N is the set of natural numbers. The relation R is defined on `NxxN` as follows (a,b) R (c, d)' a + d = b + c Prove that R is an equivalence relation.

Answer

To prove that the relation \( R \) defined on \( N \times N \) by \( (a, b) R (c, d) \) if and only if \( a + d = b + c \) is an equivalence relation, we need to show that it satisfies three properties: reflexivity, symmetry, and transitivity.

### Step 1: Proving Reflexivity
To show that \( R \) is reflexive, we need to prove that for any \( (a, b) \in N \times N \), it holds that \( (a, b) R (a, b) \).

- We check the condition:
\[ a + b = b + a \]
This is clearly true since addition is commutative.

Thus, \( (a, b) R (a, b) \) holds for all \( (a, b) \in N \times N \), proving that \( R \) is reflexive.

### Step 2: Proving Symmetry
Next, we need to show that if \( (a, b) R (c, d) \), then \( (c, d) R (a, b) \).

- Assume \( (a, b) R (c, d) \), which means:
\[ a + d = b + c \]
- We need to show that:
\[ c + b = d + a \]
- Rearranging the first equation gives:
\[ c + b = a + d \]
- This is equivalent to:
\[ d + a = c + b \]
- Therefore, \( (c, d) R (a, b) \) holds.

Thus, \( R \) is symmetric.

### Step 3: Proving Transitivity
Finally, we need to show that if \( (a, b) R (c, d) \) and \( (c, d) R (e, f) \), then \( (a, b) R (e, f) \).

- Assume:
\[ (a, b) R (c, d) \implies a + d = b + c \]
and
\[ (c, d) R (e, f) \implies c + f = d + e \]

- We need to show:
\[ a + f = b + e \]

- From the first equation, we have:
\[ a + d = b + c \quad \text{(1)} \]
- From the second equation, we can express \( d \) in terms of \( c \), \( e \), and \( f \):
\[ d = c + f - e \quad \text{(from (2))} \]

- Substituting \( d \) from (2) into (1):
\[ a + (c + f - e) = b + c \]
- Simplifying this gives:
\[ a + f - e = b \]
- Rearranging leads to:
\[ a + f = b + e \]

Thus, \( (a, b) R (e, f) \) holds, proving that \( R \) is transitive.

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

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