1
$\begingroup$

I've been working through various examples of functions that are 1-1, onto or both and came across an example that I didn't really know how to tackle.

Essentially we have a function $f(x,y) = (-1)^xy$ with a domain and codomain of $Z$.

Showing that this function isn't 1-1 is relatively straight forward through the use of a counterexample (say $f(2, 1)$, $f(4, 1)$), however, I am having some trouble showing that this function is onto.

I can see from the function definition that the function is onto (as you essentially have -1 or 1 * any integer) but don't know how to show this as an actual proof.

$\endgroup$

1 Answer 1

0
$\begingroup$

Just go back to the formal definition of a surjective function. Here you have a function $f : {\Bbb Z} \times {\Bbb Z} \to {\Bbb Z}$. This function is surjective if for each $z \in {\Bbb Z}$, there exists $(x,y) \in {\Bbb Z} \times {\Bbb Z}$ such that $f(x,y) = z$. Since $f(2,z) = (-1)^2z= z$, your function is indeed surjective.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.