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At the top of the page 20 of Rudin's book ''Principles of Mathematical Analysis'' he writes: ''The proofs (of the multiplication axioms) are so similar to the ones given in detail in Step 4 (proof of the addition axioms) that we omit them''. I tried to prove them but I got stuck in the proof of \begin{equation}\alpha \cdot {\alpha }^{-1}=1^*\end{equation} where $\alpha$ is positive cut and ${\alpha }^{-1}=\mathbb{Q}_{-}\bigcup\left\{0\right\}\bigcup\left\{t\in \mathbb{Q}:0<t<r\text{ for some }r\in \mathbb{Q}:\frac{1}{r}\notin \alpha\right\}$ is the candidate for the multiplicative inverse of $\alpha$. I have already proved that ${\alpha }^{-1}$ is a cut and $\alpha \cdot {\alpha }^{-1}\le 1^*$.

My question is how do we prove the opposite direction similarly to the proof Rudin gives for $\alpha +(-\alpha) \le 0^*$. A proof completely different to that one can be found here: Dedekind cut multiplicative inverse

Here is what I have tried thus far:

Let $p\in 1^*$. If $p\le 0$ then obviously $p\in \alpha\cdot \alpha^{-1}$.

Suppose $0<p<1$ and $q=q(p)\in \mathbb{Q}_{+}$. By the Archimedean Property of Rational numbers \begin{equation}\exists n\in \mathbb{N}:nq\in \alpha\text{ and }(n+1)q\notin \alpha\end{equation} We must find a $u \in \alpha^{-1}$ such as that $p=(nq)\cdot u$ or equivalenty, $u=\frac{p}{nq}$

In order for $u \in \alpha^{-1}$ we must have that $0<u<r$ and $\frac{1}{r}\notin \alpha$ for some rational $r$. The only reasonable choice for $r$ would be $\frac{1}{(n+1)q}$. But then, \begin{equation}u<r\Leftrightarrow \frac{p}{nq}<\frac{1}{(n+1)q}\Leftrightarrow p<\frac{n}{n+1}\end{equation} which may not be true for some values of $n$ (like $0$). Where can we derive a restriction for these values of $n$?

EDIT: Found another proof here: http://mypage.iu.edu/~sgautam/m413.33418.11f/Dedekind.pdf STill nothing similar to Rudin's...

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2 Answers 2

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Let $p\in 1^*$ with $0 < p < 1$. There exists an $n\in \mathbb N$ such that $$ p < 1 - \frac 1 {m + 1} = \frac m {m + 1} \tag{1} $$ for each $m\in \mathbb N$, $m \geq n$.

Let $r\in \alpha, r >0$ and $0 < q < r/n$. There exists an $m$ such that $m q\in \alpha$ and $(m + 1)q\notin \alpha$. Evidently we have $m \geq n$.

Inequality (1) implies $$ \frac p {mq} < \frac m {m + 1}\cdot \frac 1 {mq} = \frac 1 {(m + 1) q} $$ so $\frac p {mq} \in \alpha^{-1}$ and $$ p = (mq)\cdot \frac p {mq}. $$

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    $\begingroup$ Very very nice proof! Not exactly what I was looking for but still, this is fantastic! You made my day! Note: In the beggining you must let $0<p<1$, but that won't affect proof at all as the other case is obvious $\endgroup$
    – Nameless
    Jul 27, 2012 at 13:29
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    $\begingroup$ I assumed $0 < p < 1$, because that was the only case you had difficulties with. Anyway, now, I explicitly mentioned that condition. $\endgroup$
    – AlbertH
    Jul 28, 2012 at 11:22
  • $\begingroup$ Sorry to reopen a question decided two years ago, but precisely why is it evident that $m \geq n$? Otherwise I agree completely... $\endgroup$
    – User12345
    Aug 31, 2014 at 23:52
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    $\begingroup$ Since $p < 1$, there exists an integer $n$ great enough to satisfy $p < 1 - (n + 1)^{-1}$. Then for each $m$ greater than $n$ inequality (1) holds. $\endgroup$
    – AlbertH
    Sep 1, 2014 at 8:34
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    $\begingroup$ @User12345 $0<q<r/n$ and $mq<rm/n$ then $(m+1)q<r(m+1)/n$. If $m<n$ then $m=n-k$ for some $k\geq 1$ hence $(m+1)/n = (n-k+1)/n = 1+(1-k)/n\leq 1$. This implies $(m+1)q<r(m+1)/n\leq r$, then $(m+1)q\in\alpha$, which is contradictory. $\endgroup$ Jan 28, 2016 at 16:02
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EDIT, February 16th, 2022: I never intended this to be a perfect exposition — my aim was simply to show that Rudin's proof for addition could be translated analogously to multiplication. But now that I read over what I wrote, I see there are little inaccuracies everywhere, so I feel compelled to try to fix them. Deviations from Rudin are in brackets. A more significant gap in logic, pointed out by commenter Gary, is addressed below.

I think we can reproduce Rudin's proof completely analogously. I will try to copy Rudin's exposition nearly symbol by symbol.

Fix $\alpha \in \mathbb R^+$. Let $\beta$ be the union of $(-\infty,0]$ with $\beta^+$, the set of all $p>0$ with the following property: There exists $r>1$ such that $(1/p)/r \notin \alpha$.

We show that $\beta \in \mathbb R^+$ and $\alpha\beta = 1^*$.

