If f: N rightarrow Z is defined by f(n)=begin | vedclass

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(B) Given the function $f: N ightarrow Z$ defined as:$f(n) = \begin{cases} 2 & 	ext{if } n=3k, k \in Z \ 10 & 	ext{if } n=3k+1, k \in Z \ 0 &

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$\{1, 4, 7, \dots\}$

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(B) Given the function $f: N ightarrow Z$ defined as:$f(n) = \begin{cases} 2 & 	ext{if } n=3k, k \in Z \ 10 & 	ext{if } n=3k+1, k \in Z \ 0 & 	ext{if } n=3k+2, k \in Z \end{cases}$We want to find the set $\{n \in N: f(n) > 2\}$.Looking at the definition, $f(n) > 2$ only when $f(n) = 10$.This occurs when $n = 3k + 1$ for $k \in Z$.Since $n \in N$ (natural numbers), we consider $k \geq 0$ (assuming $N = \{1, 2, 3, \dots\}$):For $k=0, n = 3(0) + 1 = 1$.For $k=1, n = 3(1) + 1 = 4$.For $k=2, n = 3(2) + 1 = 7$.Thus, the set is $\{1, 4, 7, \dots\}$.

If $f: N \rightarrow Z$ is defined by $f(n)=\begin{cases} 2 & \text{if } n=3k, k \in Z \\ 10 & \text{if } n=3k+1, k \in Z \\ 0 & \text{if } n=3k+2, k \in Z \end{cases}$, then $\{n \in N: f(n)>2\}$ is equal to

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