Equivalence relation

Step 1: (12=2^2\cdot 3), so its set of prime factors is ({2,3}). Step 2: (18=2\cdot 3^2) also has the prime factor set ({2,3}). Step 3: Powers may differ, but this relation checks only the set of prime factors.

Step 1: Equivalence classes are either completely disjoint or exactly the same. Step 2: If two classes have a common element, that element connects both classes. Step 3: Therefore the two classes cannot remain distinct, so (B_1=B_2).

Step 1: In (A\times A), every element of (A) is related to every element of (A). Step 2: Therefore no two elements can lie in different classes. Step 3: This is the universal relation, and its single equivalence class is the whole set (A).

Step 1: (3) itself is a multiple of (3). Step 2: Its equivalence class contains exactly those elements that are multiples of (3). Step 3: In the given set, the multiples of (3) are (3) and (6), so the class is ({3,6}).

Step 1: A number (x) is related to (1.7) only when (x-1.7) is an integer. Step 2: This means (x=1.7+n), where (n\in Z). Step 3: Writing just one interval is wrong because the class contains numbers at integer distances.

Step 1: (\min(5,3)=3). Step 2: The elements for which (\min(a,3)=3) are (3,4,5). Step 3: In a relation made by equal function values, all elements with the same value lie in one class.

Step 1: The main property of equivalence classes is that two classes are either equal or disjoint. Step 2: Here ([a]\neq[b]), so they are not equal. Step 3: Therefore their intersection must be empty.

Step 1: (4) leaves remainder (0) when divided by (2). Step 2: (7) leaves remainder (1) when divided by (2). Step 3: The remainders are different, so (4) and (7) are not related and ((4,7)\notin R).

Step 1: Every student has the same date of birth as himself or herself. Step 2: If the first student has the same date as the second, the second has the same date as the first. Step 3: If two students share the same date with a third student, they share it with each other too.

Step 1: To be an equivalence relation, at least every ((a,a)) pair is needed. Step 2: Taking only these reflexive pairs still preserves symmetry and transitivity. Step 3: Therefore the equality relation is the smallest equivalence relation.

Step 1: The remainder of (3) on division by (4) is (3). Step 2: In the given set, the elements with remainder (3) are (3) and (7). Step 3: While writing an equivalence class, include only elements from the given base set.

Step 1: (3^2-5^2=9-25=-16), which is divisible by (8). Step 2: In the other options, the difference of squares is not divisible by (8). Step 3: In such questions, directly checking the difference of squares is quick and reliable.

Step 1: A number (x) is related to (\sqrt{2}) when (x-\sqrt{2}) is rational. Step 2: So (x=\sqrt{2}+q), where (q\in Q). Step 3: Classes based on rational difference contain infinitely many numbers, not just one number.

Step 1: The odd class ({1,3,5}) has (3) elements and the even class ({2,4}) has (2) elements. Step 2: The number of ordered pairs is (3^2+2^2=9+4). Step 3: Pairs between different classes are not counted, so the total is (13).

Step 1: Elements in the same block are related to each other. Step 2: (2) and (6) are together in the first block, so ((2,6)\in R). Step 3: (3) and (4) are in different blocks, so ((3,4)\notin R).

Step 1: ((-5)^2=25). Step 2: The real numbers whose square is (25) are (-5) and (5). Step 3: In equality of squares, the sign may change, so both numbers must be included.

Step 1: (\frac{a}{a}=1), which is rational, so reflexivity holds. Step 2: If (\frac{a}{b}) is rational and non-zero, then (\frac{b}{a}) is also rational. Step 3: The product of two rational ratios is rational, so transitivity also holds.

Step 1: For ((2,5)), the sum is (2+5=7). Step 2: Every pair related to it must also have sum (7). Step 3: In such relations, an equivalence class may appear as a set of points satisfying a fixed value condition.

Step 1: (1) is related to (2), and (2) is related to (3). Step 2: Transitivity requires (1) to be related to (3), and symmetry requires the reverse pair too. Step 3: (4) is only related to itself, so no new pair involving (4) is needed.

Step 1: The number of equivalence relations on a set equals the number of partitions of that set. Step 2: A set with (4) elements has (15) partitions. Step 3: In exams, remember the direct link between equivalence relations and partitions.