# 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.

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If $1<0$ then $-1>0$, hence $1=(-1)\cdot(-1)>0$.

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Why do you assume that (−1)⋅(−1)>0 ? – vondip Nov 6 '12 at 16:38
He doesn't assume it. The product of positives is positive. That's part of the definition of an ordered field. – kahen Nov 6 '12 at 16:41
I do not assume that. I assume that $1<0$. By adding $-1$ on both sides, I find $0<-1$. By multiplying with the positive(!) number $-1$ then $1>0$ follows. – Hagen von Eitzen Nov 6 '12 at 16:41
I don't know if it's called anything, but the definition goes like this: An ordered pair $(F,P)$ where $F$ is a field and $P \subset F$ satisfies (1) $0 \in P$ and (2) $x,y \in P \implies x+y \in P$ and $xy \in P$ is said to be an ordered field and $P$ is said to be the positive cone of $F$. $\quad$ It's an easy exercise to prove that if $(F,P)$ is an ordered field then $x \leq y \iff y-x \in P$ defines a total order on $F$. – kahen Nov 6 '12 at 16:45
Oh. I forgot part of the definition. You need (3) $F$ is the disjoint union of $P\setminus\{0\}$, $-P\setminus\{0\}$ and $\{0\}$. Otherwise $P$ is just a prepositive cone. – kahen Nov 7 '12 at 6:34

You can use the trivial inequality $x^2 \geq 0$ for all $x$. Prove this fact and use it to prove $1 >0$.

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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, $$a\cdot a^{-1}<0$$ $$a\cdot a^{-1}\cdot a<0\cdot a$$ $$a<0$$ A contradiction, since by the trichotomy axiom, we cannot have $a>0$ and $a<0$. Thus we must have that $1>0$.

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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

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