Functions Problem 3

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

Outline of Solution

Proof: Let (x,y) and (a,b) be elements of \(\mathbb{Z}^2\) and suppose that \(f(x,y) = f(a,b)\).

[Magic happens here]

We need to show that \((x,y)=(a,b)\).

Comments and hints

Did you start your outline by assuming \((x,y)\not =(a,b)\), hoping to conclude that \(f(x,y) \not = f(a,b)\)? Don't do that. It's not technically wrong, but it's likely to make the rest of the proof MUCH harder to write.p

Notice that the domain of f is \(\mathbb{Z}^2\). So when you pick two input values at the start of the proof, each needs to be a 2D point.

To fill in the middle part of the proof, begin by using the definition of f.

At some point in the middle, you'll need to use the fact that h is one-to-one. Look for a place where you have figured out that h(p)=h(q). The one-to-one-ness of h will allow you to conclude that p=q.