Principles of Mathematical Analysis, Dedekind Cuts, Multiplicative Inverse

Question

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...

Answer 1

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 $$ p = (mq)\cdot \frac p {mq}. $$

Answer 2

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 $\alpha\beta = 1^*$.