Equivalence relation

Step 1: Every number has the same digit sum as itself. Step 2: Equality of digit sums remains true in reverse order. Step 3: If two numbers have the same digit sum as a third, then they have the same digit sum as each other.

Step 1: The given blocks are the equivalence classes. Step 2: Their sizes are (2,3,3), so the pair count is (2^2+3^2+3^2=4+9+9). Step 3: Total pairs are (22).

Step 1: ([1]) is the class of all odd integers. Step 2: ([2]) is the class of all even integers. Step 3: No integer is both even and odd, so the intersection is empty.

Step 1: Same remainder modulo (2) means both numbers are even or both are odd. Step 2: (3) is odd and (4) is even, so they are in different classes. Step 3: Pairs from different classes are not in the relation.

Step 1: The equality relation contains only pairs of the form ((a,a)). Step 2: With (5) elements, there are (5) such pairs. Step 3: Remember that the equality relation and universal relation have different counts.

Step 1: In (A\times A), every element is related to every element. Step 2: Thus all (5) elements lie in one class. Step 3: The universal relation always has exactly one equivalence class.

Step 1: (\sin x=\sin x), so reflexivity holds. Step 2: If (\sin x=\sin y), then (\sin y=\sin x), so symmetry holds. Step 3: Transitivity of equality makes this an equivalence relation.

Step 1: Pairs inside the same class belong to the relation. Step 2: (1) is in ({1,6}), while (3) is in ({2,3,5}). Step 3: Elements from different classes are not related, so ((1,3)) will not be present.

The correct answer is D. Every equivalence relation corresponds exactly to a partition of the set into nonempty, unlabeled equivalence classes. We must partition four distinct elements into exactly two classes. There are two possible class-size patterns. For sizes 1 and 3, choose the single element; this gives 4 partitions. For sizes 2 and 2, choose two elements for one class, but choosing the other two describes the same partition, so the count is C(4,2)/2 = 3. Therefore the total is 4 + 3 = 7. Hence option B, not D, is correct. The key warning is that equivalence classes are unlabeled: swapping their names does not create a new relation.

The smallest equivalence relation containing R must be reflexive, symmetric, and transitive. The pairs (1,2) and (2,3), together with these three properties, place 1, 2, and 3 in one equivalence class. Once a class has three elements, it must contain every ordered pair among them, contributing 3² = 9 pairs, including diagonal and reverse pairs. The pair (4,5) creates a separate class {4,5}, which contributes 2² = 4 pairs. Element groups cannot be connected without adding unnecessary pairs, so the two classes remain separate in the smallest relation. Hence the total is 9 + 4 = 13, making option C correct. The values 9 and 11 omit required pairs, while 25 treats all five elements as one class and is not minimal.

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.