Understanding the Relation on Integers

The question describes a relation R defined on the set of integers, denoted by Z. The relation is given by aRb if and only if (a - b) is divisible by 5, for any integers a and b belonging to Z.

This type of relation, where the difference between two numbers is divisible by a fixed integer, is a classic example of a congruence relation. Specifically, aRb means a is congruent to b modulo 5, often written as a \equiv b \pmod{5}.

Verifying the Relation is an Equivalence Relation

For a relation to partition a set into equivalence classes, it must be an equivalence relation. An equivalence relation satisfies three properties:

• Reflexivity: For any a \in Z, is aRa? This means (a - a) must be divisible by 5. a - a = 0, and 0 is indeed divisible by any non-zero integer, including 5 (since 0 = 5 \times 0). Thus, R is reflexive.
• Symmetry: If aRb, is bRa? If aRb, then (a - b) is divisible by 5. This means a - b = 5k for some integer k. We want to check if (b - a) is divisible by 5. Note that b - a = -(a - b) = -(5k) = 5(-k). Since -k is also an integer, (b - a) is divisible by 5. Thus, R is symmetric.
• Transitivity: If aRb and bRc, is aRc? If aRb, then (a - b) is divisible by 5, so a - b = 5k_1 for some integer k_1. If bRc, then (b - c) is divisible by 5, so b - c = 5k_2 for some integer k_2. To check if aRc, we look at (a - c). We can write a - c = (a - b) + (b - c). Substituting the expressions, we get a - c = 5k_1 + 5k_2 = 5(k_1 + k_2). Since k_1 + k_2 is an integer, (a - c) is divisible by 5. Thus, R is transitive.

Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

Analyzing Statement 1: Partition into Equivalent Classes

Statement 1 says: "The relation R partitions Z into five equivalent classes".

An equivalence class of an element a \in Z, denoted by [a], is the set of all elements in Z that are related to a by the relation R. [a] = \{b \in Z \mid aRb\} = \{b \in Z \mid (a - b) \text{ is divisible by } 5\}

This means b is in the equivalence class of a if and only if a - b = 5k for some integer k, which is equivalent to a \equiv b \pmod{5}. This happens if and only if a and b have the same remainder when divided by 5.

The possible remainders when an integer is divided by 5 are 0, 1, 2, 3, and 4.

These remainders define the distinct equivalence classes:

• The class of integers with remainder 0 when divided by 5: [0] = \{..., -10, -5, 0, 5, 10, ...\}
• The class of integers with remainder 1 when divided by 5: [1] = \{..., -9, -4, 1, 6, 11, ...\}
• The class of integers with remainder 2 when divided by 5: [2] = \{..., -8, -3, 2, 7, 12, ...\}
• The class of integers with remainder 3 when divided by 5: [3] = \{..., -7, -2, 3, 8, 13, ...\}
• The class of integers with remainder 4 when divided by 5: [4] = \{..., -6, -1, 4, 9, 14, ...\}

Every integer belongs to exactly one of these five classes based on its remainder modulo 5. These five distinct classes cover the entire set of integers Z. Therefore, the relation R partitions Z into exactly five equivalence classes.

Statement 1 is correct.

Analyzing Statement 2: Property of Equivalent Classes

Statement 2 says: "Any two equivalent classes are either equal or disjoint".

This is a fundamental property of any set of equivalence classes formed by an equivalence relation on a set. For any equivalence relation R on a set S, its equivalence classes partition S. A key characteristic of a partition is that the subsets are pairwise disjoint (except for possibly being identical) and their union covers the entire set.

Let [a] and [b] be two equivalence classes. We need to show that either [a] = [b] or [a] \cap [b] = \emptyset.

Suppose [a] and [b] are not disjoint. This means there exists an element c such that c \in [a] and c \in [b].

• If c \in [a], then aRc.
• If c \in [b], then bRc.

Since R is symmetric, bRc implies cRb. Now, using transitivity, aRc and cRb implies aRb.

If aRb, then any element x \in [a] satisfies aRx. By transitivity (since aRb and aRx implies bRx), we have x \in [b]. So, [a] \subseteq [b].

Similarly, if aRb, then by symmetry bRa. Any element y \in [b] satisfies bRy. By transitivity (since bRa and bRy implies aRy), we have y \in [a]. So, [b] \subseteq [a].

Since [a] \subseteq [b] and [b] \subseteq [a], it follows that [a] = [b].

Therefore, if two equivalence classes have at least one element in common (i.e., are not disjoint), they must be the same class. This proves that any two equivalence classes are either identical or completely separate (disjoint).

Statement 2 is a fundamental property of equivalence relations and the partitions they induce. It is correct.

Conclusion

Both Statement 1, which correctly identifies the number of equivalence classes for the given relation on integers modulo 5, and Statement 2, which states a general property of equivalence classes, are correct.

Revision Table: Key Concepts

Additional Information: Equivalence Relations and Partitions

The concept of an equivalence relation is closely linked to the concept of a partition of a set. An equivalence relation on a set S creates a natural way to group elements of S into disjoint, non-empty subsets called equivalence classes. The collection of all such equivalence classes forms a partition of S.

Conversely, given any partition of a set S, we can define an equivalence relation on S where two elements are related if and only if they belong to the same subset in the partition. This establishes a one-to-one correspondence between equivalence relations on a set and partitions of that set.

In this problem, the set Z is partitioned based on the remainder modulo 5. Each 'part' of the partition is one of the five equivalence classes [0], [1], [2], [3], [4]. These parts are non-empty (e.g., 0 ∈ [0], 1 ∈ [1], etc.), their union is Z, and any two distinct parts are disjoint.