Q:

Prove that $\forall x,y,z \in \Bbb R^+:(x\cdot y)\cdot z = x\cdot (y\cdot z)$

Let $\Bbb R^+=\{x\in\Bbb R \mid x>0\}$ and $x,y\in\Bbb R^+$. We define the multiplication operation $(\cdot)$ on $\Bbb R^+$ by $$x\cdot y:=\inf\{r\cdot s\mid r,s\in\Bbb Q \text{ and } x<r  \text{ and } y<s\}$$ Prove that $$\forall x,y,z \in \Bbb R^+:(x\cdot y)\cdot z = x\cdot (y\cdot z)$$

My textbook said that the proof is straightforward, but it took me a lot of time to come up with a rigorous one. Please help me verify it. Thank you so much!

My attempt:
By definition, we have:

$x\cdot y = \inf\{r\cdot s\mid r,s\in\Bbb Q, x<r, y<s\}$
$y\cdot z = \inf\{s\cdot t\mid s,t\in\Bbb Q, y<s, z<t\}$
$(x\cdot y)\cdot z = \inf\{p\cdot t\mid p,t\in\Bbb Q, x\cdot y<p, z<t\}=\inf A$
$x\cdot (y\cdot z) = \inf\{r\cdot q\mid r,q\in\Bbb Q, x<r, y\cdot z<q\}=\inf B$

It suffices to prove that $A=B$.
Notice that $p\in \Bbb Q$ and $p>x\cdot y \iff$ $p\in \Bbb Q$ and  $p>r\cdot s$ for some $r,s\in\Bbb Q$ such that $r>x,s>y$. Let $p=r\cdot \bar s>r\cdot s$. Then $\bar s \in \Bbb Q$ and $\bar s>s>y$. Thus $p=r\cdot \bar s$ where $r,\bar s \in \Bbb Q$ such that $r>x,\bar s>y$.
Similarly, $q\in \Bbb Q$ and $q>y\cdot z \implies q=s\cdot\bar t$ where $s,\bar t \in \Bbb Q$ such that $s>y,\bar t>z$.

$a\in A \implies a=p\cdot t$ for some $p,t\in\Bbb Q,p>x\cdot y,t>z$ $\implies a=(r\cdot \bar s)\cdot t$ for some $r,\bar s,t \in \Bbb Q$ such that $r>x,\bar s>y,t>z$ $\implies a=r\cdot (\bar s\cdot t)$ for some $r,\bar s,t \in \Bbb Q$ such that $r>x,\bar s>y,t>z$ $\implies a=r\cdot q$ for some $r,q=\bar s\cdot t\in \Bbb Q$ such that $r>x,q>y\cdot z$ $\implies a\in B$.
Similarly, $a\in B\implies a\in A$.

Hence $A=B$ and thus $\inf A=\inf B$.

A:

We define the addition operation $(+)$ on $\Bbb R$ by $$x+ y:=\inf\{r+ s\mid r,s\in\Bbb Q \text{ and } x<r  \text{ and } y<s\}$$
I would like to present a proof that $\forall x,y,z \in \Bbb R^+:x\cdot (y+z) = x\cdot y +x\cdot z$.

My attempt:
First, we have some useful observations:

$m\in \Bbb Q$ and $m > x\cdot y$ $\iff m=r \cdot  s$ for some $r,s\in \Bbb Q$ such that $r>x,s>y$.
$p\in\Bbb Q$ and $p>y+z$ $\iff p=s+t$ for some $s,t\in \Bbb Q$ such that $s>y,t>z$.

By definition, we have:

$x\cdot (y+z) = \inf \{r\cdot p \mid r,p\in\Bbb Q,x<r,y+z<p\}=\inf \{r\cdot(s+t) \mid r,s,t\in\Bbb Q, x<r,y<s,z<t\}=\inf \{r\cdot s+r\cdot t \mid r,s,t\in\Bbb Q, x<r,y<s,z<t\}.$
$x\cdot y+x\cdot z = \inf \{m+n \mid m,n\in\Bbb Q,x\cdot y<m,x\cdot z<n\}=\inf \{r\cdot s+r\cdot t \mid r,s,t\in\Bbb Q, x<r,y<s,z<t\}.$

Hence $x\cdot (y+z) = x\cdot y +x\cdot z$

