Introduction to Relations

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.

Step 1: (a-a=0) is an integer, so the relation is reflexive. Step 2: If (a-b) is an integer, then (b-a) is also an integer. Step 3: The sum of two integer differences is an integer, so transitivity also holds.

Step 1: (a-a=0) is not irrational, so it is not reflexive. Step 2: If (a-b) is irrational, then (b-a) is also irrational, so it is symmetric. Step 3: (\sqrt{2}R0) and (0R\sqrt{2}) hold, but (\sqrt{2}R\sqrt{2}) fails, so it is not transitive.

Step 1: For every (a), (a^3=a^3), so it is reflexive. Step 2: Equality remains true when reversed, so it is symmetric. Step 3: If (a^3=b^3) and (b^3=c^3), then (a^3=c^3), so it is transitive.

Step 1: (a^2\le a^2) gives reflexivity. Step 2: If (a^2\le b^2) and (b^2\le c^2), then (a^2\le c^2), so it is transitive. Step 3: (1^2\le 2^2) is true but (2^2\le 1^2) is false, so it is not symmetric.

Step 1: For every (a), (|a|=|a|), so it is reflexive. Step 2: If (|a|=|b|), then (|b|=|a|), so it is symmetric. Step 3: Equality of absolute values passes through a middle element, so it is transitive.

Step 1: For real numbers, (a^2+b^2=0) is possible only when (a=0,b=0). Step 2: Thus the relation contains only ((0,0)), which keeps symmetry and transitivity true. Step 3: It is not reflexive because ((a,a)) is not present for every real (a).

Step 1: The empty relation has no pair, so no counterexample breaks symmetry or transitivity. Step 2: For non-empty (A), reflexivity needs ((a,a)), but no diagonal pair is present. Step 3: Hence it is symmetric and transitive but not reflexive.

Step 1: From ((1,2)) and ((2,3)), ((1,3)) is required and present. Step 2: No other chain creates a new required pair. Step 3: Since diagonal and reverse pairs are missing, do not call it reflexive or symmetric.

Step 1: From ((1,2)) and ((2,4)), transitivity requires ((1,4)), which is present. Step 2: ((3,3)) only requires itself again in a chain, and it is present. Step 3: All required cases are satisfied, so the relation is transitive.

Step 1: ((2,1)) and ((1,2)) are present. Step 2: Transitivity requires ((2,2)) from these two pairs. Step 3: ((1,2)) and ((2,1)) require ((1,1)), which is already present.

Step 1: ((1,2)) and ((2,3)) require ((1,3)). Step 2: ((2,3)) and ((3,4)) require ((2,4)). Step 3: Then ((1,3)) and ((3,4)) require ((1,4)), so (3) pairs are needed.

Step 1: ((1,2)) and ((2,3)) connect all three elements into one equivalence class. Step 2: In one class, every ordered pair between the elements must be present. Step 3: A class of (3) elements gives (3^2=9) pairs.

Step 1: ((1,2)) forms the class ({1,2}), and ((3,4)) forms the class ({3,4}). Step 2: Each two-element class contributes (2^2=4) ordered pairs. Step 3: Therefore the total number of pairs is (4+4=8).

Step 1: Both relations contain all diagonal pairs, so the intersection does too. Step 2: If a pair is in the intersection, its reverse is in both relations and hence in the intersection. Step 3: Transitivity also remains valid because both relations provide the required pair.

Step 1: The union of two equivalence relations may remain reflexive and symmetric. Step 2: But one pair may come from one relation and the next pair from the other, so the required third pair may be missing. Step 3: Therefore the union is not always an equivalence relation.