Equivalence relation

Step 1: (\gcd(4,6)=2). Step 2: In (A), the elements whose gcd with (6) is (2) are (2) and (4). Step 3: First compute the assigned value, then collect elements with the same value.

Step 1: Both relations contain every ((a,a)), so the intersection does too. Step 2: Symmetry and transitivity are preserved in the intersection. Step 3: The intersection of two equivalence relations is again an equivalence relation.

Step 1: (a^2\equiv a^2 \pmod{5}), so reflexivity holds. Step 2: If (a^2\equiv b^2 \pmod{5}), then (b^2\equiv a^2 \pmod{5}), so symmetry holds. Step 3: Congruence is transitive, so the third condition also holds.

Step 1: (\frac{x}{y}>0) means (x) and (y) have the same sign. Step 2: All positive numbers form one class and all negative numbers form another. Step 3: For sign-based relations, split the set into positive and negative groups.

Step 1: The odd class is ({1,3,5}) and the even class is ({2,4,6}). Step 2: Each class gives (3^2=9) ordered pairs. Step 3: Total pairs are (9+9=18).

Step 1: An equivalence relation needs all three properties. Step 2: Here transitivity is missing, so a necessary condition fails. Step 3: Even if two properties hold, always test the third one.

Step 1: All pairs inside ({1,2,5}) must be in the relation. Step 2: If ((1,3)) were in the relation, then (3) would belong to the same class. Step 3: A pair with an outside element changes the class, so it is not possible here.

Step 1: (x-x=0) is rational, so reflexivity holds. Step 2: If (x-y) is rational, then (y-x) is also rational. Step 3: The sum of two rational differences is rational, so transitivity holds.

Step 1: Divisibility of (a-b) by (2) means the numbers have the same parity. Step 2: Odd elements form ({1,3,5,7}), and even elements form ({2,4,6}). Step 3: For parity relations, separate odd and even elements first.

Step 1: For (b) to be related to (0), (0-b) must be divisible by (6). Step 2: This means (b) is a multiple of (6). Step 3: For infinite sets, write the class in general form as ({6k:k\in\mathbb{Z}}).

Step 1: (A\times A) contains every possible ordered pair. Step 2: Therefore reflexivity, symmetry, and transitivity all hold automatically. Step 3: Treat the universal relation as a basic example of an equivalence relation.

Step 1: For non-empty (A), each ((a,a)) is required for reflexivity. Step 2: The empty relation has no pair, so reflexivity fails. Step 3: When testing equivalence, check reflexivity first.

Step 1: Exactly two equivalence classes mean splitting (A) into two non-empty blocks. Step 2: The number of ways to divide (4) elements into two unnamed non-empty blocks is (7). Step 3: Counting equivalence relations is the same as counting partitions.

Step 1: (f(a)=f(a)), so reflexivity holds. Step 2: If (f(a)=f(b)), then (f(b)=f(a)), so symmetry holds. Step 3: Equality is transitive, so the relation is always an equivalence relation.

Step 1: (2^2=4). Step 2: In (A), the elements whose square is (4) are (-2) and (2). Step 3: For square-based relations, check both positive and negative values.

Step 1: (|x|=|x|), so reflexivity holds. Step 2: If (|x|=|y|), then (|y|=|x|), so symmetry holds. Step 3: By transitivity of equality, (|x|=|z|) also follows.

Step 1: (a+a=2a), so reflexivity holds. Step 2: (a+b) and (b+a) are the same, so symmetry holds. Step 3: If two pairs show the same parity, the first and third also have the same parity.

Step 1: Compute (\lfloor\frac{a}{2}\rfloor) for each element. Step 2: The values (0,1,2,3) give the classes ({1},{2,3},{4,5},{6}). Step 3: Elements giving the same value form one equivalence class.

Step 1: ((a,b)\in R) means (a) and (b) are in the same equivalence class. Step 2: Elements in the same class have equal equivalence classes. Step 3: Remember, ([a]=[b]) does not necessarily mean (a=b).

Step 1: Division by (3) gives three classes, each with (3) elements. Step 2: Each class gives (3^2=9) ordered pairs. Step 3: Total pairs are (9+9+9=27).