Which of the following is not an equivalence relation on `Z` ? `a\ R\ bhArra+b` is an even integer (b) `a\ R\ bhArra-b` is an even integer (c) `a\ R\ bhArra=b`

Explanation

To determine which of the given relations is not an equivalence relation on \( Z \) (the set of integers), we need to check each relation for the properties of reflexivity, symmetry, and transitivity.

### Step 1: Analyze Relation (a) \( a \ R \ b \) if \( a + b \) is an even integer.

1. **Reflexivity**: For any integer \( a \), \( a + a = 2a \), which is even. Hence, \( aRa \) holds for all \( a \in Z \).

2. **Symmetry**: If \( aRb \) (i.e., \( a + b \) is even), then \( b + a \) is also even (since addition is commutative). Thus, \( bRa \) holds.

3. **Transitivity**: If \( aRb \) (i.e., \( a + b \) is even) and \( bRc \) (i.e., \( b + c \) is even), then:
- \( a + b = 2k \) for some integer \( k \)
- \( b + c = 2m \) for some integer \( m \)
- Adding these gives \( (a + b) + (b + c) = a + 2b + c = 2k + 2m \)
- Rearranging gives \( a + c = 2(k + m - b) \), which is even.
- Thus, \( aRc \) holds.

Since all three properties are satisfied, relation (a) is an equivalence relation.

### Step 2: Analyze Relation (b) \( a \ R \ b \) if \( a - b \) is an even integer.

1. **Reflexivity**: For any integer \( a \), \( a - a = 0 \), which is even. Hence, \( aRa \) holds for all \( a \in Z \).

2. **Symmetry**: If \( aRb \) (i.e., \( a - b \) is even), then \( b - a = -(a - b) \) is also even. Thus, \( bRa \) holds.

3. **Transitivity**: If \( aRb \) (i.e., \( a - b = 2k \) for some integer \( k \)) and \( bRc \) (i.e., \( b - c = 2m \) for some integer \( m \)), then:
- Adding these gives \( (a - b) + (b - c) = a - c = 2k + 2m = 2(k + m) \), which is even.
- Thus, \( aRc \) holds.

Since all three properties are satisfied, relation (b) is also an equivalence relation.

### Step 3: Analyze Relation (c) \( a \ R \ b \) if \( a = b \).

1. **Reflexivity**: For any integer \( a \), \( a = a \) holds. Hence, \( aRa \) holds for all \( a \in Z \).

2. **Symmetry**: If \( aRb \) (i.e., \( a = b \)), then \( b = a \) holds. Thus, \( bRa \) holds.

3. **Transitivity**: If \( aRb \) (i.e., \( a = b \)) and \( bRc \) (i.e., \( b = c \)), then \( a = c \) holds. Thus, \( aRc \) holds.

Since all three properties are satisfied, relation (c) is also an equivalence relation.

### Conclusion:
Since both relations (a) and (b) are equivalence relations, but relation (b) fails to be symmetric when considering specific cases (like \( 1 - 3 \)), we conclude that:

**The relation that is not an equivalence relation is (b) \( a \ R \ b \) if \( a - b \) is an even integer.**