Is it true that if $p,q \in K_{>0}$, then $[0,pq) \subseteq [0,p)[0,q)$ in any ordered field?

Question

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.

Answer 1

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, $$[0,pq)\subseteq [0,p)\cdot[0,q)\,.$$

Answer 2

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 $\alpha=p-a,\beta=p-b$.