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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 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 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 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$n$ (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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    http://math.stackexchange.com/questions/156853/multiplication-inverse-for-dedekind-cut2012-07-27

2 Answers 2

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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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    @User12345 0 and mq then (m+1)q. If m 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, then $(m+1)q\in\alpha$, which is contradictory.2016-01-28
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suppose $m, then $m\le n-1$ and $m+1 \le n$. As a result,$(m+1)q\le nq and hence $(m+1)q$ belongs to $\alpha$, which contradicts the choice of $m$.

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