# 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 $$(x,y)=(a,b)$$.

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.