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.