Vtyjkyuk

Definition: Ordered field is a field $mathcal F=(BbbF,+,cdot)$ together with relation $<$ which satisfy:
$$forall x,yinBbbF: textexactly one of these 3 hold: x<yvee y<xvee x=ytagi$$
$$forall x,y,zinBbbF:x<ywedge y<z Rightarrow x<ztagii$$
$$forall x,y,zinBbbF: y<zRightarrow x+y<x+ztagiii$$
$$forall x,yin BbbF:x,y>0Rightarrow xcdot y>0tagiv$$

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 $$zy <zx. $$

Note that (i) is irrelevant for this. And I would call a field with (i) a "totally ordered" field.