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Let Z be the set of integers and aRb, where a, b ∈ Z if and only if (a - b) is divisible by 5. Consider the following statements: 1. The relation R partitions Z into five equivalent classes 2. Any two equivalent classes are either equal or disjoint Which of the above statements is/are correct?

Let Z be the set of integers and aRb, where a, b ∈ Z if and only if (a - b) is divisible by 5. Consider the following statements: 1. The relation R partitions Z into five equivalent classes 2. Any two equivalent classes are either equal or disjoint Which of the above statements is/are correct? Correct Answer Both 1 and 2

Concept:

Let R be a binary relation on a set A.

1. Reflexive: Each element is related to itself.

  • R is reflexive if for all x ∈ A, xRx.

2. Symmetric: If any one element is related to any other element, then the second element is related to the first.

  • R is symmetric if for all x, y ∈ A, if xRy, then yRx.

3. Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.

  • R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

4. R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.

Calculation:

A relation is defined on Z such that aRb ⇒ (a − b) is divisible by 5,

For Reflexive: (a, a) ∈ R.

Since, (a − a) = 0 is divisible by 5. 

Therefore, the relation is reflexive.

For symmetric: If (a, b) ∈ R ⇒ (b, a) ∈ R.

(a, b)∈ R ⇒ (a − b) is divisible by 5.

Now, (b − a) = − (a − b) is also divisible by 5.

Therefore, (b, a) ∈ R

Hence, the relation is symmetric.

For Transitive: If (a, b) ∈ R and (b, a) ∈ R ⇒ (a, c) ∈ R.

(a, b) ∈ R ⇒ (a − b) is divisible by 5.

(b, c) ∈ R ⇒ (b − c) is divisible by 5.

Then,

(a − c) = (a – b + b −c)

(a − c)  = (a − b) + (b − c)

We know that (a − b) is divisible by 5 and (b − c) is divisible by 5 then (a − c) is also divisible by 5.  Therefore, (a, c) ∈ R.

Hence, the relation is transitive.

∴ the relation is equivalent.

Now, depending upon the remainder obtained when dividing (a−b) by 5 we can divide the set Z into 5 equivalent classes and they are disjoint i.e., there are no common elements between any two classes.

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