I need to show that if $a,b\in \mathbb{R}$, then only one of the following holds: $a\in \mathbb{P}, -a\in \mathbb{P}$, or $a=0.$
By a definition in my book, if $a-b \in \mathbb{P}$, then $a>b$; and if $a-b\in\mathbb{P}\cup {\{0\}}$, then $a\geq b$.
I was going to try to set $b=0$ and use that to show that $a-0\in \mathbb{P}$ so $a\in \mathbb{P}$, meaning $a>0$.
Then $b-a = 0-a=-a\in \mathbb{P}$, meaning $a<0$.
Then $b-a=0$, meaning $a=b=0$.
Since you can't have both $a>0$, $a<0$, and $a=0$, then you can only have one of the following: $a\in \mathbb{P}, -a\in \mathbb{P}$, or $a=0.$
However I'm not sure if this is rigorous enough (note: I am taking an intro analysis course). For one, am I just allowed to set $b=0$ to make these conclusions? Next, can I just say that $b-a=0$ rather than saying that if $a-b\in\mathbb{P}\cup {\{0\}}$ and $b-a\in\mathbb{P}\cup {\{0\}}$ then $a\geq b$ and $b\geq a$ so $a=b=0$? (The above just seems easier/simpler.) Finally, for that last line, I'm not sure how I "know" that I "can't have both $a>0$, $a<0$, and $a=0$", as in if this is trivial or if it comes from a theorem (I couldn't find one in the book).
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$\begingroup$As Andrew said in his answer, based on the approach taken on your book, you cannot prove these properties of $\mathbb{P}$ in $\mathbb{R}$. The existence of $\mathbb{P} \subseteq \mathbb{R}$ satisfying the three order axioms is one of the basic properties of $\mathbb{R}$.
What you are apparently being asked to do here is to prove that for the field $K$ there exists a set $\mathbb{P}_K$ of numbers, which will be called the "positive elements of $K$," that makes $K$ into an ordered field. In fact, the exercise in the book tells you the order on $K$ should be the "order inherited from $\mathbb{R}$," which means that you should call an element of $K$ positive precisely when it is positive considered as a real number.
Therefore, define the set $\mathbb{P}_K$ to be $K \cap \mathbb{P}$. Then prove that $\mathbb{P}_K$ has the three required properties to play the role of a set of positive numbers for $K$. (Since your main problem seems to have been knowing what you needed to prove, I'll omit the proof, which is easy.)
Once these properties have been proved, for all $a, b \in K$ you define $b <_K a$ to mean $a - b \in \mathbb{P}_K$. It is easy to see that $b <_K a$ if and only if $ b < a$, so the subscript $K$ can be dropped. In other words, of $a$ and $b$, the larger number in $K$ is the same as the larger number in $\mathbb{R}$.
$\endgroup$ $\begingroup$Bartle and Sherbert define $\mathbb{P}$ as a set possessing three properties, one of which is the Trichotomy Property (p. 26). Therefore, $\mathbb{P}$ possesses the Trichotomy Property by definition.
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