Key Concepts and Formulas

• Equivalence Relation: A relation $R$ on a set $S$ is an equivalence relation if it is reflexive, symmetric, and transitive.
  • Reflexive: For all $a \in S$, $a R a$.
  • Symmetric: For all $a, b \in S$, if $a R b$, then $b R a$.
  • Transitive: For all $a, b, c \in S$, if $a R b$ and $b R c$, then $a R c$.
• Properties of Real Numbers and Natural Numbers: Commutative property of addition ($x+y=y+x$) and transitive property of equality ($a=b$ and $b=c \implies a=c$).

Step-by-Step Solution

Step 1: Analyze Relation $R_1$ for Equivalence Properties

The relation $R_1$ is defined as $a R_1 b \Leftrightarrow a^2+b^2=1$ for all $a, b \in \mathbf{R}$. We will check reflexivity, symmetry, and transitivity.

• Reflexivity Check for $R_1$: For $R_1$ to be reflexive, $a R_1 a$ must hold for all $a \in \mathbf{R}$. This means $a^2+a^2=1$, which simplifies to $2a^2=1$, or $a^2 = \frac{1}{2}$. This equation is only true for $a = \pm \frac{1}{\sqrt{2}}$. Since this condition does not hold for all real numbers (e.g., if $a=1$, $1^2+1^2=2 \neq 1$), $R_1$ is not reflexive.

• Symmetry Check for $R_1$: Assume $a R_1 b$, which means $a^2+b^2=1$. We need to check if $b R_1 a$ holds, i.e., if $b^2+a^2=1$. Due to the commutative property of addition, $a^2+b^2 = b^2+a^2$. Thus, if $a^2+b^2=1$, then $b^2+a^2=1$ is also true. So, $R_1$ is symmetric.

• Transitivity Check for $R_1$: For $R_1$ to be transitive, if $a R_1 b$ and $b R_1 c$, then $a R_1 c$ must hold. This means if $a^2+b^2=1$ and $b^2+c^2=1$, then $a^2+c^2=1$ must be true. Consider a counterexample: Let $a=1$, $b=0$, and $c=1$. Then $a R_1 b$ since $1^2+0^2 = 1+0=1$. And $b R_1 c$ since $0^2+1^2 = 0+1=1$. However, $a R_1 c$ would require $1^2+1^2=1$, which is $1+1=2 \neq 1$. Since we found a counterexample where the condition for transitivity fails, $R_1$ is not transitive.

Since $R_1$ is not reflexive and not transitive, it is not an equivalence relation.

Step 2: Analyze Relation $R_2$ for Equivalence Properties

The relation $R_2$ is defined as $(a, b) R_2(c, d) \Leftrightarrow a+d=b+c$ for all $(a, b),(c, d) \in \mathbf{N} \times \mathbf{N}$. We will check reflexivity, symmetry, and transitivity.

• Reflexivity Check for $R_2$: For $R_2$ to be reflexive, $(a, b) R_2 (a, b)$ must hold for all $(a, b) \in \mathbf{N} \times \mathbf{N}$. This means $a+b=b+a$. This is true for all natural numbers due to the commutative property of addition. Thus, $R_2$ is reflexive.

• Symmetry Check for $R_2$: Assume $(a, b) R_2 (c, d)$, which means $a+d=b+c$. We need to check if $(c, d) R_2 (a, b)$ holds, i.e., if $c+b=d+a$. Starting with $a+d=b+c$, we can rearrange it to $b+c=a+d$. Using the commutative property of addition, we can write this as $c+b=d+a$. This is precisely the condition for $(c, d) R_2 (a, b)$. Thus, $R_2$ is symmetric.

• Transitivity Check for $R_2$: For $R_2$ to be transitive, if $(a, b) R_2 (c, d)$ and $(c, d) R_2 (e, f)$, then $(a, b) R_2 (e, f)$ must hold. This means if $a+d=b+c$ and $c+f=d+e$, then $a+f=b+e$ must be true. From $a+d=b+c$, we can write $a-b = c-d$. From $c+f=d+e$, we can write $c-d = e-f$. By the transitive property of equality, since $a-b = c-d$ and $c-d = e-f$, we have $a-b = e-f$. Rearranging this equation, we get $a+f = b+e$. This is the condition for $(a, b) R_2 (e, f)$. Thus, $R_2$ is transitive.

Since $R_2$ is reflexive, symmetric, and transitive, it is an equivalence relation.

Step 3: Conclude and Select the Correct Option

We found that $R_1$ is not an equivalence relation, and $R_2$ is an equivalence relation. Therefore, only $R_2$ is an equivalence relation.

Common Mistakes & Tips

• Confusing Sets: Ensure you are applying the relation's definition to the correct set ($\mathbf{R}$ for $R_1$, $\mathbf{N} \times \mathbf{N}$ for $R_2$).
• Counterexample Strategy: For proving a relation is not an equivalence relation, finding just one property that fails with a single counterexample is sufficient.
• Algebraic Manipulation: For relations defined by algebraic conditions, carefully use properties of numbers (like commutativity, associativity, transitivity of equality) to prove or disprove the required properties.

Summary

We systematically analyzed the relations $R_1$ and $R_2$ by checking the three properties of an equivalence relation: reflexivity, symmetry, and transitivity. Relation $R_1$, defined on real numbers, failed the reflexivity and transitivity tests, thus it is not an equivalence relation. Relation $R_2$, defined on pairs of natural numbers, satisfied all three properties, confirming it is an equivalence relation.

The final answer is \boxed{C}.