In this Rudin's book, he introduces Dedekind cut at the end of chapter 1. I provide the context first. Then I describe my question carefully.
$\ $
$\ $
In the first picture, it shows the "field axioms" (A1)-(A5), (M1)-(M5), and (D). In step 4, it shows that (A1)-(A5) hold in $\mathbf{R}$. In step 6, it says that the proofs of (M1)-(M5) in $\mathbf{R^+}$ are very similar to (A1)-(A5) in $\mathbf{R}$, so these proofs are omitted. Actually, I can prove (M1)-(M4) in $\mathbf{R^+}$ by myself. But I meet some trouble to prove (M5) in $\mathbf{R^+}$, which seems not that "similar" to (A5) in $\mathbf{R}$. I rewrite the proof of (A5) in $\mathbf{R}$ and then I describe the difficults of using similar methods to prove (M5) in $\mathbf{R^+}$.
$\ $
Part I Review the proof of (A5) in $\mathbf{R}$
Fix $\alpha\in\mathbf{R}$. $\beta=\{p\ |\ p\in\mathbf{Q},\ and,\ \exists r\in\mathbf{Q^{+}},-p-r\notin\alpha\}$.
Then it is easy to proves $\beta\in\mathbf{R}$ and $\alpha+\beta\subset0^{\ast}$.
I want to review the proof of $0^{\ast}\subset\alpha+\beta$.
First, it is easy to prove that for all $w\in\mathbf{Q}^{+}$, there exists an integer $n$ such that $nw\in\alpha$ and $(n+1)w\notin\alpha$.
Fix $w\in\mathbf{Q}^{+}$. We set $p=-(n+2)w$. Then $-p-w=(n+1)w\notin\alpha$. So, $p\in\beta$. So, $nw+p\in\alpha+\beta$. So, $-2w\in\alpha+\beta$. So, for all $w\in\mathbf{Q}^{+}$, $-2w\in\alpha+\beta$.
Fix $v\in0^{\ast}$. So, $v\in\mathbf{Q}$ and $v<0$. So, $-v/2\in\mathbf{Q}^{+}$. So, $v\in\alpha+\beta$. So, for all $v\in0^{\ast}$, $v\in\alpha+\beta$. So, $0^{\ast}\subset\alpha+\beta$.
$\ $
Part 2 Try to use "similar" method to prove (M5) in $\mathbf{R}^{+}$.
Fix $\alpha\in\mathbf{R}^{+}$. $\beta=\{p\ |\ p\in\mathbf{Q},\ q\in\{q\ |\ q\in\mathbf{Q}^{+},\ and,\ \exists r\in\mathbf{Q}^{+},1/q-r\notin\alpha\},\ p\leq q\}$
Then it is easy to prove $\beta\in\mathbf{R}^{+}$ and $\alpha\beta\subset1^{\ast}$.
I try to prove $1^{\ast}\subset\alpha\beta$ similarily (but I fail).
First, it is easy to prove that all $w\in\mathbf{Q}$ satifying $w>1$, there exists an integer $n$ such that $w^{n}\in\alpha$ and $w^{n+1}\notin\alpha$.
Fix $w\in\mathbf{Q}$ satisfying $w>1$. We set $q=\frac{1}{2w^{n+1}}$. Then $1/q-w^{n+1}=w^{n+1}\notin\alpha$. So, $q\in\beta$. So, $w^{n}q\in\alpha\beta$. So, $\frac{1}{2w}\in\alpha\beta$. So, for all $w\in\mathbf{Q}$ satisfying $w>1$, $\frac{1}{2w}\in\alpha\beta$.
But by using "for all $w\in\mathbf{Q}$ satisfying $w>1$, $\frac{1}{2w}\in\alpha\beta$.", we can only prove $(\frac{1}{2})^{\ast}\subset\alpha\beta$. I fail to prove $1^{\ast}\subset\alpha\beta$ and I don't know how to prove $1^{\ast}\subset\alpha\beta$.
$\ $
If you don't have a lot of time to check my approach, I just want to ask you how to prove $1^{\ast}\subset\alpha\beta$.
Thanks for your help!