Let A = {0, 1, 2, ldots, 9}. Let R be a relat | vedclass

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(B) The set $A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ has $10$ elements.We partition $A$ into equivalence classes based on remainders modulo $3$:$C_0 = \

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Statement $I$ is incorrect but Statement $II$ is correct

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(B) The set $A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ has $10$ elements.We partition $A$ into equivalence classes based on remainders modulo $3$:$C_0 = \{0, 3, 6, 9\}$ (size $4$)$C_1 = \{1, 4, 7\}$ (size $3$)$C_2 = \{2, 5, 8\}$ (size $3$)For $(x, y) \in R$, $|x - y|$ must be a multiple of $3$, meaning $x$ and $y$ must belong to the same equivalence class.The number of elements in $R$ is $n(R) = |C_0|^2 + |C_1|^2 + |C_2|^2 = 4^2 + 3^2 + 3^2 = 16 + 9 + 9 = 34$.Thus, Statement $I$ is incorrect $(34 
eq 36)$.For Statement $II$: $1$. Reflexive: $|x - x| = 0$, which is a multiple of $3$.$2$. Symmetric: If $|x - y| = 3k$, then $|y - x| = 3k$.$3$. Transitive: If $|x - y| = 3k$ and $|y - z| = 3m$, then $|x - z| = |(x - y) + (y - z)| = 3|k \pm m|$, which is a multiple of $3$.Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation. Statement $II$ is correct.

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