On the set of integers Z,define f: Z rightarr | vedclass

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(C) Given the function $f: Z ightarrow Z$ defined by $f(n) = \begin{cases} \frac{n}{2}, & n 	ext{ is even} \ 0, & n 	ext{ is odd} \end{cases}$.$1

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surjective but not injective

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(C) Given the function $f: Z ightarrow Z$ defined by $f(n) = \begin{cases} \frac{n}{2}, & n 	ext{ is even} \ 0, & n 	ext{ is odd} \end{cases}$.$1$. Check for Injectivity (One-to-One): $A$ function is injective if $f(n_1) = f(n_2) \implies n_1 = n_2$. Consider odd integers $n_1 = 1$ and $n_2 = 3$. Then $f(1) = 0$ and $f(3) = 0$. Since $f(1) = f(3)$ but $1 
eq 3$,the function is not injective (it is many-to-one).$2$. Check for Surjectivity (Onto): $A$ function is surjective if for every $y \in Z$,there exists an $n \in Z$ such that $f(n) = y$. If we choose $y = 5$,there is no integer $n$ such that $f(n) = 5$ because even numbers map to $n/2$ (which would require $n=10$) and odd numbers map to $0$. However,the range of $f$ is the set of all integers $Z$ because for any $y \in Z$,we can choose $n = 2y$ (which is even),then $f(2y) = \frac{2y}{2} = y$. Thus,for every $y \in Z$,there exists $n = 2y$ such that $f(n) = y$. Therefore,the function is surjective.Conclusion: The function is surjective but not injective.

On the set of integers $Z$,define $f: Z \rightarrow Z$ as $f(n) = \begin{cases} \frac{n}{2}, & n \text{ is even} \\ 0, & n \text{ is odd} \end{cases}$. Then $f$ is:

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