A question about Dedekind cut in Rudin's Principles of Mathematical Analysis 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.
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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 $\mathbf{R^+}$.
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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, $0^{\ast}\subset\alpha+\beta$.
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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
$1^{\ast}\subset\alpha\beta$.
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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!
 A: The proof that follows is due to stackexchange user Jeremy Weissmann, who attempted to imitate Rudin's additive inverse proof with great success.
I have filled in key details in the form of four lemmas in the hope of easing understanding. Certainly they shorten the final proof. I assume you have already proved commutativity of multiplication of positive cuts, and that $\beta{}$ is a positive cut.
Given $\alpha{}>0^*$, define $\beta{}:=\{q|q\in{}\mathbb{Q}\,\,\,\wedge{}\,\,\,q\leq{}p\,\,\,\wedge{}\,\,\,p\in{}\mathbb{Q}\,\,\,\wedge{}\,\,\,\exists{}r(r\in{}\mathbb{Q}\,\,\,\wedge{}\,\,\,r>1\,\,\,\wedge{}\,\,\frac{1}{pr}\in{}\mathbb{Q}\setminus{}\alpha{})\}$
Claim: $1^*\subset{}\alpha{}\beta{}$
Proof:
Take $v\in{}1^*,\,\,0<v<1$. I must show $v\in{}\alpha{}\beta{}$. I can take $\alpha{}>1^*$ without loss of generality: if $\alpha{}<1^*$, then by Lemma 1 $\beta{}>1^*$. Because $\alpha{}\beta{}=\beta{}\alpha{}$ and by Lemma 4, the problem is reduced to showing that $v\in{}\beta{}\alpha{}$ and I may relabel $\alpha{}$ as $\beta{}$ and $\beta{}$ as $\alpha{}$ in the proof below.
Use Lemma 2 to write $v=jk$. Define $w=\frac{1}{j}$ and note $w>1$. Use Lemma 3 so that $w^n\in{}\alpha{}$ and $w^{n+1}\in{}\mathbb{Q}\setminus{}\alpha{}$. Then $\frac{k}{w^{n+1}}\in{}\beta{}$ (choose $r=\frac{1}{k}$ in the definition of $\beta{}$) and $v=w^n\cdot{}\frac{k}{w^{n+1}}$ proves the claim.
For $v\leq{}0$, note that $1\in{}\alpha{}$ and $v\in{}\beta{}$. Therefore $v=1\cdot{}v$ shows the claim.
Lemma 1: $\alpha{}<1^{*}\implies{}1^{*}<\beta{}$
Proof of Lemma 1: $\,\,\alpha{}<1^{*}\implies{}\exists{}q(q\in{}\mathbb{Q}^+\,\,\,\wedge{}\,\,\,q<1\,\,\,\wedge{}\,\,\,q\not\in{}\alpha{})$. Define $q':=\frac{1+q}{2}$ so $q<q'<1$ and $q'\not\in{}\alpha{}$ and $\frac{1}{q'}>1$. Set $r:=\frac{1+q}{2q}$ and note $r>1$. Then $\frac{q'}{r}\not\in{}\alpha{}$ shows $\frac{1}{q'}\in{}\beta{}$, requiring $\beta{}>1^{*}$.
Lemma 2: $v\in{}\mathbb{Q}^+\,\,\,\wedge{}\,\,\,v<1\implies{}\exists{}j\exists{}k(j,k\in{}\mathbb{Q}^+\,\,\,\wedge{}\,\,\,j,k<1\,\,\,\wedge{}\,\,\,jk=v)$
Proof of Lemma 2: Put $j=\frac{1+v}{2}$ and $k=\frac{2v}{1+v}$.
Lemma 3: $\alpha{}>1^*,\,\,\,w\in{}\mathbb{Q},\,\,\,w>1\implies{}\exists{}n(n\in{}\mathbb{N}\cup{}\{0\}\,\,\,\wedge{}\,\,\,w^n\in{}\alpha{}\,\,\,\wedge{}\,\,\,w^{n+1}\not\in{}\alpha{})$.
Proof of Lemma 3: $k\in{}\mathbb{N}\,\cup\,\{0\}\implies{}w^k=((w-1)+1)^k=\sum_{i=0}^k\binom{k}{i}(w-1)^i1^{k-i}$ $=1+k(w-1)+\ldots{}+k(w-1)^{k-1}+(w-1)^k>k(w-1)$. By the archimedean property of $\mathbb{Q}$, I can choose $k$ such that $k(w-1)>q$ for $q\in{}\alpha{}$. Then $q<k(w-1)<w^k$ thus $w^k\in{}\mathbb{Q}\setminus{}\alpha{}$. Also, $1=w^0\in{}\alpha{}$. Define $S:=\{i|i\in{}\mathbb{N}\,\,\,\wedge{}\,\,\,w^i\in{}\mathbb{Q}\setminus{}\alpha{}\}$. The above shows $S\neq{}\varnothing{}$. By the well-ordering principle, $S$ has a minimum element $m$. Set $n=m-1$. Then $w^n\in{}\alpha{}$ and $w^{n+1}\in{}\mathbb{Q}\setminus{}\alpha{}$.
Lemma 4: $\alpha{}=S$, where $S=\{q|q\in{}\mathbb{Q}\,\,\wedge{}\,\,q\leq{}p\,\,\wedge{}\,\,p\in{}\mathbb{Q}\,\,\wedge{}\,\,\exists{}r(r\in{}\mathbb{Q}\,\,\wedge{}\,\,r>1\,\,\wedge{}\,\,\frac{1}{pr}\in{}\mathbb{Q}\setminus{}\beta{})\}$.
Proof of Lemma 4:
$(\subset{})$: Choose $q\in{}\alpha{},\,\,q\neq{}0$.$\,\,\,q\in{}\alpha{}\implies{}\exists{}q'(q'\in{}\alpha{}\,\,\wedge{}\,\,q<q')$. Set $r:=\frac{q'}{q}$ and note that $r>1$. Then $\frac{1}{qr}=\frac{1}{q'}\in{}\mathbb{Q}\setminus{}\beta{}$ shows that $q\in{}S$.
$(\supset{}):$ $q\in{}S\implies{}q\leq{}p$ for $p\in{}\mathbb{Q}$ satisfying $\frac{1}{pr}\in{}\mathbb{Q}\setminus{}\beta{}$ for some $r\in{}\mathbb{Q},\,\,r>1$. $\frac{1}{pr}\in{}\mathbb{Q}\setminus{}\beta{}\implies{}\forall{}r'(r'\in{}\mathbb{Q}\,\,\wedge{}\,\,r'>1\implies{}\frac{pr}{r'}\in{}\alpha{})$. Take $r':=r$ to show that $p\in{}\alpha{}$ and therefore $q\in{}\alpha{}$.
