" Suppose that all "n in N" Of लिए."f(n)={[(n+1)/(2)," Yadi "n" Is odd "],[(n)/(2)," If "n" Is even "]

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 f:N rarr N defined by f(n)={{:((n+1)/(2) " if n is odd"),(n/2" if n is even"):} then f 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 -

Let f : N rarr N be defined by f(n) = {{:((n+1)/(2), if "n is odd"),((n)/(2),"if n is even"):} Show that f is many one and onto function.

Let N be the set of natural numbers and two functions f and g be defined as f, g : N to N such that : f(n)={((n+1)/(2),"if n is odd"),((n)/(2),"if n is even"):}and g(n)=n-(-1)^(n) . The fog is :