Question

State true or false:

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

• A. True
• B. False

Answer 1

The correct option is B False

f is a function from the set of natural numbers to set A.
Thus domain of f(n) is $nϵN$.
Now it is given that $f(2n-1)=0$ and $f\left(2n\right)=1$.
Hence
$f\left(n\right)=0$ if n is odd and
$f\left(n\right)=1$ if n is even.
Thus for all $2nϵN$ $f\left(n\right)=1$ and for all $(2n-1)ϵN$ $f\left(n\right)=0$.
Hence there are more than 1 one value of 'n' for which f(n) is 0.
Similarly there are more than 1 one value of 'n' for which f(n) is 1.
Hence f(n) is a many one function.
However f(n) successfully maps every element in the co domain $\left(\right\{0,1\left\}\right)$ to atleast one element in the domain of $\left(N\right)$.
Hence f(n) is an onto function.