Equivalence relation

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.

Step 1: (\gcd(6,9)=3). Step 2: In the given set, (\gcd(3,9)=3) and (\gcd(6,9)=3). Step 3: Elements with the same greatest common divisor lie in the same class.

Step 1: The coordinate sum of ((2,-1)) is (1). Step 2: All related points must satisfy (x+y=1). Step 3: In a same-sum relation, the equivalence class is a line.

Step 1: The value of (x-y) stays fixed in the relation. Step 2: The equation (x-y=c) represents a straight line. Step 3: Different values of (c) give parallel equivalence classes.

Step 1: If (aRb), then (a) and (b) lie in the same equivalence class. Step 2: Therefore ([a]=[b]). Step 3: The difference of equal sets is the empty set.

Step 1: Equivalence classes are either equal or disjoint. Step 2: If ([a]\setminus[b]\neq\varnothing), the classes are not equal. Step 3: Therefore their intersection is empty.

Step 1: The class sizes are (2,2,2,1). Step 2: The number of pairs is (2^2+2^2+2^2+1^2). Step 3: The total is (4+4+4+1=13).

Step 1: The condition (a=b) gives all diagonal pairs. Step 2: The condition (a+b=8) links (1,7), (2,6), (3,5), while (4) remains alone. Step 3: These form complete disjoint classes, so the relation is an equivalence relation.

Step 1: (10-3=7), which is divisible by (7). Step 2: Therefore (3) and (10) are in the same equivalence class. Step 3: The intersection of equal classes is that same class.

Step 1: (11-3=8), which is not divisible by (7). Step 2: Thus (3) and (11) lie in different equivalence classes. Step 3: Distinct equivalence classes have empty intersection.

Step 1: ((\sqrt{2}+5)-\sqrt{2}=5). Step 2: Since (5) is rational, (\sqrt{2}+5) lies in the same class. Step 3: Adding a rational number does not change the class in this relation.

Step 1: From (aRb), (\frac{a}{b}) is rational, and from (bRc), (\frac{b}{c}) is rational. Step 2: Their product is (\frac{a}{c}). Step 3: A product of rational numbers is rational, proving transitivity.

Step 1: There are (6) remainder classes, each with (2) elements. Step 2: Each class contributes (2^2=4) ordered pairs. Step 3: The total number of pairs is (6\cdot4=24).