Let f: N rightarrow N be defined by f(n) = be | vedclass

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(C) Testing for one-one: $f(1) = \frac{1+1}{2} = 1$ and $f(2) = \frac{2}{2} = 1$. Since $f(1) = f(2)$ for $1 
eq 2$, the function is not one-one (it

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Many-one and onto

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(C) Testing for one-one: $f(1) = \frac{1+1}{2} = 1$ and $f(2) = \frac{2}{2} = 1$. Since $f(1) = f(2)$ for $1 
eq 2$, the function is not one-one (it is many-one).Testing for onto: For any $y \in N$, if $y$ is odd, we can have $n = 2y-1$, then $f(2y-1) = \frac{(2y-1)+1}{2} = y$. If $y$ is even, we can have $n = 2y$, then $f(2y) = \frac{2y}{2} = y$. Thus, for every element in the codomain, there is a pre-image in the domain. Therefore, the function is onto.

Let $f: N \rightarrow N$ be defined by $f(n) = \begin{cases} \frac{n+1}{2}; & \text{if } n \text{ is odd} \\ \frac{n}{2}; & \text{if } n \text{ is even} \end{cases}$, for all $n \in N$ then $f$ is $\dots \dots \dots$

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