Let N be the set of all natural numbers,Z be | vedclass

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(D) Given the function $\sigma: N ightarrow Z$ defined as: $\sigma(n) = \begin{cases} \frac{n}{2} & 	ext{if } n 	ext{ is even} \ -\frac{n-1}{2} &

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$\sigma$ is one-one and onto

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(D) Given the function $\sigma: N ightarrow Z$ defined as: $\sigma(n) = \begin{cases} \frac{n}{2} & 	ext{if } n 	ext{ is even} \ -\frac{n-1}{2} & 	ext{if } n 	ext{ is odd} \end{cases}$ Case-$I$: If $n$ is even,let $n = 2k$ for $k \in N$. Then $\sigma(2k) = \frac{2k}{2} = k$. As $k$ takes values $1, 2, 3, \dots$,$\sigma(n)$ takes values $1, 2, 3, \dots$ (all positive integers). Case-$II$: If $n$ is odd,let $n = 2k-1$ for $k \in N$. Then $\sigma(2k-1) = -\frac{(2k-1)-1}{2} = -\frac{2k-2}{2} = -(k-1) = 1-k$. As $k$ takes values $1, 2, 3, \dots$,$\sigma(n)$ takes values $0, -1, -2, \dots$ (all non-positive integers). Combining both cases,the range of $\sigma$ is ${0, 1, -1, 2, -2, 3, -3, \dots} = Z$. Since every distinct $n \in N$ maps to a distinct integer in $Z$,the function is one-one. Since the range of $\sigma$ is equal to the codomain $Z$,the function is onto. Therefore,$\sigma$ is both one-one and onto.

Let $N$ be the set of all natural numbers,$Z$ be the set of all integers and $\sigma: N \rightarrow Z$ be defined by $\sigma(n)=\begin{cases} \frac{n}{2}, & \text{if } n \text{ is even} \\ -\frac{n-1}{2}, & \text{if } n \text{ is odd} \end{cases}$. Then,

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