Let n be a fixed positive integer. Define a relation R on I (the set of all intergers) as follows :
aRB iff n / (a - b), that is iff a - b is divisible by n. Show that R is an equivalence relation on I.

Answer

To show that the relation \( R \) defined on the set of integers \( I \) by \( aRb \) if and only if \( n \) divides \( (a - b) \) is an equivalence relation, we need to verify three properties: reflexivity, symmetry, and transitivity.

### Step 1: Prove Reflexivity
A relation \( R \) is reflexive if for every element \( a \in I \), it holds that \( aRa \).

**Proof:**
For any integer \( a \), we have:
\[ a - a = 0 \]
Since \( n \) divides \( 0 \) (as \( 0 = n \times 0 \)), it follows that:
\[ aRa \]
Thus, \( R \) is reflexive.

### Step 2: Prove Symmetry
A relation \( R \) is symmetric if whenever \( aRb \), then \( bRa \).

**Proof:**
Assume \( aRb \). This means:
\[ n \text{ divides } (a - b) \]
This implies there exists an integer \( k \) such that:
\[ a - b = nk \]
Rearranging gives:
\[ b - a = -nk \]
Since \( -nk \) is also divisible by \( n \), we have:
\[ n \text{ divides } (b - a) \]
Thus, \( bRa \). Therefore, \( R \) is symmetric.

### Step 3: Prove Transitivity
A relation \( R \) is transitive if whenever \( aRb \) and \( bRc \), then \( aRc \).

**Proof:**
Assume \( aRb \) and \( bRc \). This means:
\[ n \text{ divides } (a - b) \]
and
\[ n \text{ divides } (b - c) \]
This implies there exist integers \( k_1 \) and \( k_2 \) such that:
\[ a - b = nk_1 \]
\[ b - c = nk_2 \]
Adding these two equations gives:
\[ (a - b) + (b - c) = nk_1 + nk_2 \]
This simplifies to:
\[ a - c = n(k_1 + k_2) \]
Since \( k_1 + k_2 \) is an integer, it follows that:
\[ n \text{ divides } (a - c) \]
Thus, \( aRc \). Therefore, \( R \) is transitive.

### Conclusion
Since \( R \) is reflexive, symmetric, and transitive, we conclude that \( R \) is an equivalence relation on \( I \).

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