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An equivalence relation must satisfy three properties: Reflexivity, Symmetry, and Transitivity.
For R to be reflexive, $(x, x) \in R$ for all $x \in A$. This means $x + x = 2x$ must be divisible by 2. Since $2x$ is always divisible by 2 for any integer $x$, R is reflexive.
For R to be symmetric, if $(x, y) \in R$, then $(y, x) \in R$. This means if $x + y$ is divisible by 2, then $y + x$ must also be divisible by 2. Since $x + y = y + x$, if $x + y$ is divisible by 2, so is $y + x$. Thus, R is symmetric.
For R to be transitive, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$. This means if $x + y$ is divisible by 2 and $y + z$ is divisible by 2, then $x + z$ must also be divisible by 2. $x + y = 2m$ and $y + z = 2n$ for some integers $m$ and $n$. Adding these two equations, we get $x + 2y + z = 2m + 2n$. So, $x + z = 2m + 2n - 2y = 2(m + n - y)$. Since $x + z$ is divisible by 2, R is transitive.
Since R is reflexive, symmetric, and transitive, R is an equivalence relation.
The equivalence class [2] consists of all elements $y \in A$ such that $(2, y) \in R$. This means $2 + y$ is divisible by 2. So, $2 + y = 2k$ for some integer $k$. This implies $y = 2k - 2$, which means $y$ must be an even integer. The even integers in set A are $\{-4, -2, 0, 2, 4\}$. Therefore, the equivalence class [2] is $\{-4, -2, 0, 2, 4\}$.
Final Answer: R is an equivalence relation. [2] = {-4, -2, 0, 2, 4}