If f: Z rightarrow Z, f(x) = begin{cases} fra | vedclass

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(A) Given, $f: Z ightarrow Z$, $f(x) = \begin{cases} \frac{x}{2}, & 	ext{if } x 	ext{ is even} \ 0, & 	ext{if } x 	ext{ is odd} \end{cases}$.Fo

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onto but not one-to-one

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(A) Given, $f: Z ightarrow Z$, $f(x) = \begin{cases} \frac{x}{2}, & 	ext{if } x 	ext{ is even} \ 0, & 	ext{if } x 	ext{ is odd} \end{cases}$.For $f$ to be one-to-one, $f(x_1) = f(x_2)$ must imply $x_1 = x_2$.Consider $x_1 = 1$ and $x_2 = 3$. Both are odd, so $f(1) = 0$ and $f(3) = 0$.Since $f(1) = f(3)$ but $1 
eq 3$, $f$ is not one-to-one.For $f$ to be onto, the range must equal the codomain $Z$.If $x$ is even, let $x = 2k$ for some $k \in Z$. Then $f(2k) = \frac{2k}{2} = k$.Since $k$ can be any integer in $Z$, the range of $f$ is $Z$.Thus, $f$ is onto.Therefore, $f$ is onto but not one-to-one.

If $f: Z \rightarrow Z$, $f(x) = \begin{cases} \frac{x}{2}, & \text{if } x \text{ is even} \\ 0, & \text{if } x \text{ is odd} \end{cases}$, then $f$ is

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