Let's analyze each relation one by one:
A) R₁ = {(a, b) | -1 <= a - b ≤ 1} on ℝ (the set of real numbers)
Reflexive: For any element a in ℝ, we have -1 <= a - a = 0 ≤ 1, so (a, a) is always in R₁. Therefore, R₁ is reflexive.
Symmetric: If (a, b) is in R₁, then -1 <= a - b ≤ 1. Since subtraction is symmetric, we have -1 <= -(a - b) ≤ 1, which simplifies to -1 <= b - a ≤ 1. Thus, (b, a) is also in R₁. Therefore, R₁ is symmetric.
Antisymmetric: If (a, b) and (b, a) are both in R₁, we have -1 <= a - b ≤ 1 and -1 <= b - a ≤ 1. Combining these inequalities, we get -2 <= a - b + b - a = 0 ≤ 2. This means a - b = 0, so a = b. Therefore, R₁ is antisymmetric.
Transitive: If (a, b) and (b, c) are both in R₁, we have -1 <= a - b ≤ 1 and -1 <= b - c ≤ 1. Adding these inequalities, we get -2 <= a - b + b - c = a - c ≤ 2. Therefore, (a, c) is in R₁. Hence, R₁ is transitive.
Since R₁ is reflexive, symmetric, and transitive, it is an equivalence relation.
B) R₂ = {(1, 1), (1, 3), (1, 4), (2, 2), (2, 4), (3, 1)} on {1, 2, 3, 4}
Reflexive: All elements in {1, 2, 3, 4} appear as (x, x) pairs in R₂, so R₂ is reflexive.
Symmetric: R₂ is not symmetric because, for example, (1, 3) is in R₂, but (3, 1) is not.
Antisymmetric: R₂ is antisymmetric because there are no distinct pairs (a, b) and (b, a) where both are in R₂.
Transitive: R₂ is not transitive because, for example, (1, 3) and (3, 1) are both in R₂, but (1, 1) is not.
Since R₂ is reflexive but not symmetric, antisymmetric, or transitive, it is not an equivalence relation.
C) R₃ = {(X, Y) | X ∩ Y = ∅} on ℘({a, b, c})
Reflexive: For any set X in ℘({a, b, c}), we have X ∩ X = X, which is only empty if X is empty. Since the empty set is a subset of any set, we can conclude that (X, X) is not in R₃ for any non-empty set X. Therefore, R₃ is not reflexive.
Symmetric: R₃ is not symmetric because, for example, ({a}, ∅) is in R₃, but (∅, {a}) is not.
Antisymmetric: R₃ is antisymmetric because if (X, Y) and (Y, X) are both in R₃, then X ∩ Y = ∅ and Y ∩ X = ∅, implying X = ∅ and Y = ∅. Therefore, X = Y. However, since R₃ is not reflexive, it doesn't have any such pairs.
Transitive: R₃ is transitive because if (X, Y) and (Y, Z) are both in R₃, then X ∩ Y = ∅ and Y ∩ Z = ∅. Since the intersection of two empty sets is empty, we have X ∩ Z = (X ∩ Y) ∩ (Y ∩ Z) = ∅ ∩ ∅ = ∅. Therefore, (X, Z) is in R₃.
Since R₃ is not reflexive, not symmetric, but antisymmetric and transitive, it is not an equivalence relation.
D) R₄ = {(a, b) is a friend of} on the set of all people
Reflexive: For any person a, a is a friend of themselves. Therefore, (a, a) is in R₄ for all people a, and R₄ is reflexive.
Symmetric: If a is a friend of b, then b is also a friend of a. Therefore, if (a, b) is in R₄, then (b, a) is also in R₄. R₄ is symmetric.
Antisymmetric: R₄ is not antisymmetric because there can be cases where both (a, b) and (b, a) are in R₄, indicating mutual friendship between a and b.
Transitive: If a is a friend of b and b is a friend of c, it does not necessarily mean that a is a friend of c. Therefore, R₄ is not transitive.
Since R₄ is reflexive and symmetric but not antisymmetric or transitive, it is not an equivalence relation.
E) R₅ = {((a, b), (c, d)) | ad = bc} on ℤ* × ℤ
Reflexive: For any element ((a, b), (a, b)) in ℤ* × ℤ, we have a * b = b * a, which is true for any integers a and b. Therefore, R₅ is reflexive.
Symmetric: If ((a, b), (c, d)) is in R₅, then ad = bc. Swapping the positions, we have (c, d), (a, b)) with cd = ba. Therefore, R₅ is symmetric.
Antisymmetric: R₅ is not antisymmetric because, for example, ((1, 2), (2, 4)) and ((2, 4), (1, 2)) are both in R₅, indicating a non-trivial equivalence.
Transitive: If ((a, b), (c, d)) and ((c, d), (e, f)) are both in R₅, then ad = bc and cf = de. Multiplying these equations, we get (ad)(cf) = (bc)(de), which simplifies to (adcf) = (bcde). Therefore, ((a, b), (e, f)) is in R₅, and R₅ is transitive.
Since R₅ is reflexive, symmetric, and transitive but not antisymmetric, it is not an equivalence relation.
F) R₆ = {(a, b) | a/b ∈ ℤ} on ℕ (the set of natural numbers)
Reflexive: For any natural number a, a divided
