Introduction to Relations

Step 1: Since (1+4=5), ((1,4)) and ((4,1)) are in the relation. Step 2: Transitivity would require ((1,1)). Step 3: ((1,1)) is absent because (1+1\ne5), so transitivity fails.

Step 1: (1) and (2) are related in both directions, so they are in one class. Step 2: (3) is related only to itself. Step 3: Hence the classes are ({1,2}) and ({3}).

Step 1: In an equivalence relation, pairs are formed only within the same class. Step 2: (1) is in ({1,2}), while (3) is in ({3,4}). Step 3: Hence ((1,3)) will not be in the relation.

Step 1: (2) is related to (1), and (1) is related to (3). Step 2: Transitivity requires (2) to be related to (3). Step 3: To preserve symmetry, both ((2,3)) and ((3,2)) must be added.

Step 1: The inverse relation contains the reverse ((2,1)) of ((1,2)). Step 2: ((2,1)) is not in the original relation. Step 3: While comparing inverse and original relation, reverse each non-self pair.

Step 1: All self-pairs make it reflexive. Step 2: No reverse pair exists for distinct elements, so it is antisymmetric. Step 3: ((1,2)) and ((2,3)) require ((1,3)), which is present, so it is transitive.

Step 1: Every distinct pair has its reverse, so symmetry holds. Step 2: ((1,1)), ((2,2)), and ((3,3)) are absent. Step 3: Hence the relation is symmetric but not reflexive.

Step 1: The main chain ((1,2)) and ((2,3)) requires ((1,3)). Step 2: ((1,3)) is present. Step 3: No other chain asks for a new missing pair, so the relation is transitive.

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: All ordered pairs within each class are included. Step 2: Both classes have (2) elements, so each class gives (2^2=4) pairs. Step 3: Total pairs are (4+4=8).

Step 1: One equivalence class means every element is related to every other element. Step 2: Therefore, all pairs of (A\times A) are in the relation. Step 3: Such a relation is the universal relation.

Step 1: Singleton classes mean no two distinct elements are related. Step 2: Every element is related only to itself. Step 3: Hence the relation is the identity relation.

Step 1: Equivalence also requires transitivity. Step 2: ((1,2)) and ((2,3)) require ((1,3)). Step 3: Reflexivity gives self-pairs, but this third pair comes from transitivity.

Step 1: A partial order relation is antisymmetric. Step 2: In antisymmetry, if both ((a,b)) and ((b,a)) are present, then (a=b). Step 3: Here (1\ne2), so this situation is impossible in a partial order.

Step 1: Reflexivity and transitivity do not automatically give symmetry. Step 2: Symmetry requires the reverse of every pair. Step 3: Therefore, reverse pairs must be checked separately.

Step 1: All self-pairs are present, so reflexivity does not fail. Step 2: Both ((1,3)) and ((3,1)) are present while (1\ne3). Step 3: This definitely breaks antisymmetry.

Step 1: Symmetry needs the reverse of every non-self pair. Step 2: ((2,3)) is present, but its reverse ((3,2)) is absent. Step 3: Even one missing reverse pair prevents symmetry.

Step 1: All self-pairs make it reflexive. Step 2: Reverse pairs of distinct elements are absent, so it is antisymmetric. Step 3: Forward chains such as ((1,2),(2,3)) giving ((1,3)), and ((2,3),(3,4)) giving ((2,4)), are complete.

Step 1: A partial order requires transitivity. Step 2: ((1,3)) and ((3,4)) require ((1,4)), and ((1,2)) with ((2,4)) also requires ((1,4)). Step 3: Since ((1,4)) is missing, transitivity fails.

Step 1: Reflexivity makes all self-pairs compulsory. Step 2: Being both symmetric and antisymmetric forbids pairs between distinct elements. Step 3: Therefore, only the identity relation is possible.