Let $$R_1 = \{(a,b) \in N \times N : |a - b| \leq 13\}$$ and $$R_2 = \{(a,b) \in N \times N : |a - b| \neq 13\}$$. Then on $$N$$:

We need to check whether $$R_1$$ and $$R_2$$ are equivalence relations on $$\mathbb{N}$$.

Analysis of $$R_1 = \{(a, b) \in \mathbb{N} \times \mathbb{N} : |a - b| \leq 13\}$$:

• Reflexive: $$|a - a| = 0 \leq 13$$ $$\checkmark$$
• Symmetric: $$|a - b| = |b - a|$$ $$\checkmark$$
• Transitive: Consider $$a = 1, b = 14, c = 27$$. We have $$|1 - 14| = 13 \leq 13$$ and $$|14 - 27| = 13 \leq 13$$, but $$|1 - 27| = 26 > 13$$. So $$(1, 14) \in R_1$$ and $$(14, 27) \in R_1$$, but $$(1, 27) \notin R_1$$. $$\boldsymbol{\times}$$

$$R_1$$ is not an equivalence relation (fails transitivity).

Analysis of $$R_2 = \{(a, b) \in \mathbb{N} \times \mathbb{N} : |a - b| \neq 13\}$$:

• Reflexive: $$|a - a| = 0 \neq 13$$ $$\checkmark$$
• Symmetric: $$|a - b| = |b - a|$$ $$\checkmark$$
• Transitive: Consider $$a = 1, b = 2, c = 14$$. We have $$|1 - 2| = 1 \neq 13$$ and $$|2 - 14| = 12 \neq 13$$, but $$|1 - 14| = 13$$. So $$(1, 2) \in R_2$$ and $$(2, 14) \in R_2$$, but $$(1, 14) \notin R_2$$. $$\boldsymbol{\times}$$

$$R_2$$ is not an equivalence relation (fails transitivity).

The correct answer is Option B: Neither $$R_1$$ nor $$R_2$$ is an equivalence relation.