Introduction to Relations

Option C is correct because two elements are related exactly when they belong to the same equivalence class. The elements 1 and 3 lie in different classes, {1,2} and {3,4}, so (1,3) cannot belong to R. In contrast, (1,2), (2,1), and (4,3) connect elements within one class and therefore must belong to the relation.

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.

Option C is correct. Every equivalence class contributes all ordered pairs formed from its own elements. Class {1,2} contributes (1,1),(1,2),(2,1),(2,2), giving 4 pairs, and class {3,4} contributes another 4 pairs. Pairs joining the two different classes are excluded, so the total is 4+4=8, not 16.

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.

Step 1: The (4) diagonal pairs can be chosen or not chosen independently. Step 2: For each pair of distinct elements, there are three choices: one direction, the reverse direction, or none. There are (\frac{4\cdot 3}{2}=6) such pairs. Step 3: Hence the total number is (2^4\cdot 3^6).

Step 1: Reflexivity makes all (5) diagonal pairs compulsory. Step 2: For every pair of distinct elements, antisymmetry gives three choices. There are (\frac{5\cdot 4}{2}=10) such pairs. Step 3: Therefore the number of relations is (3^{10}).

Step 1: If symmetry and antisymmetry hold together, no off-diagonal pair between distinct elements can be included. Step 2: Only diagonal pairs are independently optional. Step 3: With (4) diagonal pairs, the total number is (2^4).

Step 1: Reflexivity requires all diagonal pairs. Step 2: Being both symmetric and antisymmetric allows no off-diagonal pair. Step 3: Thus only the identity relation is possible, so the number is (1).

Step 1: An equivalence relation partitions the set into disjoint classes. Step 2: The number of partitions of a (4)-element set is (15). Step 3: Hence the number of equivalence relations is (15).

Step 1: Exactly two classes means partitioning (4) elements into two non-empty disjoint parts. Step 2: There are (2^4-2=14) non-empty proper subsets, but each partition is counted twice. Step 3: Therefore the count is (\frac{14}{2}=7).

Step 1: An equivalence relation is determined by its partition into classes. Step 2: Choose the class of size (2) in (\binom{5}{2}=10) ways. Step 3: The remaining three elements automatically form the other class.

Step 1: We only need to choose the single two-element class. Step 2: Two elements can be chosen from (4) elements in (\binom{4}{2}=6) ways. Step 3: The remaining two elements become singleton classes.

Step 1: (7) leaves remainder (1) when divided by (6). Step 2: Its class contains all integers of the form (6k+1). Step 3: (6k+7) represents the same set, but (6k+1) is the standard simpler form.

Step 1: (a+2a=3a) is always divisible by (3), so it is reflexive. Step 2: (a+2b\equiv 0\pmod 3) implies (b+2a\equiv 0\pmod 3). Step 3: This condition is equivalent to (a\equiv b\pmod 3), so transitivity also holds and no property fails.