The function f: N rightarrow N defined by f(x | vedclass

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(C) To check if $f$ is one-one:Let $f(x_1) = f(x_2)$.If $x_1$ is odd and $x_2$ is even,then $x_1+1 = x_2-1$,so $x_2 - x_1 = 2$. Since $x_1$ is odd,$x_

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one-one and onto

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(C) To check if $f$ is one-one:Let $f(x_1) = f(x_2)$.If $x_1$ is odd and $x_2$ is even,then $x_1+1 = x_2-1$,so $x_2 - x_1 = 2$. Since $x_1$ is odd,$x_1+1$ is even. Since $x_2$ is even,$x_2-1$ is odd. This contradicts $f(x_1) = f(x_2)$.If both are odd,$x_1+1 = x_2+1 \implies x_1 = x_2$.If both are even,$x_1-1 = x_2-1 \implies x_1 = x_2$.Thus,$f$ is one-one.To check if $f$ is onto:For any $y \in N$,if $y$ is odd,we can choose $x = y+1$ (which is even),then $f(y+1) = (y+1)-1 = y$.If $y$ is even,we can choose $x = y-1$ (which is odd),then $f(y-1) = (y-1)+1 = y$.Since for every $y \in N$,there exists an $x \in N$ such that $f(x) = y$,$f$ is onto.Therefore,$f$ is one-one and onto.

The function $f: N \rightarrow N$ defined by $f(x) = \begin{cases} x+1, & x \text{ is odd} \\ x-1, & x \text{ is even} \end{cases}$ is . . . . . . .

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