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 \(f(x,y) = h(x)+h(y)\).


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.


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.