Can a rational bigger than product of two reals be written as a product of two rationals bigger than each real?

Question

Suppose you have $f,g \in \mathbb{R}$ with $f,g>0$ and $q \in \mathbb{Q}$ such that $fg<q$. How can I prove that there exist $q_1, q_2 \in \mathbb{Q}$ s.t. $q=q_1 q_2$ and $f<q_1$, $g<q_2$?

This is a part of the proof of every complete ordered field being isomorphic to $\mathbb{R}$ that every author I've found seems to leave as an exercise, but I can't get my head around it!!

Your help is very much appreciated!

Answer 1

HINT: Since $f<\frac{q}g$, you can pick a rational $q_1\in\left(f,\frac{q}g\right)$. Then $q_1g<q$; now do something even simpler to get $q_2$.

Answer 2

Following Brians's hint you can choose $q_1$ s.t. $f<q_1<\frac{q}{g}$. Then: $ fg<q_1g<q $. Taking $q_2=\dfrac{q}{q_1}$ you get $fg<q_1g<q_1q_2=q$.

Thank you!