Consider the relations `R_(1)` and `R_(2)` defined as `aR_(1)b hArr a^(2)+b^(2)=1` for all `a,b in R` and `(a,b)R_(2)(c,d) hArr ,a+d=b+c forall (a,b),(c,d) in N times N` .Then

Options

A. Neither `R_(1)` nor `R_(2)` is an equivalence relation
B. Only `R_(1)` is an equivalence relation`
C. Only `R_(2)` is an equivalence relation
D. `R_(1)` and `R_(2)` both are equivalence relations

Answer

To solve the problem, we need to analyze the two relations \( R_1 \) and \( R_2 \) given in the question.

### Step-by-Step Solution

1. **Understanding the Relation \( R_1 \)**:
The relation \( R_1 \) is defined as:
\[ a R_1 b \iff a^2 + b^2 = 1 \]
for all \( a, b \in \mathbb{R} \).

2. **Checking Reflexivity for \( R_1 \)**:
A relation is reflexive if for every element \( a \), \( a R_1 a \) holds true.
\[ a R_1 a \implies a^2 + a^2 = 1 \implies 2a^2 = 1 \implies a^2 = \frac{1}{2} \implies a = \pm \frac{1}{\sqrt{2}} \]
Since this is not true for all \( a \in \mathbb{R} \) (only for specific values), \( R_1 \) is **not reflexive**.

3. **Checking Symmetry for \( R_1 \)**:
A relation is symmetric if \( a R_1 b \) implies \( b R_1 a \).
\[ a R_1 b \implies a^2 + b^2 = 1 \implies b^2 + a^2 = 1 \implies b R_1 a \]
Hence, \( R_1 \) is **symmetric**.

4. **Checking Transitivity for \( R_1 \)**:
A relation is transitive if \( a R_1 b \) and \( b R_1 c \) imply \( a R_1 c \).
Let's assume \( a R_1 b \) and \( b R_1 c \):
\[ a^2 + b^2 = 1 \quad \text{and} \quad b^2 + c^2 = 1 \]
Adding these:
\[ a^2 + 2b^2 + c^2 = 2 \]
This does not guarantee \( a^2 + c^2 = 1 \). Therefore, \( R_1 \) is **not transitive**.

5. **Conclusion for \( R_1 \)**:
Since \( R_1 \) is not reflexive and not transitive, it is not an equivalence relation.

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6. **Understanding the Relation \( R_2 \)**:
The relation \( R_2 \) is defined as:
\[ (a, b) R_2 (c, d) \iff a + d = b + c \]

7. **Checking Reflexivity for \( R_2 \)**:
For reflexivity:
\[ (a, b) R_2 (a, b) \implies a + b = b + a \quad \text{(which is always true)} \]
Hence, \( R_2 \) is **reflexive**.

8. **Checking Symmetry for \( R_2 \)**:
For symmetry:
\[ (a, b) R_2 (c, d) \implies a + d = b + c \implies c + b = d + a \implies (c, d) R_2 (a, b) \]
Hence, \( R_2 \) is **symmetric**.

9. **Checking Transitivity for \( R_2 \)**:
For transitivity:
Assume \( (a, b) R_2 (c, d) \) and \( (c, d) R_2 (e, f) \):
\[ a + d = b + c \quad \text{(1)} \quad \text{and} \quad c + f = d + e \quad \text{(2)} \]
From (1), we can express \( d \) in terms of \( a, b, c \):
\[ d = b + c - a \]
Substituting \( d \) into (2):
\[ c + f = (b + c - a) + e \implies f = e + a - b \]
Thus, \( (a, b) R_2 (e, f) \) holds true, confirming that \( R_2 \) is **transitive**.

10. **Conclusion for \( R_2 \)**:
Since \( R_2 \) is reflexive, symmetric, and transitive, it is an **equivalence relation**.

### Final Answer:
- \( R_1 \) is not an equivalence relation.
- \( R_2 \) is an equivalence relation.

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