# Principles of Mathematical Analysis, Dedekind Cuts, Multiplicative Inverse

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 $$\alpha \cdot {\alpha }^{-1}=1^*$$ 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 $$\exists n\in \mathbb{N}:nq\in \alpha\text{ and }(n+1)q\notin \alpha$$ 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, $$u<r\Leftrightarrow \frac{p}{nq}<\frac{1}{(n+1)q}\Leftrightarrow p<\frac{n}{n+1}$$ 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...

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

• 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 Jul 27, 2012 at 13:29
• I assumed $0 < p < 1$, because that was the only case you had difficulties with. Anyway, now, I explicitly mentioned that condition. Jul 28, 2012 at 11:22
• Sorry to reopen a question decided two years ago, but precisely why is it evident that $m \geq n$? Otherwise I agree completely... Aug 31, 2014 at 23:52
• 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. Sep 1, 2014 at 8:34
• @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. Jan 28, 2016 at 16:02

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 $$1 and $$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 $$0 and $$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 $$\alpha\beta = 1^*$$.

• "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...
– Gary
Feb 15, 2022 at 19:27
• 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 Feb 16, 2022 at 0:22
• 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?
– Gary
Feb 16, 2022 at 18:06
• @Gary, done, with several other inaccuracies attended to. Feb 16, 2022 at 23:32
• 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 $\beta{}>1^{*}$.
– Gary
Feb 17, 2022 at 19:14