We have f:N→I Let x and y are two even natural numbers, and f(x)=f(y)⇒
−x
2
=
−y
2
⇒x=y ∴f(n) is one-one for even natural number. Let x and y are two odd natural numbers and f(x)=f(y)⇒
x−1
2
=
y−1
2
⇒x=y ∴f(n) is one-one for odd natural number. Hence f is one-one. Let y=
n−1
2
⇒2y+1=n This shows that n is always odd number for y∈I ......(i) and y=
−n
2
⇒−2y=n This shows that n is always even number for y∈I ......(ii) From (i) and (ii) Range of f=I= codomain ∴f is onto. Hence f is one one and onto both.