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

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

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

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.

If f:NrarrZ defined as f(n)={{:((n-1)/(2),":"," if n is odd"),((-n)/(2),":", " if n is even"):} and g:NrarrN defined as g(n)=n-(-1)^(n) , then fog is (where, N is the set of natural numbers and Z is the set of integers)

If f:NrarrZ defined as f(n)={{:((n-1)/(2),":"," if n is odd"),((-n)/(2),":", " if n is even"):} and g:NrarrN defined as g(n)=n-(-1)^(n) , then fog is (where, N is the set of natural numbers and Z is the set of integers)

If f: N rarr Z is defined by f(n) ={{:((n-1)/(2),"when n is odd"),(-n/2,"when n is even"):} then identify the type of function

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

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.

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.

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 -