Equivalence relation

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).

Step 1: All ((a,a)) pairs are present, so reflexivity holds. Step 2: ((2,1)) is present with ((1,2)), so symmetry holds. Step 3: The classes are ({1,2}), ({3}), and ({4}), so transitivity also holds.

Step 1: (|a-a|=0), so reflexivity holds. Step 2: Distance is the same in both directions, so symmetry holds. Step 3: (1R3) and (3R5), but (1R5) is false, so transitivity fails.

Step 1: (x-y=0) means (x=y). Step 2: Equality is reflexive, symmetric, and transitive. Step 3: The equality relation is the most basic example of an equivalence relation.

Step 1: Since ((1,2)) and ((2,3)) are present, transitivity connects (1) and (3). Step 2: With symmetry and reflexivity, all three elements are in one class. Step 3: One class of size (3) gives (3^2=9) pairs.

Step 1: (2) leaves remainder (2) when divided by (3). Step 2: In (A), (5) also leaves remainder (2). Step 3: Hence ([2]={2,5}).

Step 1: Division by (4) gives possible remainders (0,1,2,3). Step 2: Each remainder gives one equivalence class. Step 3: For integers modulo (n), there are usually (n) classes.

Step 1: All ordered pairs inside each class are included. Step 2: The class sizes are (2,1,3), so the count is (2^2+1^2+3^2=4+1+9). Step 3: Total pairs are (14).

Step 1: For real numbers, (x^3=y^3) implies (x=y). Step 2: So the relation behaves like equality. Step 3: Equality-based relations are reflexive, symmetric, and transitive.

Step 1: (x-x=0), so reflexivity holds. Step 2: (3R2) is true, but (2R3) is not true. Step 3: The reverse pair does not always hold, so symmetry fails.

Step 1: For reflexivity, (a+a) must be divisible by (3) for every (a). Step 2: (1+1=2), which is not divisible by (3). Step 3: One counterexample is enough to break reflexivity.

Step 1: A single equivalence class means every element is related to every other element. Step 2: Therefore all possible ordered pairs are present. Step 3: This is exactly (A\times A), the universal relation.

Step 1: Equivalence classes divide the set into separate blocks. Step 2: Distinct blocks have no common element. Step 3: Therefore the intersection of distinct classes is empty.