Question

Let $A=\{0,1\}$ and $N$ be the set of natural numbers. Then the mapping $f:N\to A$ defined by $f(2n–1)=0,f\left(2n\right)=1,\forall nϵN$, is onto.

Answer 1

Use definition of onto functions

Given: $f:N\to AA=\{0,1\}$
$f(2n–1)=0,\forall nϵN$
$\Rightarrow f\left(odd\right)=0$
and $f\left(2n\right)=1,\forall nϵN$$\Rightarrow f\left(even\right)=1$
$\Rightarrow$ Range $\left(f\right)=\{0,1\}=$ co-domain $\left(f\right)$
Therefore, $f$ is onto function

Hence, the given statement is true.