If $s \notin \alpha$ [so that $s > 0$] and $p=(1/s)/2$, then $(1/p)/2=s \notin \alpha$, hence $p \in \beta$. So $\beta^+$ is not empty [because $p > 0$]. If $0 < q \in \alpha$ [which exists because $0^* < \alpha$], then $1/q \notin \beta$. So $\beta \ne \mathbb Q$. Hence $\beta$ satisfies (I).

Pick $0 < p \in \beta$ [which exists because $\beta^+$ is nonempty], and pick $r > 1$, so that $(1/p)/r \notin \alpha$. If $0 < q < p$, then $0 < (1/p)/r < (1/q)/r$, hence $(1/q)/r \notin \alpha$. Thus $q \in \beta$, and (II) holds.

[For the proof of (III), Rudin uses the expression '$r/2$', which in multiplicative language would be '$\sqrt r$', which may not be rational. We can do just as well with rational $j,k$ satisfying $1<j,k$ and $jk = r$. For example, we can choose $j = {1+r\over 2}$ and $k = r/j = {2r \over 1+r}$, which are both greater than $1$ if $r$ is.]

Put $t = pj$. Then $t > p$, and $(1/t)/k = (1/p)/r \notin\alpha$, so that $t \in \beta$. Hence $\beta$ satisfies (III).

We have proved that $\beta \in \mathbb R$ [and, since $\beta^+$ is nonempty, that $\beta \in \mathbb R^+$].

If $0 < r \in \alpha$ and $0 < s \in \beta$, then $1/s \notin \alpha$, hence $r < 1/s$, $rs < 1$. Thus $\alpha\beta \subseteq 1^*$.

To prove the opposite inclusion, pick $v \in 1^*$, $v > 0$. [Once again, we would like to follow Rudin and set $w = 1/\sqrt v$, but $\sqrt v$ may not be rational. It suffices to take rational $j,k$ satisfying $0<j,k<1$ and $jk = v$. For example, we can choose $j = {v+1\over 2}$ and $k = v/j = {2v \over v+1}$, which are both less than $1$ if $v$ is.]

Put $w = 1/j$. Then $w > 1$, and there is a nonnegative integer $n$ such that $w^n \in \alpha$ but $w^{n+1} \notin \alpha$. (See Comment below.) [This follows from $w^n = ((w-1)+1)^n > n(w-1)$ by binomial expansion, then using the archimedean property of $\mathbb Q$, since $w-1>0$.] Put $p = k/w^{n+1}$. Then $p \in \beta$, since $(1/p)/(1/k) \notin \alpha$, and $$v = w^np \in \alpha\beta.$$ Thus $1^* \subseteq \alpha\beta$.

[Comment: This argument does not work if $w^0 = 1$ is not in $\alpha$, for in that case there is no nonnegative integer $n$ with the required property. However, in that case we can simply switch the roles of $\alpha$ and $\beta$: Indeed, if $1 \notin \alpha$, then $\alpha < 1^*$, from which it follows that $\beta > 1^*$. And there is a symmetry between $\alpha$ and $\beta$, in that $\alpha$ is the union of $(-\infty,0]$ with $\alpha^+$, the set of all $p > 0$ such that there exists $r>1$ which satisfies $(1/p)/r \notin \beta$. Hence we can switch $\alpha$ and $\beta$ in the preceding paragraph and draw the same conclusion, as claimed.]

We conclude that $\alpha\beta = 1^*$.

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  • $\begingroup$ "This follows from $w^n=((w-1)+1)^n>n(w-1)$ by binomial expansion..." Binomial expansion with negative powers would seem to rely on calculus material that Rudin cannot develop without first developing the real numbers... $\endgroup$
    – Gary
    Feb 15, 2022 at 19:27
  • $\begingroup$ I don’t think I intended to use this fact with negative values of $n$. It’s been a long time since I wrote this and don’t remember the details but near as I can tell the claim is not true if $\alpha<1^\ast$. But then $\beta>1^\ast$, and one can check that $\alpha$ has the same property with respect to $\beta$ that $\beta$ has with respect to $\alpha$, so there is complete symmetry and the argument can go through with $\alpha$ and $\beta$ switched. Or not! You tell me! :P $\endgroup$ Feb 16, 2022 at 0:22
  • $\begingroup$ My guess is the claim is true for $\alpha{}<1^*$ if we allow negative $n$ but I can't even approach that case without developing derivatives first. I believe you are correct about symmetry. Your answer helped me a lot. Would you consider modifying your answer to restrict yourself to $n\in{}\mathbb{N}$ and incorporate the symmetry argument? $\endgroup$
    – Gary
    Feb 16, 2022 at 18:06
  • $\begingroup$ @Gary, done, with several other inaccuracies attended to. $\endgroup$ Feb 16, 2022 at 23:32
  • $\begingroup$ Thanks Jeremy. For any student who will come across this and will wonder about why $\alpha{}<1^{*}\implies{}1^{*}<\beta{}\,\,$: $\,\,\alpha{}<1^{*}\implies{}\exists{}q(q\in{}\mathbb{Q}^+\,\,\,\wedge{}\,\,\,q<1\,\,\,\wedge{}\,\,\,q\not\in{}\alpha{})$. 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 $\beta{}>1^{*}$. $\endgroup$
    – Gary
    Feb 17, 2022 at 19:14

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