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