JEE Main 2022 (Online) 28th June Evening Shift

Explanation To check if $R_1$ and $R_2$ are equivalence relations, we need to verify if they are reflexive, symmetric, and transitive. For $R_1 = \{(a, b) \in N \times N : |a - b| \le 13\}$: Reflexive: For any $a \in N$, $|a - a| = 0 \le 13$. So, $(a, a) \in R_1$. Thus, $R_1$ is reflexive. Symmetric: If $(a, b) \in R_1$, then $|a - b| \le 13$. This implies $|b - a| \le 13$, so $(b, a) \in R_1$. Thus, $R_1$ is symmetric. Transitive: Consider $a = 2, b = 11, c = 19$. Then $|2 - 11| = 9 \le 13$, so $(2, 11) \in R_1$. Also, $|11 - 19| = 8 \le 13$, so $(11, 19) \in R_1$. However, $|2 - 19| = 17 > 13$, so $(2, 19) \notin R_1$. Thus, $R_1$ is not transitive. Since $R_1$ is not transitive, $R_1$ is not an equivalence relation. For $R_2 = \{(a, b) \in N \times N : |a - b| \ne 13\}$: Reflexive: For any $a \in N$, $|a - a| = 0 \ne 13$. So, $(a, a) \in R_2$. Thus, $R_2$ is reflexive. Symmetric: If $(a, b) \in R_2$, then $|a - b| \ne 13$. This implies $|b - a| \ne 13$, so $(b, a) \in R_2$. Thus, $R_2$ is symmetric. Transitive: Consider $a = 13, b = 3, c = 26$. Then $|13 - 3| = 10 \ne 13$, so $(13, 3) \in R_2$. Also, $|3 - 26| = 23 \ne 13$, so $(3, 26) \in R_2$. However, $|13 - 26| = 13$, so $(13, 26) \notin R_2$. Thus, $R_2$ is not transitive. Since $R_2$ is not transitive, $R_2$ is not an equivalence relation. Therefore, neither $R_1$ nor $R_2$ is an equivalence relation.