Let R_{1} = {(a, b) in N times N : |a - b| le | vedclass

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(B) For $R_{1} = \{(a, b) \in N 	imes N : |a - b| \leq 13\}$:$(i)$ Reflexive: $|a - a| = 0 \leq 13$,so $(a, a) \in R_{1}$. It is reflexive.(ii) Symme

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Neither $R_{1}$ nor $R_{2}$ is an equivalence relation.

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(B) For $R_{1} = \{(a, b) \in N 	imes N : |a - b| \leq 13\}$:$(i)$ Reflexive: $|a - a| = 0 \leq 13$,so $(a, a) \in R_{1}$. It is reflexive.(ii) Symmetric: If $|a - b| \leq 13$,then $|b - a| = |-(a - b)| = |a - b| \leq 13$. So $(b, a) \in R_{1}$. It is symmetric.(iii) Transitive: Let $(1, 10) \in R_{1}$ and $(10, 20) \in R_{1}$. Here $|1 - 10| = 9 \leq 13$ and $|10 - 20| = 10 \leq 13$. However,$|1 - 20| = 19 
ot\leq 13$. Thus,$(1, 20) 
otin R_{1}$. $R_{1}$ is not transitive,hence not an equivalence relation.For $R_{2} = \{(a, b) \in N 	imes N : |a - b| 
eq 13\}$:$(i)$ Reflexive: $|a - a| = 0 
eq 13$. So $(a, a) \in R_{2}$. It is reflexive.(ii) Symmetric: If $|a - b| 
eq 13$,then $|b - a| = |a - b| 
eq 13$. So $(b, a) \in R_{2}$. It is symmetric.(iii) Transitive: Let $(1, 14) \in R_{2}$ and $(14, 1) \in R_{2}$. Here $|1 - 14| = 13$ and $|14 - 1| = 13$. Wait,$(1, 14) 
otin R_{2}$ and $(14, 1) 
otin R_{2}$. Let's test $(1, 2) \in R_{2}$ and $(2, 15) \in R_{2}$. $|1 - 2| = 1 
eq 13$ and $|2 - 15| = 13$. Since $(2, 15) 
otin R_{2}$,this is not a valid counterexample. Consider $(1, 2) \in R_{2}$ and $(2, 14) \in R_{2}$. $|1 - 2| = 1 
eq 13$ and $|2 - 14| = 12 
eq 13$. But $|1 - 14| = 13$,so $(1, 14) 
otin R_{2}$. Thus,$R_{2}$ is not transitive,hence not an equivalence relation.Therefore,neither $R_{1}$ nor $R_{2}$ is an equivalence relation.

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 $N$:

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