Equivalence relation

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.

Step 1: Every number has the same digit sum as itself. Step 2: Equality of digit sums remains true in reverse order. Step 3: If two numbers have the same digit sum as a third, then they have the same digit sum as each other.

Step 1: The given blocks are the equivalence classes. Step 2: Their sizes are (2,3,3), so the pair count is (2^2+3^2+3^2=4+9+9). Step 3: Total pairs are (22).

Step 1: ([1]) is the class of all odd integers. Step 2: ([2]) is the class of all even integers. Step 3: No integer is both even and odd, so the intersection is empty.

Step 1: Same remainder modulo (2) means both numbers are even or both are odd. Step 2: (3) is odd and (4) is even, so they are in different classes. Step 3: Pairs from different classes are not in the relation.

Step 1: The equality relation contains only pairs of the form ((a,a)). Step 2: With (5) elements, there are (5) such pairs. Step 3: Remember that the equality relation and universal relation have different counts.

Step 1: In (A\times A), every element is related to every element. Step 2: Thus all (5) elements lie in one class. Step 3: The universal relation always has exactly one equivalence class.

Step 1: (\sin x=\sin x), so reflexivity holds. Step 2: If (\sin x=\sin y), then (\sin y=\sin x), so symmetry holds. Step 3: Transitivity of equality makes this an equivalence relation.

Step 1: Pairs inside the same class belong to the relation. Step 2: (1) is in ({1,6}), while (3) is in ({2,3,5}). Step 3: Elements from different classes are not related, so ((1,3)) will not be present.