# Functions Problem 4

Recap of definitions: $$h:\mathbb{Z}\rightarrow \mathbb{Z}$$ is known to be onto. $$f:\mathbb{Z}^2\rightarrow \mathbb{Z}$$ is defined by $$f(x,y) = h(x)+h(y)$$.

### Solution

Proof: Let k be an integer.

Since h is onto, we can find integers p and q such that h(p) = 0 and h(q) = k. Now, consider f(p,q). $$f(p,q) = h(p) + h(q) = 0 + k = k$$.

So, for an arbitrary choice of k, we found a point (p,q) that mapped onto k. This means that f is onto.