Equivalence relation

Correct answer: A. Identity relation

Step 1: The numbers (1,2,3,4,5,6) have distinct remainders modulo (7). Step 2: So no two different elements are related. Step 3: Only the pairs ((a,a)) remain, making it the identity relation.

Correct answer: A. Its classes are ({1,5},{2,4},{3})

Step 1: The condition (a=b) gives all diagonal pairs. Step 2: Divisibility of (a+b) by (6) links (1) with (5), and (2) with (4). Step 3: These form complete separate classes, so the relation is an equivalence relation.

Correct answer: A. (B) contains elements from more than one distinct equivalence class

Step 1: An equivalence class is a complete block, not any larger chosen subset. Step 2: If (B) contains ([a]) plus elements from another class, it is not a single class. Step 3: You cannot cut or arbitrarily combine classes to make a new class.

Correct answer: A. ((2,6))

Step 1: All elements of the same equivalence class are related to one another. Step 2: (2) and (6) both lie in ({1,2,6}). Step 3: Therefore ((2,6)) must belong to the relation.

Correct answer: A. (6)

Step 1: In the intersection, the difference must be divisible by both (2) and (3). Step 2: Hence the difference is divisible by (\operatorname{lcm}(2,3)=6). Step 3: Congruence modulo (6) has (6) equivalence classes on integers.

Correct answer: A. Transitivity fails

Step 1: (|1-2|\leq 1) and (|2-3|\leq 1), so (1R2) and (2R3). Step 2: But (|1-3|=2), so (1R3) is false. Step 3: Therefore transitivity fails, so it is not an equivalence relation.

Correct answer: A. Not an equivalence relation because reflexivity fails

Step 1: (R) contains every ((a,a)) because each number has the same remainder as itself. Step 2: The complement contains no ((a,a)) pair. Step 3: Without reflexivity, the complement is not an equivalence relation.

Correct answer: A. (15)

Step 1: The classes are ({1,5,7}), ({2,4}), ({3}), and ({6}). Step 2: The number of ordered pairs is (3^2+2^2+1^2+1^2). Step 3: The total is (9+4+1+1=15).

Correct answer: A. ({1,3,5},{2,4,6})

Step 1: The square of a number has the same parity as the number. Step 2: (a^2+b^2) is even only when (a) and (b) have the same parity. Step 3: Hence the classes are the odd and even numbers.

Correct answer: A. (5)

Step 1: (4^2=16), which gives remainder (7) modulo (9). Step 2: (5^2=25), which also gives remainder (7). Step 3: Equal square remainders place the integers in the same equivalence class.

Correct answer: A. (41)

Step 1: The classes are ({1,2,4,5,7,8}), ({3,6}), and ({9}). Step 2: The pair count is (6^2+2^2+1^2). Step 3: Thus the total is (36+4+1=41).

Correct answer: A. ({-4,2})

Step 1: ((-4)^2+2(-4)=8). Step 2: (x^2+2x=8) gives (x^2+2x-8=0), so ((x+4)(x-2)=0). Step 3: Hence (x=-4) or (x=2), so the class is ({-4,2}).

Correct answer: A. ({5,6,7,8})

Step 1: (\min(7,5)=5). Step 2: For (5,6,7,8), the minimum value with (5) is (5). Step 3: Elements with the same function value form one equivalence class.

Correct answer: A. (5,1)

Step 1: Class sizes must add to (6), and their squares must add to (31). Step 2: Testing the listed patterns gives (26,20,18,) and (14). Step 3: None of the given options gives (31), so no listed pattern is possible.

Correct answer: A. ({3})

Step 1: The remainders must match modulo (2) and modulo (5). Step 2: This is like matching remainders modulo (10), and within the set only (3) has that pair of remainders. Step 3: Combine both modulo conditions carefully.

Correct answer: A. Transitivity fails

Step 1: (0R8) because (0-8=-8) is divisible by (8). Step 2: (8R20) because (8-20=-12) is divisible by (12). Step 3: But (0-20=-20) is divisible by neither (8) nor (12), so transitivity fails.

Correct answer: A. (15)

Step 1: Subsets with the same number of elements are in one class. Step 2: The number of (2)-element subsets of a (6)-element set is (\binom{6}{2}=15). Step 3: Use combinations for subset-size equivalence classes.

Correct answer: A. (3)

Step 1: The classes are not divisible by (2): ({1,3,5,7}), divisible by (2) but not (4): ({2,6}), and divisible by (4): ({4,8}). Step 2: These are three non-empty classes. Step 3: Count the non-empty groups formed by the combined statuses.

Correct answer: A. Non-empty intersections of classes of (R) and (S)

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.

Correct answer: A. (6)

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.

Correct answer: A. ({x\in\mathbb{Z}:x\equiv 2 \pmod{12}})

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}).

Correct answer: A. ((1,5),(5,1))

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.

Correct answer: A. ({1,6},{2,5},{3,4})

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.

Correct answer: A. (12)

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).

Correct answer: A. ({2n\pi:n\in\mathbb{Z}})

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.