Given $\mathbb{F}$ is an ordered field, prove 10.

Question

Given $\mathbb{F}$ is an ordered field, prove 1>0.

geerky42 · Answer

This may help: http://math.stackexchange.com/questions/231438/how-do-i-prove-that-1-0-in-an-ordered-field

anonymous · Answer

That's where I don't quite understand. How can you use some wrong assumption to prove something right?

geerky42 · Answer

Maybe to prove a contradiction

I guess we assumed something wrong to be true then we point out contradiction, hence original statement  (1>0) is true.

geerky42 · Answer

"if 1<0, then -1>0, then so is (-1)(-1) = 1 by Positive scaling. But 1 is not positive (allegedly). This is a contradiction. So 1 is positive after all. QED"

geerky42 · Answer

Sh*t man, I don't ever know what ordered field is...

What class are you taking?

geerky42 · Answer

@RolyPoly

anonymous · Answer

Introduction to Math. Analysis

geerky42 · Answer

aha, it's far away from where I am now, I think.

Does my explanation help somehow?

anonymous · Answer

I understand that we are trying to prove it by contradiction but still, I am not able to comprehend it.

geerky42 · Answer

yeah. me too ha...

Sorry, at least I tried.

geerky42 · Answer

Maybe @ganeshie8 @agent0smith can help?

anonymous · Answer

It's ok. Thanks for your help!

anonymous · Answer

Ordered fields have two important axioms (along with the usual field axioms):
$$(1)~~~~\text{If }a,b,c\in\mathbb{F}\text{ with }a\le b,\text{ then }a+c\le b+c\
(2)~~~~\text{If }0\le a\text{ and }0\le b,\text{ then }0\le ab$$
The contradiction suggested by geerky uses the second axiom to prove there is a contradiction.

anonymous · Answer

Hmm... What I am learning is a bit different, but essentially the same.
For (2),  $\text{if } a,b > 0\text{, then }a+b>0\text{ and }a*b >0$

anonymous · Answer

I don't know what's wrong with me, but seems my problem is solved. Thanks everyone! Proposition 1: If F is an ordered field and $a \in \mathbb{F}$ with $a\ne0$, precisely one of a>0 or -a<0 holds [proof omitted] Suppose 1<0, by proposition 1, we have -1>0 By O2, we have (-1) * (-1) > 0 And by some tedious proof, we have (-1) * (-1) = 1 Hence, we have 1>0, which contradicts to our assumption. Since $1\ne0$, hence, we have 1>0.