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

Determine which of $$R_1$$ and $$R_2$$ is an equivalence relation.

An equivalence relation must be reflexive, symmetric, and transitive.

Checking $$R_1$$: $$aR_1b \Leftrightarrow a^2 + b^2 = 1$$ for $$a, b \in \mathbb{R}$$.

Reflexive? We need $$aR_1a$$, i.e., $$a^2 + a^2 = 1 \Rightarrow 2a^2 = 1 \Rightarrow a = \pm\frac{1}{\sqrt{2}}$$. This fails for other values of $$a$$ (e.g., $$0^2 + 0^2 = 0 \neq 1$$). Not reflexive.

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

Checking $$R_2$$: $$(a,b)R_2(c,d) \Leftrightarrow a + d = b + c$$ for $$(a,b),(c,d) \in \mathbb{N} \times \mathbb{N}$$.

Reflexive? $$(a,b)R_2(a,b) \Leftrightarrow a + b = b + a$$. True always. Reflexive.

Symmetric? If $$(a,b)R_2(c,d)$$, then $$a+d = b+c$$, which means $$c+b = d+a$$, so $$(c,d)R_2(a,b)$$. Symmetric.

Transitive? 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: $$a+d+c+f = b+c+d+e$$, which simplifies to $$a+f = b+e$$, so $$(a,b)R_2(e,f)$$. Transitive.

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

The correct answer is Option B: Only $$R_2$$ is an equivalence relation.