Equivalence relation

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.

The direct answer is {2,5}, so option A is correct. The equivalence class [2] contains all elements x of A that are congruent to 2 modulo 3. Congruence means that x−2 is divisible by 3, or that x and 2 leave the same remainder on division by 3. Test the elements one by one: 2−2=0, which is divisible by 3, and 5−2=3, which is also divisible by 3. Thus 2 and 5 belong to [2]. For 1, the difference is −1; for 3 it is 1; and for 4 it is 2. None of these is divisible by 3, so they do not belong. Option B, {1,4}, is the class of 1; option C, {3}, is the class of 3; and option D incorrectly combines different classes. A useful memory cue is to group numbers by their remainders modulo 3: {1,4}, {2,5}, and {3}.

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.