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`.
Prove that f is many-one, onto function.

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.

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.

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: NvecN be defined by f(n)={(n+1)/2,\ "if"\ n\ "i s\ o d d"n/2,\ \ "if\ "n\ "i s\ e v e n"\ \ for\ a l l\ n\ \ N} Find whether the function f is bijective.

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.

Find the sequence of the numbers defined by a_(n)={{:(1/n,"when n is odd"),(-1/n,"when n is even"):}

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

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 to N defined as f(x)=x^(2)AA x in N, then f is :

Let N be the set of 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) . Then, fog is