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

Answer: (B)

Only R2 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.