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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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Let $p\in 1^*$, $0 < p < 1$, it exists $n\in \mathbb N$ 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 a $m$ 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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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 –  Nameless Jul 27 '12 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. –  AlbertH Jul 28 '12 at 11:22
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