Is it true that if $p,q \in K_{>0}$, then $[0,pq) \subseteq [0,p)[0,q)$ in any ordered field? It seems possible that the following holds:

Conjecture. Let $K$ denote an ordered field.
If $p,q \in K_{>0}$, then $[0,pq) \subseteq [0,p)[0,q).$
More precisely: If $p,q \in K_{>0}$, then for all $x \in [0,pq)$, there exists $\alpha \in [0,p), \beta \in [0,q)$ such that $x = \alpha\beta$.

I'm specifically interested in the cases $K = \mathbb{R}$ and $K = \mathbb{Q}$, but it should be true irregardless.
Ideas, anyone? I guess we should look for expressions $\alpha(x,p,q)$ and $\beta(x,p,q)$ involving the field operations that satisfy $x = \alpha(x,p,q) \cdot \beta(x,p,q).$ My best guess is something like $(p/q)\sqrt{x}$ and $(q/p)\sqrt{x}$, but of course we're not allowed to take square roots in an arbitrary ordered field. I also don't think these expressions respect the inequalities.
 A: Let $\mathbb{K}$ be an ordered field.  We shall prove that 
$$[0,p)\cdot[0,q)=[0,pq)$$
for all $p,q\in\mathbb{K}_{>0}$.
If $x\in [0,p)$ and $y\in [0,q)$, then we have
$$p-x>0\text{ and }q-y>0\,.$$
Since $p>0$ and $y\geq 0$, 
$$p(q-y)>0\text{ and }(p-x)y\geq 0\,.$$
Therefore, $$pq-xy=p(q-y)+(p-x)y>0\,.$$
This shows that $$[0,p)\cdot[0,q)\subseteq [0,pq)\,.$$
Now, suppose that $z\in[0,pq)$.  If $z=0$, then clearly $z=0\cdot 0$ with $0\in[0,p)$ and $0\in[0,q)$.  Suppose now that $z>0$.  Define $f:(0,p)\to \mathbb{K}$ via
$$f(x):=\frac{z}{x}$$
for all $x\in(0,p)$.  Because $z>0$ and $x>0$ for all $x\in(0,p)$, we know that the image $I$ of $f$ is contained in $\mathbb{K}_{>0}$.  If $I\cap(0,q)$ is empty, then $I\subseteq [q,\infty)$.  This means
$$\frac{z}{x}\geq q\text{ or }x \leq \frac{z}{q}$$
for all $x\in(0,p)$.  Consequently, $(0,p)$ is a subset of $\left(0,\dfrac{z}{q}\right]$.  (If this were not true, then $\dfrac{z}{q}\in (0,p)$ so $\dfrac{z}{q}<\dfrac{1}{2}\,\left(\dfrac{z}{q}+p\right)<p$, leading to a contradiction.)    This means
$$p\leq \dfrac{z}{q}\text{ or }z\geq pq\,.$$
This contradicts the assumption that $z\in[0,pq)$.  Therefore, $I\cap(0,q)$ is nonempty, whence there exist $u\in (0,p)$ and $v\in(0,q)$ such that $f(u)=v$.  This means $$\frac{z}{u}=v\text{ or }uv=z\,.$$
Hence, $z\in(0,p)\cdot(0,q)$.  Ergo,
$$[0,pq)\subseteq [0,p)\cdot[0,q)\,.$$
A: Suppose $p<q$.


*

*Then $\alpha = \frac{x}{q} \in [0,p)$ and $\beta = p \in [0,q)$.


Otherwise $p=q$.


*

*Then $x\in[0,p^2)$. Taking $\alpha = x/p\in[0,p)$ is ok, but then $\beta = p \not\in [0,p)$. 
We can fix that. 
Define 


*

*$a=p-\alpha\in (0,p]$

*$b=p-\beta\in  (0,p]$
so that $x=(p-a)(p-b)\in[0,p^2)$. Furthermore call $c=p^2-x\in(0,p^2)$, where we exclude the case $x=0$ for being trivial. Then we want to find a solution in $a,b\in(0,p)$ of
$$
p^2 - (p-a)(p-b) = c
$$
or in other words
$$
ab - (a+b)p + c = 0.
$$
For each $c$ select $a$ such that $c-ap>0$, for example $a=\tfrac{c}{2p}\in(0,p)$, and then we have
$$
b = \tfrac{c-ap}{p-a}\in(0,p)
$$
So we have found a solution in $a$ and $b$ which translates to a soltuion in $\alpha=p-a,\beta=p-b$.
