Prove that if $x > 0$ then $1/x > 0$ using axioms of real numbers 
Prove that if $x > 0$ then $1/x > 0$ using axioms of real numbers

This is how I attempted to solve this problem. However, I am not entirely sure whether the contradiction I got is enough to finish the proof, as none of the axioms explicitly states that $1>0$:
(1) Assume that $1/x < 0$.  (2) Multiply both sides of the inequality by $x > 0$ 
(3) $1<0$ - Contradiction, therefore the assumption was false.
 A: $$1>0$$
therefore
$$x\cdot\frac1x>0$$
Because $x>0$, $\frac1x>0$, via total ordering on $\mathbb R$.
A: Your argument seems fine, up to the gap you mention: Why is $1 > 0$? To avoid confusion about the axioms, I will use the definition of an ordered field from Wikipedia, which has the two axioms:

*

*$a < b \implies a+c < b+c$ for any $c$

*If $a > 0$ and $b > 0$, then $ab > 0$.

Let's start with the following claim:
$$a < 0 \iff -a > 0.$$
Proof: The implication "$\implies$"can be seen by adding $c = -a$ to the inequality and using axiom 1. The converse can be seen by adding $a$ and axiom 1.
From this we can deduce: For any $a \neq 0$ we have
$$a^2 > 0.$$
Proof: If $a > 0$ this follows directly from axiom 2. If $a < 0$, we have $-a > 0$, so applying axiom 2 to $-a$ gives
$$0 < (-a)(-a) = a^2$$
In particular we deduce $1 = 1^2 > 0$.
A: Because there's no explicit answer here, here's one method:
"Suppose $a \in \mathbb{F} ^{+}$, where $\mathbb{F}$ is a field. Since $a \neq 0$, we have that $1/a \in \mathbb{F}$ exists by the multiplicative inverse axiom of fields. Note it should be obvious that $1/a \neq 0$ (why?).
Hence, using the property $$x\in \mathbb{F} \setminus \{0\} \; \; \text{ implies } \; \; x^{2}>0,$$(like in the comments) we get $$(1/a)^{2}>0.$$It follows that $$a \cdot (1/a)^{2} > a \cdot 0,$$or equivalently, $1/a>0$. (Note the $1/a<0$ case is similar). 
QED."
A: It looks like the problem is to show that $1\gt 0$.
We know that either $1\gt 0$ or $-1\gt 0$ by the trichotomy axioms for ordering. Note that $1\ne 0$ otherwise we could easily show that all numbers are equal to $0$.
If $-1\gt 0$ then, since $-1$ and $x$ are both positive, we get that $-x\gt 0$ by the multiplicative closure axiom for positive numbers.
We can invoke trichotomy once again to point out that either $x$ or $-x$ are positive. Since $-x\gt 0$, $x$ is not positive, giving us a contradiction.
Therefore, $1\gt 0$.
A: Let's work in any ordered field $F$, so there exists a subset $P$ with the following properties:

*

*$F$ is the disjoint union of $P$, $\{0\}$ and $-P=\{-a:a\in P\}$;

*if $a,b\in P$, then $a+b\in P$;

*if $a,b\in P$, then $ab\in P$.

Given such a subset $P$ we can define $a<b$ for $b-a\in P$ and prove this is an order relation compatible with the field operations. Conversely, given such an order relation, $P=\{a\in F:0<a\}$ satisfies the properties above and $a<b$ if and only if $b-a\in P$.
Fact 1. If $a\in F$, $a\ne0$, then $a^2\in P$
Proof. If $a\in P$, then $a^2\in P$ by property 3. If $a\in -P$, then $-a\in P$ and $a^2=(-a)^2\in P$.
Fact 2. $1=1^2\in P$
Fact 3. If $a\in P$, then $1/a\in P$.
Proof. Suppose, to the contrary, that $1/a\notin P$. Then $-1/a\in P$ and
$$
-1=-\frac{1}{a}a\in P
$$
which contradicts $1\in P$.
