Introduction to Relations

Step 1: The universal relation contains both ((1,2)) and ((2,1)). Step 2: Since (1\ne2), antisymmetry fails. Step 3: On a set with two or more elements, the universal relation is usually not antisymmetric.

Step 1: On a one-element set, the only pair is ((1,1)). Step 2: This pair satisfies all required conditions. Step 3: On a singleton set, many relation properties can hold together.

Step 1: ((1,2)) and ((2,3)) are in the relation. Step 2: Transitivity requires ((1,3)). Step 3: Since ((1,3)) is missing, the relation is not transitive.

Step 1: For (A={1,2,3}), (A\times A) has (9) pairs. Step 2: The given relation contains all (9) pairs. Step 3: A relation containing all possible pairs is the universal relation.

Step 1: The class ({1,4}) gives (2^2=4) pairs. Step 2: The singleton classes ({2}) and ({3}) give (1+1) pairs. Step 3: Total pairs are (4+1+1=6).

Step 1: One class means all elements are related to one another. Step 2: A class with (4) elements gives (4^2=16) ordered pairs. Step 3: Hence the relation is universal and has (16) pairs.

Step 1: Singleton classes mean each element is related only to itself. Step 2: Therefore, only four self-pairs are formed. Step 3: This is the identity relation and it has (4) pairs.

Step 1: Reflexivity gives self-pairs, and symmetry gives reverse pairs. Step 2: ((1,2)) and ((2,3)) require ((1,3)) by transitivity. Step 3: To keep symmetry, ((3,1)) is also needed.

Step 1: Find pairs whose sum is (5). Step 2: The pairs are ((1,4)), ((2,3)), ((3,2)), and ((4,1)). Step 3: Therefore, there are (4) pairs.

Step 1: If (a+b=5), then (b+a=5), so it is symmetric. Step 2: No self-pair exists because (2a=5) is not possible in the set. Step 3: ((1,4)) and ((4,1)) would require ((1,1)), which is absent, so it is not transitive.

Step 1: All self-pairs ((1,1)), ((2,2)), and ((3,3)) are included. Step 2: Pairs with difference (1), namely ((1,2)), ((2,1)), ((2,3)), and ((3,2)), are also included. Step 3: Total pairs are (3+4=7).

Step 1: (|a-a|=0), so reflexivity holds. Step 2: (|a-b|=|b-a|), so symmetry holds. Step 3: ((1,2)) and ((2,3)) are present but ((1,3)) is absent, so transitivity fails.

Step 1: Even difference means the two elements have the same parity. Step 2: The odd class ({1,3}) gives (4) pairs and the even class ({2,4}) gives (4) pairs. Step 3: Total pairs are (8).

Step 1: From (1), we get four pairs with (1,2,3,4). Step 2: From (2), we get three pairs; from (3), two pairs; from (4), one pair. Step 3: Total pairs are (4+3+2+1=10).

Step 1: From (1), pairs with (2,3,4) are formed. Step 2: From (2), pairs with (3,4) are formed, and from (3), one pair with (4) is formed. Step 3: Total pairs are (3+2+1=6).

Step 1: (a<a) is never true, so it is not reflexive. Step 2: If (a<b) and (b<c), then (a<c), so it is transitive. Step 3: Do not expect symmetry in a directed order relation.

Step 1: For every (a), (a\le a), so it is reflexive. Step 2: If (a\le b) and (b\le a), then (a=b), so it is antisymmetric. Step 3: If (a\le b) and (b\le c), then (a\le c), so it is a partial order relation.

Step 1: (1) and (2) are related in both directions. Step 2: (3) and (4) are also related in both directions. Step 3: Hence the partition is ({{1,2},{3,4}}).

Step 1: Within the same equivalence class, all ordered pairs are present. Step 2: (3) and (4) are in the same class ({3,4}). Step 3: Hence ((3,4)) must belong to the relation.

Step 1: Composition (R\circ R) is checked through a chain of two pairs. Step 2: For ((1,3)) to appear, there must be some middle element (b) with (1) related to (b) and (b) related to (3). Step 3: In composition questions, look for the middle element.

Step 1: All self-pairs are present, so the relation is reflexive. Step 2: There are no reverse non-self pairs, and all forward chains are complete, so it is antisymmetric and transitive. Step 3: ((1,2)) is present but ((2,1)) is absent, so the statement that it is symmetric is false.