Equivalence relation

Step 1: (\lfloor -1.2\rfloor=-2). Step 2: Real numbers with floor value (-2) lie in ([-2,-1)). Step 3: For negative numbers, be careful that the floor goes to the smaller integer.

Step 1: From (aRb) and (bRc), transitivity gives (aRc). Step 2: From (aRc) and (cRd), transitivity gives (aRd). Step 3: In an equivalence relation, all elements connected in a chain lie in the same class.

Step 1: The three classes share at least one common element. Step 2: If two equivalence classes have non-empty intersection, they are equal. Step 3: Applying this to all three classes gives ([a]=[b]=[c]).

Step 1: (\max(2,4)=4). Step 2: For (1,2,3,4), the value of (\max(a,4)) is (4). Step 3: A relation based on equal function value groups all elements with the same value.

Step 1: The possible values of (\max(a,4)) are (4,5,6). Step 2: The classes are ({1,2,3,4},{5},{6}). Step 3: The number of distinct function values gives the number of classes.

Step 1: The distance of ((3,4)) from the origin is (\sqrt{3^2+4^2}=5). Step 2: The point ((0,5)) also has distance (5). Step 3: All points with the same distance from the origin belong to the same class.

Step 1: Having the same distance from the origin means having a fixed radius. Step 2: All points with a fixed radius form a circle centered at the origin. Step 3: In geometric relations, equivalence classes often appear as shapes.

Step 1: (R\subseteq S) means every pair related by (R) is also related by (S). Step 2: Thus (S) may merge more elements together. Step 3: Each (R)-class is contained in some (S)-class.

Step 1: (S) merges the two (R)-classes ({1,2}) and ({3,4}) into a larger class. Step 2: Therefore every pair inside an (R)-class remains inside an (S)-class. Step 3: Hence (R\subseteq S).

Step 1: In (R\cap S), two elements stay together only if they are together in both relations. Step 2: Here no two distinct elements are together in both partitions. Step 3: Therefore all classes are singleton classes.

Step 1: ((\sqrt{2}+\sqrt{3}+7)-(\sqrt{2}+\sqrt{3})=7). Step 2: Since (7) is rational, it is in the same equivalence class. Step 3: Adding a rational number does not change the class in a rational-difference relation.

Step 1: (\frac{-6\sqrt{5}}{-2\sqrt{5}}=3). Step 2: (3) is a positive rational number, so (-6\sqrt{5}) is in the same class. Step 3: Here the ratio must be positive rational, not just similar-looking.

Step 1: For every (a), (a^3\equiv a^3 \pmod{7}), so reflexivity holds. Step 2: Equality of remainders is symmetric. Step 3: Equality of remainders is transitive, so the relation is an equivalence relation.

Step 1: (2^2=4), so the class contains elements whose square has remainder (4) modulo (5). Step 2: (2,3,7,8) have square remainder (4) modulo (5). Step 3: Compare square remainders, not the numbers themselves.

Step 1: The square remainders that appear are only (0,1,4). Step 2: The classes are ({5},{1,4,6},{2,3,7,8}). Step 3: The number of distinct square remainders gives the number of classes.

Step 1: Subsets with the same size belong to the same class. Step 2: The number of (3)-element subsets of a (5)-element set is (\binom{5}{3}=10). Step 3: Use combinations in subset-cardinality questions.

Step 1: ({1,3}\cap{1,2}={1}), so the count is (1). Step 2: Choose exactly one element from ({1,2}) in (2) ways and any subset of ({3,4}) in (4) ways. Step 3: The class size is (2\cdot4=8).

Step 1: From ((1,2)) and ((2,4)), transitivity requires ((1,4)). Step 2: Symmetry then requires ((4,1)). Step 3: Then (1,2,4) become one complete equivalence class.

Step 1: All diagonal pairs are present, so the relation is reflexive. Step 2: ((1,2)) has ((2,1)), and ((3,4)) has ((4,3)). Step 3: The classes ({1,2}) and ({3,4}) are complete, so transitivity holds.

Step 1: Inside one equivalence class, every element is related to every element. Step 2: A class with (4) elements contributes (4^2) ordered pairs. Step 3: Hence this class contributes (16) pairs.

Step 1: (a\in[b]) means (a) is related to (b), so ([a]=[b]). Step 2: (b\in[c]) gives ([b]=[c]). Step 3: Therefore all three equivalence classes are equal.

Step 1: (6) is a multiple of (2), but not a multiple of (4). Step 2: In the given set, (2) has the same status. Step 3: Matching both statuses gives the class ({2,6}).

Step 1: In (R\cap S), numbers must have the same remainder modulo (3) and modulo (5). Step 2: Since (3) and (5) are coprime, this is the same as congruence modulo (15). Step 3: Hence the class of (7) is the integers congruent to (7) modulo (15).

Step 1: The sum of squares of class sizes gives the number of ordered pairs. Step 2: (3^2+3^2+1^2=9+9+1=19). Step 3: Therefore (3,3,1) is a possible class-size pattern.

Step 1: The class sizes must add to (5), and their squares must add to (13). Step 2: (3+2=5) and (3^2+2^2=9+4=13). Step 3: Thus the class sizes can be (3,2).