Introduction to Relations

Step 1: (A\times A) has (n^2) ordered pairs. Step 2: Every relation can be any subset of these (n^2) pairs. Step 3: Therefore, the total number of relations is (2^{n^2}).

Step 1: (A\times A) has (n^2) pairs. Step 2: A reflexive relation must contain the (n) diagonal pairs. Step 3: The remaining (n^2-n) pairs are optional, so the count is (2^{n^2-n}).

Step 1: The (n) diagonal pairs can be chosen independently. Step 2: The off-diagonal pairs form (\frac{n(n-1)}{2}) reverse-pair groups. Step 3: Total independent choices are (n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}), giving (2^{\frac{n(n+1)}{2}}).

Step 1: Reflexivity makes all (n) diagonal pairs compulsory. Step 2: For symmetry, only the off-diagonal reverse-pair groups are independent. Step 3: There are (\frac{n(n-1)}{2}) such groups, so the count is (2^{\frac{n(n-1)}{2}}).

Step 1: From ((1,2)) and ((2,3)), transitivity requires ((1,3)), which is present. Step 2: Other possible chains either have the required pair or do not form a chain. Step 3: Since ((3,3)) is missing, do not call it reflexive.

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.