# Proving multiplication is compatible with ordering considering an ordered field.

Definition: Ordered field is a field $\mathcal F=(\Bbb{F},+,\cdot)$ together with relation $<$ which satisfy: $$\forall x,y\in\Bbb{F}: \text{exactly one of these 3 hold: }x<y\vee y<x\vee x=y\tag{i}$$ $$\forall x,y,z\in\Bbb{F}:x<y\wedge y<z \Rightarrow x<z\tag{ii}$$ $$\forall x,y,z\in\Bbb{F}: y<z\Rightarrow x+y<x+z\tag{iii}$$ $$\forall x,y\in \Bbb{F}:x,y>0\Rightarrow x\cdot y>0\tag{iv}$$

One of the consequences of this should be that if $x<y$ and $z>0$ then $xz<yz$ and if $z<0$ then $xz>yz$. I'm quite unsure where to start the proof of this. My attempt goes like this:

First assume that $x>y>0$ we can say that $xz>0$ and $yz>0$ now this goes $xz+yz>0$ all i can make of this that $xz>-yz$ which is quite trivial, because right side is negative and left is positive. In fact I'm struggling to somehow involve the multiplication, since there is no property that would allow me to "multiply both of an inequality". Help please.

Take $x <y$, $z>0$. Using (iii), you get $y-x>0$. Now (iv) gives $$(y-x)z>0,$$ which is $$yz-xz>0,$$ and now with (iii) you get $$yz>xz.$$
When $z <0$, use (iii) to get $-z>0$ , and now by the above $$(y-x)(-z)>0,$$ and you unravel this to $$zy <zx.$$
• Well, i didn't say $<$ is a total order, i just said it's a relation satisfying these properties. I could say explicitly that there is a total order $\leq$ with iii) and iv) and that's basically equivalent. In MY definition, i can define $a\leq b\Leftrightarrow a<b \vee a=b$ But I think that may differ in various books and stuff. – Michal Dvořák Aug 26 '18 at 13:29
• You are missing my point. I'm talking about trichotomy (i), not about the difference between $<$ and $\leq$. – Martin Argerami Aug 26 '18 at 13:33
• @GEdgar: if I hear "ordered field" I assume partially ordered. And I would call $\mathbb C$ a "common (non-trivial) example". – Martin Argerami Aug 26 '18 at 13:41