Consider the relations $R_1$ and $R_2$ defined as \[a R_1 b \iff a^2 + b^2 = 1 \quad \text{for all } a, b \in \mathbb{R},\]and \[(a, b) R_2 (c, d) \iff a + d = b + c \quad \text{for all } (a, b), (c, d) \in \mathbb{N} \times \mathbb{N}.\]Then:
- Only R1 is an equivalence relation
- Only R2 is an equivalence relation
- R1 and R2 both are equivalence relations
- Neither R1 nor R2 is an equivalence relation
The Correct Option is B
Solution and Explanation
To determine if the given relations \( R_1 \) and \( R_2 \) are equivalence relations, we check the properties of reflexivity, symmetry, and transitivity.
Checking \( R_1 \)
Given: \( aR_1b \iff a^2 + b^2 = 1 \) for all \( a, b \in R \).
- Reflexivity: For a relation to be reflexive, \( aR_1a \) should hold for all \( a \in R \). This implies:
\[ a^2 + a^2 = 1 \implies 2a^2 = 1 \implies a = \pm \frac{1}{\sqrt{2}} \]
Since this condition does not hold for all \( a \in R \), \( R_1 \) is not reflexive.
- Symmetry: For a relation to be symmetric, if \( aR_1b \) holds, then \( bR_1a \) should also hold.
Given \( a^2 + b^2 = 1 \), the equation is symmetric in \( a \) and \( b \). Thus, \( R_1 \) is symmetric.
- Transitivity: For a relation to be transitive, if \( aR_1b \) and \( bR_1c \) hold, then \( aR_1c \) should also hold. Consider:
\[ a^2 + b^2 = 1 \quad \text{and} \quad b^2 + c^2 = 1 \]
This does not imply \( a^2 + c^2 = 1 \). Hence, \( R_1 \) is not transitive.
Since \( R_1 \) fails the reflexivity and transitivity properties, it is not an equivalence relation.
Checking \( R_2 \)
Given: \( (a, b)R_2(c, d) \iff a + d = b + c \) for all \( (a, b), (c, d) \in \mathbb{N} \times \mathbb{N} \).
- Reflexivity: For any pair \( (a, b) \), we have:
\[ a + b = b + a \]
Hence, \( R_2 \) is reflexive.
- Symmetry: If \( (a, b)R_2(c, d) \), then \( a + d = b + c \). This implies:
\[ c + b = d + a \]
Thus, \( R_2 \) is symmetric.
- Transitivity: If \( (a, b)R_2(c, d) \) and \( (c, d)R_2(e, f) \), then:
\[ a + d = b + c \quad \text{and} \quad c + f = d + e \]
Adding these equations:
\[ a + d + c + f = b + c + d + e \implies a + f = b + e \]
Therefore, \( (a, b)R_2(e, f) \), and \( R_2 \) is transitive.
Since \( R_2 \) satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation.
Conclusion: Only \( R_2 \) is an equivalence relation.
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