Proof that $\frac{1}{2} > 0$ using order properties of $\mathbb R$ Let $\mathbb P$ be a non-empty subset of $\mathbb R$ such that for all $a, b \in \mathbb P$,


*

*$a + b \in \mathbb P$

*$ab \in \mathbb P$

*Exactly one of the following holds: 
$a \in \mathbb P$, $-a \in \Bbb P$, $a = 0$.


I already know that $n > 0$ for all $n\in \mathbb N$. How do I show that $\frac{1}{2} > 0$, or, in general, $\frac{1}{n} > 0$? My initial attempt at doing this was to assume by contradiction that $\frac{1}{2} < 0$. Then by definition, $-\frac{1}{2} \in \mathbb P$. My question is, can I assume that $-\frac{1}{2} - \frac{1}{2} = -1$ if I'm basing this on the algebraic properties of $\mathbb R$ where $\mathbb R$ is treated as a field?
For instance, although it's obvious that $1*0 = 0$ in common sense math, in field theory, it is proven rigorously (a.k.a the annihilator property). Does a similar idea apply for other numbers in the field, for instance, the number $\frac{1}{2}$?
Thanks in advance.
 A: You can prove that for any $a\neq 0$ that $a^2 \in \Bbb{P}$. From here you can note that $1/4 = (1/2)^2$ and $1/4 + 1/4 = 1/2$
A: Thank you for a well-asked question! I believe your strategy is sound, but it can be simplified and generalized:
Let $m,n\in\Bbb N$; we will show that $\frac mn\in\Bbb P$.
Assume that $\frac mn \notin \Bbb P$. Then by trichotomy (and nonzeroness), $-\frac mn \in \Bbb P$. Since $n\in \Bbb N$ we know that $n\in\Bbb P$; by multiplicative closure, $(-\frac mn)n = -m\in\Bbb P$. But $m\in\Bbb N$ and thus $m\in\Bbb P$, contradicting trichotomy.
A: I think the basic two axiom that this boils down to is if $a<b$ and $n > 0$ then $an < bn$.
Given $n > 0$ and the three options $\frac 1n < 0$ or $\frac 1n = 0$ or $\frac 1n > 0$ we geo $1=\frac 1n*n < 0*n =0$ (you claim you have proven that both $1 > 0$ and $0*n=0$, right? So this is a contradiction) or $1=\frac 1n*n = 0*n = 0$ (which is false by definition of a non-trivial field $1\ne 0$) or $1 =\frac 1n*n > 0*n =0$ (which is consistent.)
.....
$-1\in \mathbb P$ then $(-1)*(-1)=1\in \mathbb P$ which is a contradiction.  (Assuming you have proven that $(-a)(-b) = ab$.)
So $-1 \not \in \mathbb P$ and $1\ne 0$ so $1\in \mathbb P$ and bey induction $\mathbb N\subset P$.
If any $-\frac 1n\in \mathbb P$ then $(-\frac 1n)n =-1\in \mathbb P$ which is a contradiction and as $\frac 1n \ne 0$ we have $\frac 1n\in \mathbb P$. and thus $\mathbb Q^+\subset \mathbb P$ and $\mathbb Q^-\cap \mathbb P = \emptyset$.
And $0\not \in \mathbb P$ as $0=0$.
The only thing is to determine whether it is possible for a negative  irrational to be in $\mathbb P$.
==== old answer ====
First proof that $x^2 \ge 0$ so $1 = 1^2 = (-1)^2 \ge 0$ but as $1 \ne 0$ than $1>0$.
Then $1 > 0$ so $2 =1+1 > 0+1 > 0+0$.
Then prove that if $a > 0$ and $c > 0$ that $\frac ac > 0$ so $1 > 0$ and $2>0$ so $\frac 12 >0$.
.....
To do these you have an axiom that $a > b$ and $c > 0$ implies $ac > bc$.
You will have to prove that $c < 0 $ and $a > b$ implies $ac < bc$.... and to prove that you will have to prove $(-a)*b = -(ab)$ and then $a > 0$ means $-a < 0$.  (Which you can do with the axiome that $a < b$ means $a+c < b+c$ for all $c$.
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Prop 1:  If $a > 0$ then $-a < 0$.  Pr:  $a>0\implies a+(-a)>0 + (-a)\implies 0 > -a$.
Prop 2:  $-(-a) = a$. Proof:  $0 = a + (-a)$ and $0 = -(-a) + (-a)$ so $a+(-a)= -(-a) + (-a)$ so $a + (-a) + a = -(-a) + (-a) + a$ so $a = -(-a)$.
Prop 3: $0*b = 0$.  Proof:  $0*b = (0+0)b = 0*b + 0*b$ so $0 = 0*b +(-0*b) = 0*b + 0*b +(-0*b)= 0*b$.
Prop 4:  $(-a)b = -(ab)$  Pf:  $ab+ (-a)b = (a+(-a))*b = 0*b=0$ so $(-a)b =-(ab)$.
Can you continue:
Prop 5: $a > b$ and $c < 0$ then $ac < bc$. Pf: ?????
Prop 6: if $a > 0$ then $a^2 > 0$. If $a < 0$ then $a^2 > 0$. And if $a=0$ then $a^2 =0$ and so $a^2 \ge 0$ with equality holding if and only if $a=0$.  Pf:  ???????
Prop 8: if $a > b$ and $c > 0$ then $\frac ac > \frac bc$. Pf:??????
Prop 9:  $1 > 0$ Pf:?????
Prop 10: $2 > 0 $ Pf:?????
....
or as per WaveX suggestion
$\frac 12 = (\frac 12)^2 + (\frac 12)^2$.
A: Let $\mathbb P=\{x\in \mathbb R: x\gt0\}$. Then $\mathbb P$ satisfies the conditions. And $\forall n \in \mathbb N, \ n\in \mathbb P$.
Now suppose $-\frac{1}{2}\in \mathbb P$. Thus $(-\frac{1}{2})+(-\frac{1}{2})=-1\in\mathbb P$. Which is a contradiction since $n\in \mathbb P,\ \forall n\in \mathbb N\implies 1\in \mathbb P$ . ($-1\notin \mathbb P$).
Thus, $\frac{1}{2} \in \mathbb P$, since for any $x\in \mathbb R$, $x\gt0$ or $x\lt0$ or $x=0$ and $\frac{1}{2}\neq0$.
You can extend this argument for $\frac{1}{n}$. assume $-\frac{1}{n}\in \mathbb P$ then it implies $-\frac{1}{n}\cdot n=-1\in \mathbb P$ which is a contradiction.
A: Prove that if $a,b > 0$, then $\frac{a}{b} > 0$: suppose $\frac{a}{b} \neq 0$, then either $\frac{a}{b} = 0$ or $-\frac{a}{b} > 0$. If $\frac{a}{b} = 0$, then it follows that $a = 0$, contradicting $a > 0$. If $-\frac{a}{b} > 0$, then it follows that $-\frac{a}{b}b > 0 \Rightarrow -a > 0$, contradicting $a > 0$ as well.
