The function f: N rightarrow Z defined by f(n | vedclass

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(C) To check if the function $f: N \rightarrow Z$ is one-one and onto,we analyze the mapping:$1$. One-one check:If $n$ is even,$f(n) = \frac{n}{2}$. F

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one-one and onto

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(C) To check if the function $f: N \rightarrow Z$ is one-one and onto,we analyze the mapping:$1$. One-one check:If $n$ is even,$f(n) = \frac{n}{2}$. For $n \in \{2, 4, 6, \dots\}$,the values are $f(2)=1, f(4)=2, f(6)=3, \dots$,which maps to the set of positive integers $\{1, 2, 3, \dots\}$.If $n$ is odd,$f(n) = -\left(\frac{n-1}{2}\right)$. For $n \in \{1, 3, 5, \dots\}$,the values are $f(1)=0, f(3)=-1, f(5)=-2, \dots$,which maps to the set of non-positive integers $\{0, -1, -2, \dots\}$.Since every distinct input $n \in N$ produces a distinct output in $Z$,the function is one-one.$2$. Onto check:For any integer $y \in Z$,if $y > 0$,we can choose $n = 2y$ (which is even),so $f(2y) = \frac{2y}{2} = y$. If $y \le 0$,we can choose $n = -2y + 1$ (which is odd),so $f(-2y+1) = -\left(\frac{-2y+1-1}{2}\right) = -(-y) = y$. Since every $y \in Z$ has a pre-image in $N$,the function is onto.Therefore,the function is one-one and onto.

The function $f: N \rightarrow Z$ defined by $f(n) = \begin{cases} \frac{n}{2} & , n \text{ is even} \\ -\left(\frac{n-1}{2}\right) & , n \text{ is odd} \end{cases}$ is . . . . . . .

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