how do I prove that $1 > 0$ in an ordered field? I've started studying calculus. 
As part of studies I've encountered a question.
How does one prove that $1 > 0$?
I tried proving it by contradiction by saying that $1 < 0$,
but I can't seem to contradict this hypothesis.
Any help will be welcomed.
 A: Let x $\in$ R and $ x>0 $.
Now, $x.1 = x >0 $ $\implies$ $ x^{-1}.x.1 >x^{-1}. 0 $ $\implies 1>0 $ [$\because a.0 = 0 $ $ \forall $a $\in $ R $]$
A: Previous answers are not complete.
Axioms of ordered set give us only $\forall x \le 0 \ y \le 0 \ \ \ \ xy \ge 0$.
And we can get only $0 \le 1$.
Now we need to proof the $0 \ne 1$.
$x = x \cdot 1 = x \cdot (1 + 0) = x \cdot 1 + x \cdot 0 \rightarrow x \cdot 0 = 0$
Suppose that $0 = 1$. By the axiom $\forall x \in \mathbb{F} \ \ \ \ x \cdot 1 = x $. But $x \cdot 0 = 0$ by above. Contradiction.
$0 \ne 1 \\ 0 < 1$
QED
A: If $1<0$ then  $-1>0$, hence $1=(-1)\cdot(-1)>0$.
A: You can use the trivial inequality $x^2 > 0$ for all $x\neq 0$. Prove this fact and use it to prove $1 >0$.
A: Let’s consider the question in the ordered field of real numbers.
By the trichotomy axiom of inequality, only one of the following is true:
$$ 1=0 $$
$$ 1<0 $$
$$ 1>0 $$
Now, by the nontriviality axiom of the real numbers, $ 1\ne 0 $, so we’ve ruled out the first possibility.
Now suppose $ 1<0 $. Then for $ a\in R $, $ a>0 $, by the multiplication axioms of an ordered field,
\begin{align}
 a\cdot a^{-1}&=1<0 \\
&\Downarrow\\
 a\cdot a^{-1}\cdot a&<0\cdot a \\
&\Downarrow\\
 a&<0
\end{align}
A contradiction, since by the trichotomy axiom, we cannot have $ a>0 $ and $ a<0
$. Thus we must have that $ 1>0 $.
A: It is not provable that $0$ unequals $1$. It is part of the field-axioms.
A: Suppose $0 \ge 1$ and let $x > 0$. We have:
$$0 \ge 1$$
$$0 \cdot x \ge 1 \cdot x$$
$$0 \ge x$$ A contradiction with our assumption that  $x > 0$
A: Using that
$$ x \neq 0 \Rightarrow x^2 > 0 $$
leads to
$$ 1^2 > 0 $$
and thereby
$$ 1 > 0 $$
