Description for Correct answer:
We have \( \Large f \left(n\right) = \bigg\{ \frac{n-1}{2},\ when \ n \ is odd\ , {-\frac{n}{2},\ when\ n\ is\ even} \)

and \( \Large f : N \rightarrow I \) where N is the set of natulal numbers and I is the set of integers.

Let \( \Large x,\ y\ \epsilon\ N \) and both are even.

then \( \Large f \left(x\right) = f \left(y\right) \)

\( \Large -\frac{x}{2} = -\frac{y}{2} => x = y \)

Again \( \Large x,\ y\ \epsilon\ N \)

then \( \Large f \left(x\right)=f \left(y\right)\frac{x-1}{2}=\frac{y-1}{2} =>x=y \)

i.e. function is one-one.

Since, each negative integer is an image of even natural; and each positive integer is an image of odd natural number so function is onto.

Hence, function is one-one onto.