diagonal pairs, so keep the two ideas separate.

Step 1: Antisymmetry only forbids two-way pairs between distinct elements. Step 2: Here ((1,2)) is present but ((2,1)) is not, and ((1,3)) is present but ((3,1)) is not. Step 3: Diagonal pairs do not violate antisymmetry.

Step 1: ((1,2)) and ((2,1)) show that (1) and (2) are in the same class. Step 2: (3) is related only to itself, so it forms the separate class ({3}). Step 3: While forming equivalence classes, group elements connected by the relation.

Step 1: Reflexivity makes the three self-pairs compulsory. Step 2: There are (3) unordered pairs of distinct elements, and each pair-group may be included or excluded. Step 3: Therefore, the total number is (2^3).

Step 1: Symmetry asks for reverse pairs between distinct elements. Step 2: Antisymmetry forbids both directions between distinct elements, so no non-self pair can be included. Step 3: Only the three self-pairs are free, giving (2^3) relations.

Step 1: Reflexivity makes the three self-pairs compulsory. Step 2: For each unordered pair of distinct elements, there are three choices: one direction, the other direction, or none. Step 3: There are (3) such pairs, so the number is (3^3).

Step 1: Transitivity needs ((a,c)) from ((a,b)) and ((b,c)). Step 2: ((1,2)) and ((2,3)) require ((1,3)), which is missing. Step 3: For minimum addition, add only the required missing pair.

Step 1: ((1,2)) and ((2,3)) require ((1,3)). Step 2: ((2,3)) and ((3,4)) require ((2,4)), and then ((1,3)) and ((3,4)) require ((1,4)). Step 3: In long chains, also check pairs created during closure.

Option A is correct. The relation already contains every diagonal pair, so it is reflexive, and every listed off-diagonal pair has its reverse, so it is symmetric. Transitivity requires (1,3) from (1,2) and (2,3), and symmetry then requires (3,1). After adding both, all pairs of A×A are present, making the relation an equivalence relation.

Step 1: In an equivalence relation, all ordered pairs within each class are included. Step 2: Each class has (2) elements, so each contributes (2^2=4) pairs. Step 3: Total pairs are (4+4=8).

Step 1: The class ({1,2,3}) gives (3^2=9) ordered pairs. Step 2: The class ({4}) gives (1^2=1) pair. Step 3: Total pairs are (9+1=10).

Step 1: Dividing (1) by (3) gives remainder (1). Step 2: Dividing (4) by (3) also gives remainder (1). Step 3: Therefore, the class of (1) is ({1,4}).

Step 1: Dividing (3) by (4) gives remainder (3). Step 2: Dividing (7) by (4) also gives remainder (3). Step 3: Elements with the same remainder lie in the same equivalence class.