Question

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

• A. True
• B. False

Answer 1

The correct option is A True

Let $A=0,1$ and N the set of all natural numbers. Then the mapping $f:N⟶A$ defines by $f(2n-1)=0,$ $f\left(2x\right)=1$ $\forall$ $n$ $ϵ$ $N$ is many-one onto. The above statement is absolutely true.