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:

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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.

Answer

Only R1 is an equivalence relation

Answer

Only R2 is an equivalence relation

Answer

R1 and R2 both are equivalence relations

Answer

Neither R1 nor R2 is an equivalence relation

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:

The Correct Option is B

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