Proving onto of a two variable function

Question

So I know how to prove a function is onto if it has 1 variable. But this one has two and I'm confused about how to approach it.

$f: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ such that for any $(x,y) \in \mathbb{Z} \times \mathbb{Z}$

$f(x,y) = ax + by$

You want to prove $f$ is surjective, is that it? It's not true if $(a,b)=(0,0)$. And it's not injective either. — Git Gud, Mar 02 '14 at 17:00
So you're saying there's a restriction on a and b such that they cannot be equal to zero? — E 4 6, Mar 02 '14 at 17:07
Do you want to prove it is onto? Or do you want to prove it is 1-1? — user76568, Mar 02 '14 at 17:09
@Dror sorry I just realized my typo in the post. I want to prove it's onto. — E 4 6, Mar 02 '14 at 17:10

score 1 · Answer 1 · answered Mar 02 '14 at 17:17

1

If $a$ and $b$ are coprime then there are $\alpha\in\mathbb Z$ and $\beta\in \mathbb Z$ such that $1=\alpha a+\beta b$, then for $z\in \mathbb Z$ $z=z\alpha a+z\beta b = f(z\alpha,z\beta)$.

answered Mar 02 '14 at 17:17

alex

795

score 1 · Answer 2 · answered Mar 02 '14 at 17:18

To prove that a function $f:A\to B$ is onto, we need to show that for every $b \in B$, there exists an $a \in A$ such that $f(a) = b$. In this case, we need to show that for every $z \in \mathbb{Z}$, the equation $$ f(x,y) = z \implies ax + by = z $$ has a solution with $(x,y) \in \mathbb{Z \times Z}$.

Proving onto of a two variable function

2 Answers2

Linked

Related