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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?

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    HInt. Is the domain the real numbers or the integers? – 2017-02-23
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    How did you define the $f$? That function $f$ must be define from a set into another set. – 2017-02-23

2 Answers 2

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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 $$f: \mathbb{R}\to \mathbb{R},\quad f(x) = x^2$$ is neither one-to-one ($1$ and $-1$ both map to $1$) or onto (nothing maps to e.g. $-1$).

The function $$g:\mathbb{[0,\infty)}\to [0,\infty),\quad g(x) = x^2$$ 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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    Oh yeah I totally forgot to check the domain, thank you! – 2017-02-23
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Check the domain of the function in both the questions as a prerequisite.

The function $f (n)=n^3$ is onto if $f (n): \mathbb R \to \mathbb R$. If it helps, see here.