Proving trichotomy and transitivity from the definition of an ordered field

Question

I'm reading a set of notes (and can provide the link if anyone is interested) which attempt to build up the properties of an ordered field. After defining a field based on its axioms, it defines an ordered field as follows:

A field is ordered if there is a subset $P$ of $F$ such that

(1) $P \cap (-P) = \emptyset$

(2) $P \cup \{0\} \cup (-P) = F$

(3) For $a,b \in P$ we have $a + b \in P$ and $ab \in P$.

It then defines the notation:

Let $F$ be an ordered field. We write $a < b$ or $b > a$ or $b - a \in P$.

It asserts that trichotomy follows immediately, i.e., that for $a,b \in F$, exactly one of $a < b$, $a = b$, or $b < a$ holds. I believe I can see how, and my attempt is as follows.

By (1), $P$ and $-P$ are disjoint, and $0 \not \in P, -P$, for if $0 \in P$, $0 = -0 \in P$ and the intersection is non-zero, which is a contradiction. Suppose for $a,b \in F$ we have $a < b$. Then $b - a \in P$. So $b - a \not \in -P$, meaning that $a - b \not \in P$, so $a \ngtr b$. Furthermore, as $b - a \in P$, $b - a \neq 0$, so $a - b \neq 0$.

The last step seems like a jump because though I can say $b - a \in P$ and $0 \not \in P$, $b - a \neq 0$, this doesn't tell me immediately that $b \neq a$, though I suppose I'm free, in a field, to add to both sides and say if $b - a = 0$, then $(b - a) + a = 0 + a$ so $b = a$.

I believe there is a way to "shortcut" this proof by only proving a single implication, which I'm not fully sure of how to do. If someone has a way to prove this with only a single implication, I would appreciate a hint on it, though the remaining cases, I believe, are proved similarly, and I'm ok with moving on to transitivity for the moment.

The document then says:

You can. prove now, for example, that given elements $a$, $b$, and $c$ of an ordered field, such that $a > b$, then $a + c > b + c$ and if $c > 0$, $ac > bc$.

I'm not fully sure of how to prove this, but here is my attempt. If $a > b$, then $a - b \in P$. I know that the sum of two elements of $P$ is another element of $P$. To show $a + c > b + c$, I should show $(a + c) - (b + c) \in P$ (I think because this is a definition, I can take it as a biconditional). But $$ a - b = (a - c) + (c - b) = (a + c) - (b + c) $$ by rearrangement. So because $a - b \in P$, $(a + c) - (b + c) \in P$, so $a + c > b + c$.

Now we assume $c > 0$, which I assume I can take as $c \in P$ (though this wasn't "defined," per se). The product of two elements of $P$. is an element of $b$, so $(a - b) \cdot c \in P$. Distributing, I get $ac - bc \in P$, which means $ac > bc$.

I would appreciate any feedback on these proofs.

Answer 1

I think you're making the first proof harder to follow than it needs to be. You already know that $P, -P, \{0 \}$ divide Gaul (well, your field $F$) into three disjoint parts. Therefore, you know that $b-a$ falls into exactly one of those parts. We know that $b-a=0$ occurs exactly when $b=a$, and in that case $0=b-a=a-b \notin P$, so $b=a \iff \lnot(a \lt b \lor b \lt a)$.

Otherwise, either $b-a \in P$ or $b-a \in -P$. This latter alternative occurs exactly when $-(b-a)=a-b \in P$. That proves trichotomy because now we know that for $b \neq a, b-a$ and $a-b$ can't both occur in $P$, but one of them must occur in $P$.

The remainder of your proof looks fine.