Let Z denote the set of integers. Define f: Z | vedclass

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(D) To check if $f$ is one-one: $A$ function is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$. Consider $x_1 = 1$ and $x_2 = 3$. Both are odd intege

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neither one-one nor onto

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(D) To check if $f$ is one-one: $A$ function is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$. Consider $x_1 = 1$ and $x_2 = 3$. Both are odd integers. Thus,$f(1) = 0$ and $f(3) = 0$. Since $f(1) = f(3)$ but $1 
eq 3$,the function is not one-one.To check if $f$ is onto: $A$ function is onto if for every $y \in Z$,there exists an $x \in Z$ such that $f(x) = y$. Consider $y = 5$. For $f(x) = 5$,we need $\frac{x}{2} = 5 \implies x = 10$. Since $10 \in Z$,$f(10) = 5$. However,consider $y = 3$. If $x$ is even,$\frac{x}{2} = 3 \implies x = 6$. If $x$ is odd,$f(x) = 0$. There is no $x$ such that $f(x) = 3$ because $3$ is not an even number divided by $2$ and it is not $0$. Thus,the function is not onto.Therefore,$f$ is neither one-one nor onto.

Let $Z$ denote the set of integers. Define $f: Z \rightarrow Z$ by $f(x) = \begin{cases} \frac{x}{2}, & x \text{ is even} \\ 0, & x \text{ is odd} \end{cases}$. Then $f$ is:

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