Showing the trichotomy property for the integers

Question

I am new to proof writing. The book I am reading leaves it to the reader to show that:

Given $r,s \in \mathbb{Z}$, exactly one of the following is true: $r<s$, $s<r$, or $r=s$.

My thought is to do the following:

1. Assume r is in P
2. Use the cases that s is in P, $\{0\}$, and (-P) to show that those are the 3 options
3. Assume r is in $0$
4. Use the cases that s is in P, $\{0\}$, and (-P) to show that those are the 3 options
5. Assume r is in (-P)
6. Use the cases that s is in P, $\{0\}$, and (-P) to show that those are the 3 options

Is this a valid proof?

Answer 1

Yes, if you prove the result for all nine cases, you'll have a valid proof -- assuming that it has already been proven that $\mathbb Z = -P \cup \{0\} \cup P$. If that's how your book defined $\mathbb Z$, then you're good. If you defined $\mathbb Z$ in some other way, then $\mathbb Z = -P \cup \{0\} \cup P$ becomes a theorem that needs a proof of its own.

As Peter suggests in the comments, if you have already established even more facts about $\mathbb Z$, then there are more elegant proofs available. For example, if you know that $r<s$ iff $r-s<0$, then you can just inspect the sign of $r-s$.