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CONTEXT: Question made up by uni lecturer.

How do you prove that a function mapping from a Cartesian product of the integers to the integers is onto?

The function is $f:Z$x$Z$ to $Z$ where $f((m,n))=2m-n$.

I'm pretty sure the function is onto, since I can't think of an integer that can't be written in terms of $2m-n$ where $m$ and $n$ are integers, but am unsure of how to show this.

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  • $\begingroup$ You can take $m=0$, so this is really fairly straightforward. Slightly more challenging: is $f((m,n)) = 2m-3n$ surjective? $\endgroup$
    – Jane Doé
    May 3 '19 at 1:44
  • $\begingroup$ @JaneDoé doesn't surjective and onto mean the same thing? $\endgroup$
    – Ruby Pa
    May 3 '19 at 1:57
  • $\begingroup$ @RubyPa yes it is. $\endgroup$
    – Tomislav
    May 3 '19 at 2:04
  • $\begingroup$ Sorry, yes, "surjective" and "onto" mean the same thing, so much so that I switch to saying "surjective" without thinking! The point is that it is slightly harder to see that $f((m,n)) = 2m-3n$ is onto. $\endgroup$
    – Jane Doé
    May 3 '19 at 2:04
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Hint: Set $m=0$. $~~~~~~~~~~~~$

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$f((n,n))=2n-n=n,$ so any $n\in\mathbb Z$ has $(n,n) \in \mathbb Z \times \mathbb Z$ in its pre-image.

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  • $\begingroup$ This is an alternative to Jane Doé's suggestion and could be adapted to solve the slightly more challenging question. $\endgroup$ May 3 '19 at 2:20

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