Equivalence relation

Step 1: The relation (a>b) is not reflexive because (a>a) is false. Step 2: It is also not symmetric, since (5>3) does not imply (3>5). Step 3: Strict order relations are generally not equivalence relations.

Step 1: (a^2-a^2=0), so the relation is reflexive. Step 2: If (a^2-b^2) is divisible by (3), then (b^2-a^2) is also divisible by (3). Step 3: Equality of square remainders modulo (3) is transitive.

Step 1: (2^2=4) and (5^2=25). Step 2: (25-4=21), which is divisible by (3). Step 3: For options, subtract the squares and test divisibility.

Step 1: Equivalence classes are either equal or completely disjoint. Step 2: Since ([a]\neq[b]), they are not equal. Step 3: Hence their intersection is empty.

Step 1: All elements in the same equivalence class are related to one another. Step 2: (4) and (7) both lie in ({1,4,7}). Step 3: Therefore ((4,7)) is definitely in the relation.

Step 1: A person may be considered acquainted with himself or herself, so reflexivity can hold. Step 2: If one person is acquainted with another, the reverse is also true, so symmetry holds. Step 3: But acquaintance need not pass through a third person, so transitivity can fail.

Step 1: (6) is greater than (4). Step 2: The elements greater than (4) are (5,6,7). Step 3: In a grouping relation, the class of an element is the group containing it.

Step 1: (1) is not prime. Step 2: In the given set, (4) and (6) are also not prime. Step 3: The relation creates two classes: prime and non-prime elements.

Step 1: The condition (a=b) gives four diagonal pairs. Step 2: The condition (a+b=5) gives non-diagonal pairs ((1,4),(4,1),(2,3),(3,2)). Step 3: The total is (4+4=8) pairs.

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.