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

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

If NN be the set natural numbers, then prove that, the mapping f: NN rarr NN defined by , f(n) =n-(-1)^(n) is a bijection.

If f:N rarr N is defined by f(n)=n-(-1)^(n), then

Let f : N rarr N be defined by f(x)=x^(2)+x+1, x in N . Then f is

Let f:N rarr N be defined by f(x)=x^(2)+x+1,x in N. Then is f is

Let f: N to N be defined by f(x) = x-(1)^(x) AA x in N , Then f is