Let $f:N \to N$ be defined by $f(n) = \left\{ \begin{array}{l}\cfrac{{n + 1}}{2},\;\;if\;\;n\;\;is\;\;odd\\\cfrac{n}{2},\;\;\;\;\;\;\;if\;\;n\;\;is\;\;even\end{array} \right.$ for all $n \in N.$ State whether the function f is bijective f is bijective. Justify your answer.

Injectivity

Here, $f(1) = \cfrac{{1 + 1}}{2} = 1,\;\;\;f(2) = \cfrac{2}{2} = 1,\;\;\;f(3) = \cfrac{{3 + 1}}{2} = 2,\;\;\;f(4) = \cfrac{4}{2} = 2$
Thus $f(2k - 1) = \cfrac{{(2k - 1) + 1}}{2} = k\;\;\;and\;\;\;f(2k) = \cfrac{{2k}}{2} = k$

$\Rightarrow$ $f(2k - 1) = f(2k),\;\;where\;\;k \in N$

But, $2k - 1 \ne 2k,$ where $k \in N \Rightarrow f$ is not one-one.

Surjectivity

But, f is onto because range of f $=$ N

$\Rightarrow$ f is onto.

Hence, f is not bijective.