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?

Salvador Whitehead

Salvador Whitehead

Answered question

2022-12-01

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?

Answer & Explanation

Arturo Hogan

Arturo Hogan

Beginner2022-12-02Added 13 answers

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.

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