Question

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:

• A. Only R1 is an equivalence relation
• B. Only R2 is an equivalence relation
• C. R1 and R2 both are equivalence relations
• D. Neither R1 nor R2 is an equivalence relation

Answer 1

The Correct Option is B

Given the relations, let's evaluate whether \( R_1 \) and \( R_2 \) are equivalence relations by checking the properties: reflexivity, symmetry, and transitivity.

Relation \( R_1 \):

• Reflexivity: For \( a R_1 a \), we need \( a^2 + a^2 = 1 \), which simplifies to \( 2a^2 = 1 \). However, this doesn't hold for all \( a \in \mathbb{R} \). Thus, \( R_1 \) is not reflexive.
• Symmetry: Suppose \( a R_1 b \), which implies \( a^2 + b^2 = 1 \). Clearly, if \( a^2 + b^2 = 1 \), then \( b^2 + a^2 = 1 \), so \( b R_1 a \). Hence, \( R_1 \) is symmetric.
• Transitivity: Suppose \( a R_1 b \) and \( b R_1 c \). We have \( a^2 + b^2 = 1 \) and \( b^2 + c^2 = 1 \). Adding these gives \( a^2 + 2b^2 + c^2 = 2 \). This doesn't imply \( a^2 + c^2 = 1 \) unless \( b^2 = 0 \), which fails for all \( a, b, c \). Thus, \( R_1 \) is not transitive.

\( R_1 \) fails the reflexivity and transitivity tests, so it is not an equivalence relation.

Relation \( R_2 \):

• Reflexivity: We need to check if \( (a, b) R_2 (a, b) \). In this case, \( a + b = b + a \), which is true for all \( (a, b) \in \mathbb{N} \times \mathbb{N} \). Hence, \( R_2 \) is reflexive.
• Symmetry: If \( (a, b) R_2 (c, d) \), then \( a + d = b + c \). This implies \( c + b = d + a \), so \( (c, d) R_2 (a, b) \). Therefore, \( R_2 \) is symmetric.
• Transitivity: Suppose \( (a, b) R_2 (c, d) \) and \( (c, d) R_2 (e, f) \). Then \( a + d = b + c \) and \( c + f = d + e \). Adding these gives \( a + f = b + e \). Hence, \( (a, b) R_2 (e, f) \), showing \( R_2 \) is transitive.

\( R_2 \) satisfies reflexivity, symmetry, and transitivity, so it is an equivalence relation.

Based on these analyses, the correct answer is: Only \( R_2 \) is an equivalence relation.

Answer 2

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.