Consider the relations $R_1$ and $R_2$ defined as $aR_1b \Leftrightarrow a^2 + b^2 = 1$ for all $a, b \in R$ and $(a, b)R_2(c, d) \Leftrightarrow a+ d = b + c$ for all $(a, b), (c, d) \in N \times N$. Then:
Correct Answer:
C
Explanation
To determine if the given relations $R_1$ and $R_2$ are equivalence relations, we need to check whether each of them satisfies the three defining properties of an equivalence relation: reflexivity, symmetry, and transitivity.
For $R_1$:
Reflexivity: For $R_1$ to be reflexive, $aR_1a$ must hold for all $a \in R$. This means $a^2 + a^2 = 1$, or $2a^2 = 1$, which implies $a^2 = \frac{1}{2}$. Since this is not true for every real number $a$, $R_1$ is not reflexive.
Symmetry: If $aR_1b$, then $a^2 + b^2 = 1$. This also means $b^2 + a^2 = 1$, so $bR_1a$. Thus, $R_1$ is symmetric.
Transitivity: If $aR_1b$ and $bR_1c$, then $a^2 + b^2 = 1$ and $b^2 + c^2 = 1$. Adding these, we get $a^2 + 2b^2 + c^2 = 2$. There is no guarantee that $a^2 + c^2 = 1$. Therefore, $R_1$ is not transitive.
$R_1$ is not an equivalence relation.
For $R_2$:
Reflexivity: For any $(a, b) \in N \times N$, $a + b = b + a$. Therefore, $(a, b)R_2(a, b)$, and $R_2$ is reflexive.
Symmetry: If $(a, b)R_2(c, d)$, then $a + d = b + c$. This can be reordered to $c + b = d + a$, so $(c, d)R_2(a, b)$. Thus, $R_2$ is symmetric.
Transitivity: If $(a, b)R_2(c, d)$ and $(c, d)R_2(e, f)$, then $a + d = b + c$ and $c + f = d + e$. Adding the two equations, we get $a + d + c + f = b + c + d + e$. Simplifying, we get $a + f = b + e$, so $(a, b)R_2(e, f)$.
$R_2$ is reflexive, symmetric, and transitive; therefore, it is an equivalence relation.
Therefore, only $R_2$ is an equivalence relation.