Equivalence relation

The equivalence class [2] consists of every element x in A that is congruent to 2 modulo 3; equivalently, x − 2 must be divisible by 3. Check the elements of A: 2 − 2 = 0, which is divisible by 3, and 5 − 2 = 3, which is also divisible by 3. For 1, the difference is −1; for 3, it is 1; and for 4, it is 2, none of which is divisible by 3. Therefore [2] = {2,5}. The set {1,4} is the class of 1, while {3} is the class of 3 within A. Since different congruence classes partition A, the whole set cannot be [2]. Hence option A is correct.

There is a one-to-one correspondence between equivalence relations on a set and partitions of that set. Therefore the question asks for the number of partitions of 4 distinct elements into exactly 2 nonempty, unlabeled blocks, namely the Stirling number of the second kind S(4,2). It can be calculated by listing block-size possibilities: a 1+3 split gives C(4,1) / 2 = 2 partitions, and a 2+2 split gives C(4,2) / 2 = 3 partitions. Thus the total is 2 + 3 = 5, not 7. However, the supplied correct option B (7) is mathematically inconsistent; the valid answer 5 is absent. This item must be revised before passing.

The smallest equivalence relation containing R must be reflexive, symmetric, and transitive. The pairs (1,2) and (2,3) connect 1, 2, and 3 into one equivalence class, so all 3² = 9 ordered pairs within {1,2,3} must be included. The pair (4,5) connects 4 and 5 into a second class, contributing 2² = 4 ordered pairs. No pair can connect these two classes, because that would impose a larger relation than necessary. Therefore the total number of ordered pairs is 9 + 4 = 13, so option C is correct.

Because every element of A is positive, the equation a^2 = b^2 implies a = b; the alternative a = −b cannot occur within this set for two positive elements. Hence R is exactly {(1,1),(2,2),(3,3),(4,4)}, the identity relation on A. It is reflexive because every (a,a) is included, symmetric because (a,b) implies a=b and therefore b=a, and transitive because a=b and b=c imply a=c. A relation that is reflexive, symmetric, and transitive is an equivalence relation, so option A is correct. Option B is incomplete, option C ignores the other properties, and option D is false.

For an equivalence relation R, the equivalence class of an element a is [a] = {x in A : xRa}. Reflexivity ensures that a belongs to [a], so every class is non-empty and all elements of A are covered. If two classes have even one common element, symmetry and transitivity show that their representatives are related, so the two classes are actually identical. Thus distinct classes are disjoint, while their union is A. These are precisely the defining features of a partition of A. Therefore option A is correct. Classes need not be singletons; for example, congruence modulo 2 gives classes of several elements. They are not empty, and distinct classes do not overlap.

The relation already contains all diagonal pairs, so it is reflexive. It is also symmetric because (1,2) and (2,1), as well as (2,3) and (3,2), are both present. The missing requirement is transitivity. Since (1,2) and (2,3) belong to R, transitivity requires (1,3). Also, because (3,2) and (2,1) are present, it requires (3,1). Equivalently, symmetry would require the reverse of (1,3) once that pair is added. After inserting both pairs, every pair among {1,2,3} is present, so the resulting universal relation is certainly an equivalence relation. Adding only one pair leaves either transitivity or symmetry violated. Thus option A is necessary and sufficient.