Salvador Whitehead

Answered

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?

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

Expert

2022-12-02Added 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 $f:\mathbb{R}\to \mathbb{R},\phantom{\rule{1em}{0ex}}f(x)={x}^{2}$

is neither one-to-one ($1$ and $-1$ both map to $1$) or onto (nothing maps to e.g. $-1$).

The function

$g:\mathbb{[}\mathbb{0}\mathbb{,}\mathrm{\infty}\mathbb{)}\to [0,\mathrm{\infty}),\phantom{\rule{1em}{0ex}}g(x)={x}^{2}$

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.

For example the function $f:\mathbb{R}\to \mathbb{R},\phantom{\rule{1em}{0ex}}f(x)={x}^{2}$

is neither one-to-one ($1$ and $-1$ both map to $1$) or onto (nothing maps to e.g. $-1$).

The function

$g:\mathbb{[}\mathbb{0}\mathbb{,}\mathrm{\infty}\mathbb{)}\to [0,\mathrm{\infty}),\phantom{\rule{1em}{0ex}}g(x)={x}^{2}$

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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