# 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.

### Comments

Notice that the final proof is very short. The hard work was figuring
out how we were going to pick the input (p,q). The scratch work for
doing that is often longer than the final proof.

The last, wrap-up paragraph is good style but not required. It's most
helpful when writing for an audience (e.g. you folks) who are still
beginners with the concept of onto.