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 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
In order for $u \in \alpha^{-1}$ we must have that $0
EDIT: Found another proof here: http://mypage.iu.edu/~sgautam/m413.33418.11f/Dedekind.pdf
STill nothing similar to Rudin's...
Principles of Mathematical Analysis, Dedekind Cuts, Multiplicative Inverse
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real-analysis
elementary-set-theory
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0http://math.stackexchange.com/questions/156853/multiplication-inverse-for-dedekind-cut – 2012-07-27
2 Answers
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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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0Very 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
– 2012-07-27
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1I assumed $0 < p < 1$, because that was the only case you had difficulties with. Anyway, now, I explicitly mentioned that condition. – 2012-07-28
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0Sorry to reopen a question decided two years ago, but precisely why is it evident that $m \geq n$? Otherwise I agree completely... – 2014-08-31
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1Since $p < 1$, there exists an integer $n$ great enough to satisfy $p < 1 - (n + 1)^{-1}$. Then for each $m$ greater than $n$ inequality (1) holds. – 2014-09-01
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0@User12345 $0
$m=n-k$
for some $k\geq 1$ hence $(m+1)/n = (n-k+1)/n = 1+(1-k)/n\leq 1$. This implies $(m+1)q$(m+1)q\in\alpha$, which is contradictory. – 2016-01-28
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suppose $m
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0This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient [reputation](http://math.stackexchange.com/help/whats-reputation) you will be able to [comment on any post](http://math.stackexchange.com/help/privileges/comment). – 2014-09-19