Introduction to Relations

Step 1: Diagonal pairs are present, so reflexivity holds. Step 2: Each non-diagonal pair has its reverse, so symmetry holds. Step 3: From ((1,2)) and ((2,3)), ((1,3)) is required but missing; hence transitivity fails.

Step 1: The condition demands a third pair from two connected pairs. Step 2: This is exactly the definition of transitivity. Step 3: In exams, whenever you see ((a,b)) and ((b,c)), immediately check for ((a,c)).

Step 1: (|a-b|) being even means (a) and (b) have the same parity. Step 2: Thus one class is ({1,3}) and the other is ({2,4}). Step 3: In such questions, identify the hidden grouping first.

Step 1: (2) leaves remainder (2) when divided by (3). Step 2: In (A), the numbers with remainder (2) are (2) and (5). Step 3: While forming an equivalence class, include only elements from the given set.

Step 1: (a-a=0) is not odd, so it is not reflexive. Step 2: If (a-b) is odd, then (b-a) is also odd, so it is symmetric. Step 3: (1R2) and (2R3) hold, but (1R3) fails because the difference is even.

Step 1: All diagonal 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: No chain creates a missing required pair, so it is transitive.

Step 1: Since (R) and (S) are reflexive, every ((a,a)) belongs to both. Step 2: Pairs common to both remain in the intersection. Step 3: Hence all diagonal pairs remain in (R\cap S), so it is reflexive.

Step 1: If a pair is in (R\cup S), it lies in (R) or (S). Step 2: In whichever relation it lies, symmetry gives its reverse in the same relation. Step 3: Therefore the reverse pair also lies in the union.

Step 1: Transitivity needs a third pair from two connected pairs. Step 2: In a union, one pair may come from (R) and the other from (S), so the needed third pair may be absent. Step 3: Do not assume the union of transitive relations is transitive.

Step 1: ((1,2)) and ((2,3)) form a chain. Step 2: Transitivity then requires ((1,3)). Step 3: In such questions, match the middle element and connect the first element to the third.

Step 1: Symmetry requires ((b,a)) whenever ((a,b)) is present. Step 2: Since ((1,2)) is present, ((2,1)) must be added. Step 3: Diagonal pairs are their own reverses, but the missing reverse pair is the key issue here.

Step 1: Reflexivity requires ((1,1),(2,2),(3,3)). Step 2: None of these diagonal pairs is present in the given relation. Step 3: Therefore, all three diagonal pairs must be added.

Step 1: If (a+b=5), then (b+a=5) also holds. Step 2: So whenever ((a,b)) is in the relation, ((b,a)) is also in it. Step 3: But pairs such as ((1,1)) are absent, so it is not reflexive.

Step 1: (a-a=0), which is not greater than (0), so it is not reflexive. Step 2: (3R2) holds but (2R3) does not, so it is not symmetric. Step 3: If (a>b) and (b>c), then (a>c), so it is transitive.

Step 1: Every number has the same parity as itself, so the relation is reflexive. Step 2: If (a) and (b) have the same parity, then (b) and (a) also have the same parity. Step 3: Same parity passes through a middle element, so the relation is transitive.

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.