What conclusion can we draw? Let $f \colon Z \to \{-1, 1\}$, where $Z$ denotes the set of integers, be defined by $ f(n) = 1$ if $n$ is even and $f(n) = -1$ if $n$ is odd. 
Then we can easily show that $f(m+n) = f(m) \cdot f(n)$ for all $m$, $n \in Z$. 
What is the most revealing conclusion that we can draw about the integers? 
I know that this mapping is the same as the mapping $g \colon Z \to \{-1,1\}$ defined by $g(n)= (-1)^n$ for all $n \in Z$. 
 A: About all we can draw from this are the following rules:$\DeclareMathOperator{\even}{even}\DeclareMathOperator{\odd}{odd}$
\begin{align}\even+\even&=\even\\\even+\odd&=\odd\\\odd+\odd&=\even
\end{align}
A: What you have found is called a homomorophism between $(\mathbb{Z},+)$ and $(\{\pm1\},\times)$, but I'm not sure if you know what that means yet. These kinds of things are studied in a branch of mathematics called abstract algebra.
A homormophism is a special type of function that respects a particular property. In Algebra, we're mostly interested in homomorphisms that preserve some algebraic structures that arise from the binary operations on two groups. In your example, your homomorphism is preserving the laws of parity on integers and transforms it to the language of signs in $\{\pm 1\}$.
Another example that you might've possibly seen is the map $\exp: \mathbb{R} \to \mathbb{R}^+$ which has the property $\exp(x+y)=\exp(x).\exp(y)$. Or the inverse map which is $\log: \mathbb{R}^+ \to \mathbb{R}$ which has the property $\log(x.y) = \log(x) + \log(y)$.
