Answer
We prove by contradiction. Suppose that $3|(3n+2)$ for some integer $n$. Then by the definition of "divides," $3n+2=3k$ for some integer $k$. Subtracting $3k$ and $2$ from both sides, we get $3n-3k=-2$. Dividing both sides by $3$, we get $n-k=-\frac{2}{3}$. But this is a contradiction, since the integers are closed under substraction, but $-\frac{2}{3}$ is not an integer. Hence, our assumption must be false, and we conclude that, for all integers $n$, $3n+2$ is not divisible by $3$.
Work Step by Step
Recall that the closure properties of the integers state that the sum, difference, and product of integers is always an integer.