Equivalence relation

Step 1: (a-b) being even means the two integers have the same parity. Step 2: This relation is reflexive, symmetric, and transitive. Step 3: Order-type relations often fail symmetry.

Step 1: (9) gives remainder (4) on division by (5). Step 2: In the given set, (4) and (9) both have remainder (4). Step 3: In a modulo relation, the class is decided by the common remainder.

Step 1: All elements in the same equivalence class are related to one another. Step 2: (2) and (8) both belong to ({2,5,8}). Step 3: Therefore ((2,8)) must be in the relation.

Step 1: In an equivalence relation, related elements belong to the same class. Step 2: Since ((x,y)\in R), the classes of (x) and (y) are equal. Step 3: Do not treat equivalence classes of related elements as different.

Step 1: From (a\leq b) and (b\leq c), we get (a\leq c), so the relation is transitive. Step 2: But (2\leq 3) does not imply (3\leq 2), so symmetry fails. Step 3: Transitivity alone is not enough for equivalence.

Step 1: Every number has the same number of digits as itself. Step 2: If two numbers have the same number of digits, the reverse statement is also true. Step 3: This equality of digit count is transitive, so the relation is an equivalence relation.

Step 1: (145) has three digits. Step 2: Therefore its class is the set of all three-digit natural numbers. Step 3: Apply the defining property directly to the chosen element.

Step 1: Every line is identical to itself, so reflexivity holds. Step 2: If (l) is parallel to (m), then (m) is parallel to (l). Step 3: Lines with the same direction preserve transitivity, so the relation is an equivalence relation.

Step 1: (2) is less than (4). Step 2: The elements less than (4) are (1,2,3), so the class of (2) is ({1,2,3}). Step 3: In grouping relations, first identify the group of the given element.

Step 1: (1) and (5) are mutually related, so they form one class. Step 2: (2) and (4) form another class, while (3) is related only to itself. Step 3: Identify connected groups from the given ordered pairs.

Step 1: Reflexivity requires ((1,1),(2,2),(3,3)). Step 2: Only the first option contains all three diagonal pairs. Step 3: Check all diagonal pairs first when testing reflexivity.

Step 1: Symmetry requires ((b,a)) whenever ((a,b)) is present. Step 2: The first option has both ((1,2)) and ((2,1)), and ((3,3)) reverses to itself. Step 3: If the reverse of a non-diagonal pair is missing, symmetry fails.

Step 1: From ((1,2)) and ((2,3)), transitivity requires ((1,3)). Step 2: The first option contains this required pair. Step 3: In transitivity, use the middle element to identify the needed start-to-end pair.

Step 1: The condition (a=b) relates every element to itself. Step 2: The extra condition relates (1) and (2) to each other. Step 3: Hence (1,2) form one class, while (3) and (4) remain singleton classes.

Step 1: A class of size (m) contributes (m^2) ordered pairs. Step 2: The total is (2^2+2^2+1^2=4+4+1). Step 3: Therefore the relation has (9) pairs.

Step 1: ([a]) is the set of all elements related to (a). Step 2: ([a]=A) means every element of (A) is related to (a). Step 3: Use the definition of an equivalence class for such conclusions.

Step 1: Reflexivity is required in an equivalence relation. Step 2: Reflexivity gives (aRa), so (a\in[a]). Step 3: Therefore no equivalence class is empty.

Step 1: Having the same remainder modulo (2) means having the same parity. Step 2: Odd elements (1,3,5) and even elements (2,4,6) form two separate classes. Step 3: A modulo (2) relation splits a set into odd and even classes.

Step 1: Equivalence classes form a partition of the set. Step 2: In a partition, distinct classes do not overlap. Step 3: If two classes overlap, they are not distinct; they are equal.

Step 1: In (A\times A), every element is related to every element. Step 2: Hence all elements fall into one large equivalence class. Step 3: A universal relation on a non-empty set has only one class.