Let `f : N to N : f (n) = {underset((n)/(2), " when n is even ")( (1)/(2) (n+1) , " when n is odd ")`
Then f is

Let f: N to N be defined as f(n) = (n+1)/(2) when n is odd and f(n) = (n)/(2) when n is even for all n in N . State whether the function f is bijective. Justify your answer.

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 -

If f_(n) = {{:( (f_(n-1))/(2) " when " f_n-1 "is an even number"), (3* f_(n-1) + 1 " when " f_(n-1) " is an odd number "):} and f_(1) = 3 , then f_(5) =

Let f:NtoN defined by : f(n)={:{((n+1)/(2) "if n is odd"),((n)/(2) "if n is even"):}

Let f: N to 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 NN be the set of natural numbers and f: NN uu {0} rarr NN uu [0] be definedby : f(n)={(n+1 " when n is even" ),(n-1" when n is odd " ):} Show that ,f is a bijective mapping . Also that f^(-1)=f