Let `z` denote the set of all integers. Define : `f: z -> z` by : `f(x)= { x/2, (x is even), 0, (x is odd)` Then f is

Let Z denote the set of all integers and f: Z to Z defined by f(x) = {{:(x+2, ("x is even")),(0, ("x is odd")):} , Then f is:

Define f:Z rarrZ by f(x)={{:(x//2,"(x is even)"),(0,"(x is odd)"):} then f is

On the set of integers Z, define f : Z to Z as f(n)={{:((n)/(2)",",,"n is even."),(0",",,"n is odd."):} Then, f is

Let f : Z to Z be given by f(x) = {((x)/(2) if"is even"),(0 if" is odd"):} Then f is

On the set Z of all integers define f : Z-(0) rarr Z as follows f(n)= {(n/2, n \ is \ even) , (2/0 , n \ is \ odd):} then f is

If function f:ZtoZ is defined by f(x)={{:((x)/(2), "if x is even"),(0," if x is odd"):} , then f is

Let R be the relation in the set Z of all integers defined by R= {(x,y):x-y is an integer}. Then R is

Let Z denote the set of integers, then (x in Z,|x-3|lt4cap{x inZ:|x-4|lt5} =

Let f(x) = x[x], x in z, [.] is GIF then "f"^(')(x) =

Let Z be the set of integers and f:Z rarr Z is a bijective function then