Multiplicative Inverse all between 0 and 1

Question

Why do all "the" mutliplicative inverses in an ordered field stick in a bunch between 0 and 1?

It's given that $x>y>0 \Rightarrow 0<x^{-1}<y^{-1}$ and the 1 is its own inverse. So you get that the inverses of elements $e>1$ fall into $(0,1)$.

Why can't there be elements between 0 and 1 that are not "real inverse elements" (inverse elements of elements $e>1$)?

Answer 1

Because

$$e>1\Rightarrow e^{-1}e>e^{-1}\cdot 1\Rightarrow 1>e^{-1}.$$

Answer 2

Because $0<x<1$ implies that $x=y^{-1}$ with $y:=x^{-1}$ (which is $>1$)

Answer 3

The inverse relation is shaped like a hyperbole: $$ yx=1; $$ If you imagine the plot you understand why you can't obtain what you asked.

Moreover, I would like to demonstrate in this way what you said:

Let's take $x>1;\ x\in \mathbb{R}$, i.e. $$x=1+y,\ \mbox{for some}\ y\in \mathbb{R^+}$$ Now, if you have $x_1,x_2>1;\ x_1,x_2\in \mathbb{R}$ it holds: $$x_1=1+y_1,\ \mbox{for some}\ y_1\in \mathbb{R^+}$$ $$x_2=1+y_2,\ \mbox{for some}\ y_2\in \mathbb{R^+}$$ Hence: $$ x_1x_2=(1+y_1)(1+y_2)=1+y_1+y_2+y_1y_2, $$ and, by definition: $$ y_1+y_2+y_1y_2>0. $$ Therefore: $$ x_1x_2>1. $$