Introduction to Relations

Step 1: Self-pairs are of the form ((a,a)). Step 2: A relation with only such pairs is the identity relation when every element has its self-pair. Step 3: Identify identity relation by the absence of pairs with different elements.

Step 1: An empty relation contains no ordered pair. Step 2: (\varnothing) denotes the empty set. Step 3: If even one pair is present, the relation is not empty.

Step 1: A universal relation contains all pairs of (A\times A). Step 2: For (A={1,2}), four pairs are possible, and the first option has all four. Step 3: Match both the count and the pairs before choosing.

Step 1: It has all self-pairs, so it is reflexive. Step 2: Every pair has its reverse, and all possible pairs are included, so transitivity also holds. Step 3: A universal relation on a set can be an example of an equivalence relation.

Step 1: Symmetry is necessary for an equivalence relation. Step 2: ((1,2)) is present but ((2,1)) is missing, so symmetry fails. Step 3: Do not forget to check all three properties for equivalence relation.

Step 1: If a relation is exactly equal to (A\times A), it contains all pairs. Step 2: A relation containing all pairs is universal. Step 3: When (R=A\times A) is given, identify it as universal immediately.

Step 1: A relation on (A) is any subset of (A\times A). Step 2: Therefore, if (R\subseteq A\times A), then (R) is a relation on (A). Step 3: Always think of a relation as a set of ordered pairs.

Step 1: In an identity relation, every element is related to itself. Step 2: Since (2) is in the set, ((2,2)) will be included. Step 3: Pairs with different elements are not included in the identity relation.

Step 1: In (A\times A), elements of (A) appear in both positions. Step 2: Therefore, total pairs are (5\times5=25). Step 3: Use (n^2) for counting pairs in (A\times A).

Step 1: In a reflexive relation, every element must have its self-pair. Step 2: If any element is left out, reflexivity is not complete. Step 3: Remember reflexivity as self-relation for all elements.

Step 1: (A\times B) has (3\times2=6) ordered pairs. Step 2: Each subset forms a relation, so the total number of relations is (2^6). Step 3: In such questions first count the pairs in the Cartesian product.

Step 1: (A\times A) has (3^2=9) pairs. Step 2: For reflexivity, (3) diagonal pairs are fixed and the remaining (6) pairs are optional. Step 3: Hence the number is (2^6).

Step 1: The three diagonal pairs can be chosen or left independently. Step 2: The three off-diagonal reverse pairs can also be chosen or left. Step 3: There are (6) independent choices, so the number is (2^6).

Step 1: Reflexivity fixes all three diagonal pairs. Step 2: There are (3) off-diagonal reverse pairs, and each pair is optional. Step 3: Hence the total number of such relations is (2^3).

Step 1: (a-a=0) is even, so the relation is reflexive. Step 2: If (a-b) is even, then (b-a) is also even, and adding two even differences gives an even difference. Step 3: Since all three properties hold, it is an equivalence relation.

Step 1: For every (a), (a\le a) is true, so the relation is reflexive. Step 2: From (a\le b) and (b\le c), we get (a\le c), so it is transitive. Step 3: (1\le2) is true but (2\le1) is false, so it is not symmetric.

Step 1: (a<a) is never true, so the relation is not reflexive. Step 2: If (a<b) and (b<c), then (a<c), so it is transitive. Step 3: Relations based on less than are generally not symmetric.

Step 1: Every natural number divides itself, so the relation is reflexive. Step 2: If (a\mid b) and (b\mid c), then (a\mid c), so it is transitive. Step 3: (2\mid4) is true but (4\mid2) is false, so it is not symmetric.

Step 1: (a+a=2a) is always even, so the relation is reflexive. Step 2: If (a+b) is even, then (b+a) is also even, so it is symmetric. Step 3: Same parity carries through the chain, so it is transitive.

Step 1: (a-a=0) is divisible by (3), so the relation is reflexive. Step 2: If (a-b) is divisible by (3), then (b-a) is also divisible by (3). Step 3: Adding divisible differences again gives a divisible difference, so it is an equivalence relation.