Let `A = {0, 1}` and the set of all natural numbers.Then the mapping `f: N rarr A` defined by `f(2n-1) = 0, f(2n) = 1, AA n in N,` is

Answer

N/a

Explanation

True
Given, A = {0, 1}
`f(2n-1)=0, f(2n)=1, AA n in N`
So, the mapping `f:N to A ` is onto.