Let `A={1,\ 2,\ 3,\ ,\ 9}` and `R` be the relation on `AxxA` defined by `(a ,\ b)R\ (c ,\ d)` if `a+d=b+c` for all `(a ,\ b),\ (c ,\ d) in AxxA` . Prove that `R` is an equivalence relation and also obtain the equivalence class [(2, 5)].

Explanation

To prove that the relation \( R \) defined on \( A \times A \) by \((a, b) R (c, d)\) if \( a + d = b + c \) is an equivalence relation, we need to show that it is reflexive, symmetric, and transitive.

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

**Proof:**
For any \( (a, b) \in A \times A \):
\[ a + b = b + a \]
Thus, \( (a, b) R (a, b) \) holds true. Therefore, \( R \) is reflexive.

### Step 2: Prove Symmetry
A relation \( R \) is symmetric if whenever \( (a, b) R (c, d) \), it also holds that \( (c, d) R (a, b) \).

**Proof:**
Assume \( (a, b) R (c, d) \). This means:
\[ a + d = b + c \]
Rearranging gives:
\[ c + b = d + a \]
Thus, \( (c, d) R (a, b) \) holds true. Therefore, \( R \) is symmetric.

### Step 3: Prove Transitivity
A relation \( R \) is transitive if whenever \( (a, b) R (c, d) \) and \( (c, d) R (e, f) \), it also holds that \( (a, b) R (e, f) \).

**Proof:**
Assume \( (a, b) R (c, d) \) and \( (c, d) R (e, f) \). This gives us:
1. \( a + d = b + c \) (1)
2. \( c + f = d + e \) (2)

From (1), we can express \( d \) as:
\[ d = b + c - a \]
Substituting \( d \) into (2):
\[ c + f = (b + c - a) + e \]
Simplifying gives:
\[ f = e + a - b \]
Now we need to show that:
\[ a + f = b + e \]
Substituting \( f \):
\[ a + (e + a - b) = b + e \]
This simplifies to:
\[ 2a - b = b \]
Thus, \( (a, b) R (e, f) \) holds true. Therefore, \( R \) is transitive.

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

### Step 4: Find the Equivalence Class \([(2, 5)]\)
To find the equivalence class \([(2, 5)]\), we need to find all pairs \((c, d)\) such that:
\[ 2 + d = 5 + c \]
Rearranging gives:
\[ d = c + 3 \]
Now we will find all pairs \((c, d)\) where \( c \) and \( d \) are in the set \( A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \).

**Finding pairs:**
- If \( c = 1 \), then \( d = 1 + 3 = 4 \) → Pair: \( (1, 4) \)
- If \( c = 2 \), then \( d = 2 + 3 = 5 \) → Pair: \( (2, 5) \)
- If \( c = 3 \), then \( d = 3 + 3 = 6 \) → Pair: \( (3, 6) \)
- If \( c = 4 \), then \( d = 4 + 3 = 7 \) → Pair: \( (4, 7) \)
- If \( c = 5 \), then \( d = 5 + 3 = 8 \) → Pair: \( (5, 8) \)
- If \( c = 6 \), then \( d = 6 + 3 = 9 \) → Pair: \( (6, 9) \)

Thus, the equivalence class \([(2, 5)]\) is:
\[ \{(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)\} \]