Question

Let $A=\{0,1\}$ and $N$ the set of all natural numbers. Then show that the mapping $f:N\to A$ defined by $f(2n-1)=0,f\left(2n\right)=1\forall n\in N$ is many-one onto.

Answer 1

Given $f:N\to A$ such that
$f(2n-1)=0$ and $f\left(2n\right)=1$
So, image of all even natural numbers is $1$ and image of all odd natural numbers is $0.$
Thus, distinct elements of $N$ has same image in $A.$
Hence, $f$ is many-one.
Since, $R\left(f\right)=\{0,1\}$
which is equal to co-domain $A.$
Hence, $f$ is onto.