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

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

If f:N to N is defined by. f(n)={((n+1)/(2)", if n is odd"),((n)/(2)", if n is even"):} for all n in N . Find whether the function f is bijective.

check that f : N- N defined by f(n)={(n+1)/2,(if n is odd)),(n/2,(if n is even)) is one -one onto function ?

Let f : N->N be defined by f(n)={(n+1)/2,"if n is odd " n/2,"if n is even " for all n in N . State whether the function f is bijective. Justify your answer.

Let : f: N rarr N be defined by f(n)={{:((n+1)/2," if n is odd"),(n/2," if n is even"):} for all n inN . State whether the function f is bijective . Justify your answer.

Let f: N->N be defined0 by: f(n)={(n+1)/2, if n is odd (n-1)/2, if n is even Show that f is a bijection.

The function f:N-{1}rarr N , defined by f(n)= the highest prime factor of "n" ,is

A function f from the set of natural numbers N to the set of integers Z is defined by f(n)={((n-1)/(2),"when n is odd"),(-(n)/(2),"when n is even"):} Then f(n) is -