Introduction to Relations

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.

Step 1: The union contains all diagonal pairs and the needed reverse pairs. Step 2: It also contains ((1,2)) and ((2,3)). Step 3: Transitivity requires ((1,3)), which is missing, so it is not an equivalence relation.

Step 1: The inverse relation reverses the direction of every pair. Step 2: Symmetry means the relation behaves the same after reversing pairs. Step 3: Hence (R^{-1}) is symmetric exactly when (R) is symmetric.

Step 1: A reflexive (R) contains every ((a,a)). Step 2: The inverse of ((a,a)) is again ((a,a)). Step 3: Therefore all diagonal pairs remain in (R^{-1}).

Step 1: Since (R) is reflexive, its inverse is also reflexive. Step 2: Since (R) is symmetric, (R^{-1}=R). Step 3: Because (R) is transitive, the same property remains in its inverse.

Step 1: If ((a,b)) and ((b,c)) are in (R^{-1}), then ((b,a)) and ((c,b)) are in (R). Step 2: Transitivity of (R) gives ((c,a)). Step 3: Reversing it gives ((a,c)) in (R^{-1}), so (R^{-1}) is transitive.