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

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(B) Given $f: Z ightarrow N$ defined by $f(n) = \begin{cases} 2n, & 	ext{if } n > 0 \ 1, & 	ext{if } n = 0 \ -2n-1, & 	ext{if } n < 0 \end{case

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onto but not one-one

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(B) Given $f: Z ightarrow N$ defined by $f(n) = \begin{cases} 2n, & 	ext{if } n > 0 \ 1, & 	ext{if } n = 0 \ -2n-1, & 	ext{if } n For $n > 0$, $f(n) \in \{2, 4, 6, 8, \dots\}$.For $n = 0$, $f(0) = 1$.For $n  0$. Then $f(n) = -2(-k) - 1 = 2k - 1$. As $k$ takes values $1, 2, 3, \dots$, $f(n)$ takes values $1, 3, 5, 7, \dots$.Combining these, the range of $f$ is $\{1, 2, 3, 4, \dots\} = N$. Since Range = Codomain, $f$ is onto.Now, check for one-one: $f(0) = 1$ and $f(-1) = -2(-1) - 1 = 2 - 1 = 1$.Since $f(0) = f(-1)$ but $0 
eq -1$, the function is not one-one.Therefore, $f$ is onto but not one-one.

If $f: Z \rightarrow N$ is defined by $f(n) = \begin{cases} 2n, & \text{if } n > 0 \\ 1, & \text{if } n = 0 \\ -2n-1, & \text{if } n < 0 \end{cases}$, then the function $f$ is:

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