Part 1: additive inverse and $\lt$

In a ring with a total order, for any $x$ we have: if $0 \lt x$ then $-x \lt 0$; if $x \lt 0$ then $0 \lt -x$.

ACTIVITY
Prove the above theorem! Remember, your only givens are the ring axioms and the above axioms for a total order for a ring.

Part 2: Positive, negative and zero

With the definition of < in place, we can finally define something we think of as an obvious an intrinsic part of the integers: "positive" and "negative". So, first:

In a ring with a total order, for any $x$, exactly one of $x \lt 0$, $x = 0$, $0 \lt x$ is true.

ACTIVITY
Prove the above theorem!
Note 1: This is called the "trichotemy law".
Note 2: So the only things you may take as givens are the ring axioms and the axioms for "<" from above.
Note 3: Your proof should have four parts:

  1. Prove that at least one of $x \lt 0$, $x = 0$, $0 \lt x$ is true
  2. Prove that if $x=0$ then $x \nless 0$ and $0 \nless x$. Note: remember that $a \nless b$ is short-hand for $\neg (a \lt b)$.
  3. Prove that if $x \lt 0$ then $x \neq 0$ and $0 \nless x$. Note: remember that $a \nless b$ is short-hand for $\neg (a \lt b)$.
  4. Prove that if $0 \lt x$ then $x \neq 0$ and $x \nless 0$. Note: remember that $a \nless b$ is short-hand for $\neg (a \lt b)$.