Equivalence relation

Step 1: In the identity relation, every element is related only to itself. Step 2: Therefore it contains only ((1,1),(2,2),(3,3),(4,4)). Step 3: Keep the identity relation distinct from the universal relation.

Step 1: The universal relation contains all possible ordered pairs from the set. Step 2: For (A={1,2}), there are (2^2=4) pairs. Step 3: The option with all four pairs is the universal relation.

Step 1: The universal relation is (A\times A). Step 2: If (A) has (6) elements, then (A\times A) has (6^2=36) pairs. Step 3: In a universal relation, every element is related to every element.

Step 1: The identity relation contains only pairs of the form ((a,a)). Step 2: For (6) elements, there are (6) such diagonal pairs. Step 3: In an identity relation, the number of pairs equals the number of elements.

Step 1: Classes of an equivalence relation cover the whole set. Step 2: Distinct classes do not overlap. Step 3: Therefore every equivalence relation gives a partition of the set.

Step 1: Each block of a partition is treated as an equivalence class. Step 2: Any two elements in the same block are related. Step 3: Elements from different blocks are not related.

Step 1: In the relation formed from a partition, pairs are taken from within the same block. Step 2: (2) and (5) are together in the block ({2,5}). Step 3: Therefore ((2,5)) belongs to the relation.

Step 1: (1) is in ({1,6}), while (2) is in ({2,5}). Step 2: They are in different blocks, so they are not related. Step 3: No pair is formed between different classes.

Step 1: Every integer is either even or not even, so it is related to itself in its group. Step 2: Being in the same group is symmetric. Step 3: The relation divides integers into even and odd classes, so it is an equivalence relation.

Step 1: Reflexivity requires (aRa) for every integer (a). Step 2: A number cannot be one even and one odd with itself. Step 3: Therefore reflexivity fails, so the relation is not an equivalence relation.

Step 1: A sum is even when the two numbers have the same parity. Step 2: The odd class ({1,3}) gives (2^2=4) pairs, and the even class ({2,4}) gives (2^2=4) pairs. Step 3: The total is (8) pairs.

Step 1: Reflexivity requires (aRa). Step 2: (a-a=0), and (0) is not irrational. Step 3: Thus reflexivity fails; all three properties are needed for equivalence.

Step 1: ((\sqrt{3}+2)-\sqrt{3}=2). Step 2: Since (2) is rational, (\sqrt{3}+2) belongs to the same equivalence class. Step 3: Subtracting the two numbers is the quickest check.

Step 1: Reflexivity requires (aRa) for every (a). Step 2: (a+a=2a) is always even, not odd. Step 3: Therefore the odd-sum relation is not reflexive.

Step 1: If one person is friends with a second and the second is friends with a third, the first need not be friends with the third. Step 2: Therefore friendship is not transitive. Step 3: Always test transitivity separately for equivalence.

Step 1: The classes are ({1,4},{2,5},{3,6}). Step 2: Each class has size (2), so each contributes (2^2=4) pairs. Step 3: The total is (4+4+4=12) pairs.

Step 1: (x\in[y]) means (x) is related to (y). Step 2: Related elements have the same equivalence class. Step 3: Hence ([x]=[y]).

Step 1: Two equivalence classes are either equal or disjoint. Step 2: If ([a]=[b]), their union is the same single class. Step 3: The union of distinct classes is generally not one equivalence class.

Step 1: Elements not greater than (2) are (1,2). Step 2: Elements greater than (2) are (3,4,5). Step 3: The relation forms a partition using these two groups.

Step 1: The set ({1,2}) has (2) elements. Step 2: The subsets of (S) with (2) elements are ({1,2},{1,3},{2,3}). Step 3: To form the equivalence class, choose subsets with the same number of elements.