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

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(D) To check for one-one: $A$ function is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$. Consider $f(1) = 0$ and $f(3) = 0$. Since $f(1) = f(3)$ but

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neither one-one nor onto

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(D) To check for one-one: $A$ function is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$. Consider $f(1) = 0$ and $f(3) = 0$. Since $f(1) = f(3)$ but $1 
eq 3$, the function is not one-one.To check for onto: $A$ function is onto if for every $y \in Z$, there exists $x \in Z$ such that $f(x) = y$. For any odd integer $y$ (where $y 
eq 0$), there is no $x \in Z$ such that $f(x) = y$, because the range of $f$ only contains $0$ and even integers divided by $2$. 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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