Consider the relations $R_1$ and $R_2$ defined as $a R_1 b \Leftrightarrow a^2+b^2=1$ for all $a, b \in \mathbf{R}$ and $(a, b) R_2(c, d) \Leftrightarrow$ $a+d=b+c$ for all $(a, b),(c, d) \in \mathbf{N} \times \mathbf{N}$. Then :

<p>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.</p> <p>Let's start by analysing $R_1$:</p> <p>Reflexivity: A relation $R$ on a set $S$ is reflexive if every element is related to itself, that is, for every $a \in S$, the pair $(a,a)$ is in $R$. In the case of $R_1$, for an arbitrary $a \in \mathbf{R}$, we need to check if $a R_1 a$ holds, which translates to checking if $a^2 + a^2 = 1$. This would mean $2a^2 = 1$ or $a^2 = \frac{1}{2}$. Since this is not true for every real number $a$, $R_1$ is not reflexive.</p> <p>Symmetry: A relation $R$ is symmetric if whenever $a R b$, then $b R a$. For $R_1$, if $a^2 + b^2 = 1$, then it is also true that $b^2 + a^2 = 1$. Thus, $R_1$ is symmetric.</p> <p>Transitivity: A relation $R$ is transitive if whenever $a R b$ and $b R c$, then $a R c$. For $R_1$, suppose $a R_1 b$ and $b R_1 c$, this means $a^2 + b^2 = 1$ and $b^2 + c^2 = 1$. However, if we add these two, 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.</p> <p>Conclusion: Relation $R_1$ is not reflexive or transitive, and hence it is not an equivalence relation.</p> <p>Now let's consider $R_2$:</p> <p>Reflexivity: For any $(a, b) \in \mathbf{N} \times \mathbf{N}$, it's clear that $a + b = b + a$, which is just a reiteration of the commutative property of addition. Therefore, $(a, b) R_2 (a, b)$, and $R_2$ is reflexive.</p> <p>Symmetry: If $(a, b) R_2 (c, d)$, meaning $a + d = b + c$, then by reordering the terms we can similarly have $c + b = d + a$, which means $(c, d) R_2 (a, b)$, so $R_2$ is symmetric.</p> <p>Transitivity: Suppose $(a, b) R_2 (c, d)$ and $(c, d) R_2 (e, f)$, i.e., $a + d = b + c$ and $c + f = d + e$, we want to show that $(a, b) R_2 (e, f)$. Adding the two equations, we get $a + d + c + f = b + c + d + e$. By rearranging and simplifying, we get $a + f = b + e$, thus, $(a, b) R_2 (e, f)$.</p> <p>So $R_2$ is reflexive, symmetric, and transitive, and therefore it is an equivalence relation.</p> <p>Conclusion: According to the above examination, only $R_2$ is an equivalence relation. Thus, the correct answer is:</p> <p>Option C: Only $R_2$ is an equivalence relation.</p>