$(f_1\,{\preceq}\,f_2)\,{\land}\,0_F\,{\preceq}\,f_3{\implies}f_1f_3\,{\preceq}\,f_2f_3$

Question

How do I prove that in totally ordered fields $(f_1\,{\preceq}\,f_2)\,{\land}\,0_F\,{\preceq}\,f_3{\implies}f_1f_3\,{\preceq}\,f_2f_3$ is true?

I am pretty sure I've seen this theorem used, although during a search on the Internet I still couldn't find any proof of it.

Answer 1

Let me recall from Wikipedia that:

A field $(F, +, \cdot)$ together with a total order $\preceq$ on $F$ is an ordered field if the order satisfies the following properties for all $a, b$ and $c$ in $F$:
i) if $a \preceq b$ then $a + c \preceq b + c$, and
ii) if $0_F \preceq a$ and $0_F \preceq b$ then $0_F \preceq a \cdot b$.

Suppose that $f_1\preceq f_2$ and $0_F\preceq f_3$.

By i) we have $$0_F=f_1-f_1 \preceq f_2-f_1$$
By ii) we have $$0_F=0_F\cdot f_3 \preceq (f_2-f_1)\cdot f_3=f_2\cdot f_3- f_1\cdot f_3$$
By i) we have $$f_1\cdot f_3 =0_F+f_1\cdot f_3 \preceq f_2\cdot f_3- f_1\cdot f_3+f_1\cdot f_3 =f_2\cdot f_3$$