Mathematics - Functions Question with Solution | TestHub

$f:N\to 1$ $f(1)=0,f(2)=-1,f(3)=-1,f(4)=-2$ $f(5)=2$, and $f(6)=-3$ so on.
In this type of function every element of set $\mathrm{A}$ has unique image in set $\mathrm{B}$ and there is no element left in set $\mathrm{B}$. Hence $\mathrm{f}$ is one-one and onto function.