Equivalence relation

Step 1: (5) leaves remainder (5) on division by (6). Step 2: In the given set, (11) also leaves remainder (5) on division by (6). Step 3: While writing an equivalence class, include only elements from the base set with the same remainder.

Step 1: (-15=-21+6), so its remainder is taken as (6). Step 2: All integers with the same remainder are related to (-15). Step 3: For negative numbers also, write the standard remainder between (0) and (6).

Step 1: A number (x) is related to (\frac{3}{4}) only when (x-\frac{3}{4}) is an integer. Step 2: Thus (x=\frac{3}{4}+n), where (n\in Z). Step 3: This class contains numbers with the same fractional part, not a full interval.

Step 1: (3^2=9), which leaves remainder (2) modulo (7). Step 2: (4^2=16) also leaves remainder (2) modulo (7). Step 3: Therefore (3) and (4) are in the same equivalence class, so the class is ({3,4}).

Step 1: (2\cdot\frac{5}{4}=\frac{5}{2}), so the floor value is (2). Step 2: (\lfloor 2x\rfloor=2) means (2\le 2x<3). Step 3: Hence (1\le x<\frac{3}{2}), so the correct class is ([1,\frac{3}{2})).

Step 1: (\lfloor 2\cdot\frac{5}{4}\rfloor=\lfloor \frac{5}{2}\rfloor=2). Step 2: (\lfloor 2x\rfloor=2) means (2\le 2x<3). Step 3: Dividing by (2) gives (1\le x<\frac{3}{2}), so the class is ([1,\frac{3}{2})).

Step 1: (\operatorname{lcm}(4,6)=12). Step 2: In the given set, only (4) has least common multiple (12) with (6). Step 3: Only elements with the same value lie in one class, so the class is ({4}).

Step 1: The rank of a (3\times 3) matrix can be (0,1,2,) or (3). Step 2: Matrices with the same rank lie in the same equivalence class. Step 3: Therefore there are (4) equivalence classes.

Step 1: (2^4=16). Step 2: Over real numbers, (x^4=16) gives (x=2) and (x=-2). Step 3: With an even power, the sign can change while the value remains the same, so both elements are included.

Step 1: The real part of (2+3i) is (2). Step 2: All complex numbers related to it must have real part (2). Step 3: The imaginary part can be any real number, so the class is ({2+yi:y\in R}).

Step 1: (\lceil \frac{5}{2}\rceil=3). Step 2: (\lceil \frac{6}{2}\rceil=3), so (6) is in the same class. Step 3: In ceiling function questions, carefully check where the next integer value begins.

Step 1: For the relation, the difference of the two numbers must be rational. Step 2: (\sqrt{2}-\sqrt{3}) is not rational, so this pair is not in the relation. Step 3: In the other options, the difference is rational, so they are in the relation.

Step 1: For the given polynomial, (p(1)=1^2+1=2). Step 2: The relation checks equality of value at (1). Step 3: Hence all polynomials whose value at (1) is (2) lie in this class.

Step 1: Three equivalence classes are formed: ({1,4}), ({2,5}), and ({3,6}). Step 2: Each class has (2) elements, so each contributes (2^2=4) pairs. Step 3: Total ordered pairs are (4+4+4=12).

Step 1: Each class of size (n) contributes (n^2) ordered pairs. Step 2: Here the total is (2^2+2^2+4^2=4+4+16). Step 3: Therefore the relation contains (24) ordered pairs.

Step 1: Elements in the same block are related to each other. Step 2: (a) and (b) are in different blocks, so ((a,b)) is not in the relation. Step 3: Within a class, ordered pairs in both directions are included, so the other pairs are valid.

Step 1: For (1), (\ln 1=0). Step 2: (\frac{\ln 1}{\ln 1}=\frac{0}{0}) is undefined. Step 3: When reflexivity fails, the relation cannot be an equivalence relation.

Step 1: For real numbers, (a^2+b^2=0) only when (a=0) and (b=0). Step 2: For (1R1), (1^2+1^2=2), so (1R1) is false. Step 3: Not every element is related to itself, so it is not an equivalence relation.

Step 1: Reflexivity would require (a+a) to be prime for every (a). Step 2: For (a=2), (2+2=4), which is not prime. Step 3: Therefore the relation is not reflexive on all elements and cannot be equivalence.

Step 1: (|a-b|) being divisible by (3) means the numbers have the same remainder modulo (3). Step 2: Remainder (1) gives ({1,4}), remainder (2) gives ({2,5}), and remainder (0) gives ({3}). Step 3: Elements with the same remainder form one equivalence class.

Step 1: (18=2\cdot 3^2), so its greatest prime factor is (3). Step 2: The relation checks only the greatest prime factor. Step 3: Therefore the class contains all natural numbers whose greatest prime factor is (3).

Step 1: For ((4,1)), (4-1=3). Step 2: Every pair related to it must have first component minus second component equal to (3). Step 3: Hence the equivalence class is ({(x,y):x-y=3}).

Step 1: The (x)-coordinate of ((3,-2)) is (3). Step 2: All points in the same class must have (x)-coordinate (3). Step 3: The (y)-coordinate can be any real number, so the class is a vertical line.

Step 1: The distance of ((6,8)) from the origin is (\sqrt{6^2+8^2}=10). Step 2: All points related to it must also have distance (10). Step 3: All points at the same distance from the origin form a circle centered at the origin.

Step 1: (1) is related to (2), and (2) is related to (3). Step 2: Transitivity requires (1) to be related to (3). Step 3: To preserve symmetry, both ((1,3)) and ((3,1)) must be added.