Equivalence relation

Step 1: In (R\cap S), two elements stay together only if they are together in both (R) and (S). Step 2: This creates intersections of blocks from the two partitions. Step 3: The non-empty intersections become the new classes.

Step 1: Intersect the blocks of the two partitions. Step 2: Each non-empty intersection is a singleton: ({1},{2},{3},{4},{5},{6}). Step 3: Thus (R\cap S) has (6) classes, like the identity relation.

Step 1: In the intersection, the difference must be divisible by both (4) and (6). Step 2: This is congruence modulo (12). Step 3: (14) has remainder (2) modulo (12), so the class is (x\equiv 2 \pmod{12}).

Step 1: From ((1,2)) and ((2,5)), transitivity requires ((1,5)). Step 2: Symmetry then requires ((5,1)). Step 3: Then (1,2,5) become one complete equivalence class.

Step 1: The condition (a=b) gives all diagonal pairs. Step 2: The condition (a+b=7) links (1) with (6), (2) with (5), and (3) with (4). Step 3: These form three separate complete classes.

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

Step 1: (\cos 0=1). Step 2: The solutions of (\cos x=1) are (x=2n\pi), where (n\in\mathbb{Z}). Step 3: In trigonometric relations, include all angles with the same function value.

Step 1: The divisors of (6) are (1,2,3,6), so (\tau(6)=4). Step 2: (8) and (10) also have (4) positive divisors. Step 3: Elements with the same divisor count form one equivalence class.

Step 1: The divisor-count values appearing are (1,2,3,4). Step 2: The classes are ({1}), ({2,3,5,7}), ({4,9}), and ({6,8,10}). Step 3: The number of distinct (\tau)-values gives the number of classes.

Step 1: Since ([a]\cap[b]\neq\varnothing), the two equivalence classes are equal. Step 2: Hence ([a]=[b]) follows directly. Step 3: The union condition is extra; the key clue is the non-empty intersection.

Step 1: (8) is divisible by (4), but not by (3). Step 2: In the given set, (4) has the same status. Step 3: Matching both divisibility statuses puts them in the same class.

Step 1: The distance of (-2) from (1) is (3). Step 2: The real numbers at distance (3) from (1) are (-2) and (4). Step 3: A distance-based equivalence class contains points equally far from the center.

Step 1: For one-digit numbers, the digit sum is the number itself. Step 2: (8) gives remainder (2) modulo (3), as do (2,5,8). Step 3: The condition is based on the remainder of the digit sum.

Step 1: The number of equivalence relations equals the number of partitions of the set. Step 2: A set with (5) elements has (52) partitions. Step 3: In such counting questions, count partitions rather than pairs directly.

Step 1: The class of (1) is fixed as ({1}). Step 2: The remaining four elements ({2,3,4,5}) can be partitioned freely. Step 3: Four elements have (15) partitions, so the answer is (15).

Step 1: Keep (1) and (2) together, but keep (3) out of that block. Step 2: Each of (4,5) may or may not join the block of (1,2), and the remaining objects are partitioned. Step 3: Counting these cases gives (7) partitions.

Step 1: (R\subseteq S) means every pair of (R) is also in (S). Step 2: Therefore (aRb) implies (aSb). Step 3: The converse need not hold because (S) may be a larger relation.

Step 1: (A\times A) has (6^2=36) ordered pairs. Step 2: (R) contains (3^2+2^2+1^2=14) pairs. Step 3: Therefore (36-14=22) pairs are not in (R).

Step 1: A difference divisible by (10) is automatically divisible by (5). Step 2: So requiring both conditions leaves the stronger modulo (10) condition. Step 3: Hence (R\cap S) is congruence modulo (10).

Relation R consists of integer pairs whose difference is a multiple of 5, while S consists of pairs whose difference is a multiple of 10. Every multiple of 10 is also a multiple of 5, so S is a subset of R. The union of a relation with its subset is the larger relation itself: R∪S=R. Therefore the union is congruence modulo 5, making option A correct; modulo 10 would describe S, not the union.