Equivalence relation

Step 1: Integers related to (15) must differ from (15) by a multiple of (7). Step 2: (15) gives remainder (1) on division by (7), so all integers with remainder (1) are in the same class. Step 3: First identify the remainder to find the equivalence class quickly.

Step 1: (4) gives remainder (1) on division by (3). Step 2: (1,4,7) all have remainder (1), so they form the same class. Step 3: In a finite set, include only elements that belong to the given set.

Step 1: (1) and (4) are related to each other, so they form one class. Step 2: (2) and (3) are related only to themselves, so they form singleton classes. Step 3: Group mutually related elements together.

Step 1: Within each equivalence class, every element is related to every element. Step 2: ({p,q,r}) contributes (3^2=9) pairs and ({s}) contributes (1^2=1) pair. Step 3: The total number of pairs is (9+1=10).

Step 1: Division by (8) gives remainders (0,1,2,3,4,5,6,7). Step 2: Integers with the same remainder lie in the same equivalence class. Step 3: Although there are infinitely many integers, there are (8) distinct classes.

Step 1: (a^3=a^3), so the relation is reflexive. Step 2: If (a^3=b^3), then (b^3=a^3), so it is symmetric. Step 3: If (a^3=b^3) and (b^3=c^3), then (a^3=c^3), so it is transitive.

Step 1: The square of (1) is (1). Step 2: In the given set, both (-1) and (1) have square (1). Step 3: In square-equality relations, a number and its negative often belong to the same class.

Step 1: (|1-4|=3), which is divisible by (3). Step 2: Therefore ((1,4)) belongs to the relation. Step 3: Find the absolute difference and then test divisibility.

Step 1: Equivalence classes are either equal or disjoint. Step 2: If (a) and (b) are not related, they cannot be in the same class. Step 3: Hence their classes are disjoint.

Step 1: A partition has non-empty blocks that do not overlap. Step 2: ({1,3,5}) and ({2,4}) are disjoint and cover the whole set. Step 3: Equivalence classes always form a valid partition.

Step 1: The pairs ((2,1)) and ((1,3)) are in the relation. Step 2: Transitivity requires ((2,3)), but it is missing. Step 3: While checking equivalence, use two connected pairs to identify the required third pair.

Step 1: (a-a=0) is an integer, so the relation is reflexive. Step 2: If (a-b) is an integer, then (b-a=-(a-b)) is also an integer. Step 3: The sum of two integer differences is also an integer, so transitivity holds.

Step 1: (\frac{5}{2}-\frac{1}{2}=2). Step 2: Since (2) is an integer, (\frac{5}{2}) and (\frac{1}{2}) are in the same equivalence class. Step 3: In such questions, subtract the two numbers.

Step 1: For (a+b) to be even, the two numbers must have the same parity. Step 2: Since (2) is even, the elements related to (2) are (2,4,6). Step 3: An even-sum relation creates even and odd classes.

Step 1: Every element belongs to some equivalence class. Step 2: Distinct equivalence classes are disjoint and together cover (A). Step 3: Hence the collection of all equivalence classes is a partition of (A).

Step 1: The relation from a partition contains pairs only within the same block. Step 2: (2) and (3) are in different blocks. Step 3: Cross-block pairs are not included, so ((2,3)) is not in the relation.

Step 1: Equivalence classes must cover the whole set. Step 2: ({1}) and ({2,3,4}) are disjoint and cover all of (A). Step 3: Blocks that overlap or miss elements cannot be equivalence classes.

Step 1: Every element is in the same group as itself, so reflexivity holds. Step 2: Being in the same group is symmetric. Step 3: The two classes are ({3,6}) and ({1,2,4,5}), so the relation is an equivalence relation.

Step 1: (5) is not a multiple of (3). Step 2: Therefore its class contains all elements that are not multiples of (3). Step 3: To find a class, first identify the group of the given element.

Step 1: The statement says that the reverse of every related pair is also related. Step 2: This is the definition of symmetry. Step 3: An equivalence relation also needs reflexivity and transitivity.