Why is f ( n ) = n 3 not an onto function?I was doing an example in a book where it asked which of these functions are one to one, the answer in the back said for f ( n ) = n 3 that it is a one to one function. Then it asked which of the functions from the previous example are onto and f ( n ) = n 3 was not included in the list of onto functions.In a later example, a question asked which of these functions is a bijection, the answer included f ( x ) = x 3 This is confusing because doesn't a function have to be both an onto and one to one to be a bijection? Why would the book say it was not a onto in a previous example yet declare it to be a bijection? Is the book wrong?

Question

Why is   f  (  n  )  =      n    3   not an onto function?I was doing an example in a book where it asked which of these functions are one to one, the answer in the back said for   f  (  n  )  =      n    3   that it is a one to one function. Then it asked which of the functions from the previous example are onto and   f  (  n  )  =      n    3   was not included in the list of onto functions.In a later example, a question asked which of these functions is a bijection, the answer included   f  (  x  )  =      x    3  This is confusing because doesn&#039;t a function have to be both an onto and one to one to be a bijection? Why would the book say it was not a onto in a previous example yet declare it to be a bijection? Is the book wrong?

Arturo Hogan · Accepted Answer

You can never attempt to answer if a function is one-to-one or onto without first knowing the domain and codomain.For example the function   f  :      R    →      R    ,    f  (  x  )  =      x    2  is neither one-to-one (  1 and   −  1 both map to   1) or onto (nothing maps to e.g.   −  1).The function  g  :      [    0    ,    ∞    )    →  [  0  ,  ∞  )  ,    g  (  x  )  =      x    2  is both one-to-one and onto.We have only changed the domain and codomain, and this changed the properties of one-to-one and onto.