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 $$f(x,y) = (h(x) + h(y), h(x) - h(y))$$

Solution

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

By the definition of f, $$f(x,y) = f(a,b)$$ implies that $$(h(x) + h(y), h(x) - h(y)) = (h(a) + h(b), h(a) - h(b))$$.

So $$h(x) + h(y) = h(a) + h(b)$$ and $$h(x) - h(y) = h(a) - h(b)$$

Adding these two equations together, we find that $$2h(x) = 2h(a)$$. So h(x) = h(a). Substituting this into $$h(x) + h(y) = h(a) + h(b)$$, we get $$h(x) + h(y) = h(x) + h(b)$$. So then h(y) = h(b).

Now, we know that h(x) = h(a) and h(y) = h(b). Since h is one-to-one, this means that x=a and y=b. So $$(x,y)=(a,b)$$, which is what we needed to show.