# Multiplication inverse for dedekind cut

Let $\alpha \in P_R$ be a cut. Since there exists a cut that is not $\{q\in Q\mid q<r\}=r^*$ for every $r\in Q$, $\alpha$ doesn't need to be of the form $r^*$.

Let $$\gamma= 0^* \cup \{0\} \cup \{q\in P_Q\mid\text{ there exists }r\in P_Q\text{ such that }r>q\text{ and }1/r \notin \alpha\}\;.$$

I have proved that $\alpha \gamma$ is a subset of $1^*$. I dont't know how to prove $1^*$ is a subset of $\alpha \gamma$. Help

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What’s your definition of the product of two cuts? – Brian M. Scott Jun 11 '12 at 7:19
$\alpha \beta$ = {$p\in Q$|There exists $0<s \in \alpha$ and $0<t \in \beta$ such that $p≦st$} – Katlus Jun 11 '12 at 7:31
It's a definition for positive reals – Katlus Jun 11 '12 at 7:32
Thanks; I’ll give it some thought. – Brian M. Scott Jun 11 '12 at 7:33
It's equivalent to $0^* \cup$ {st | $0≦s \in \alpha$ and $0≦t \in \beta$} – Katlus Jun 11 '12 at 7:34

We have $\alpha\gamma=0^*\cup\{0\}\cup\{st:0<s\in\alpha\text{ and }0<t\text{ and }\exists r>t(1/r\notin\alpha)\}$. For positive $t$ the condition that there is some $r>t$ such that $1/r\notin\alpha$ is equivalent to the condition that there is a positive $r<1/t$ such that $r\notin\alpha$, i.e., that $\alpha\subsetneqq(1/t)^*$.
Let $q\in 1^*$; clearly $q\in\alpha\gamma$ if $q\le 0$, so assume that $0<q<1$. Suppose that $q\notin\alpha\gamma$. Then for every positive $s\in\alpha$ and $t$ such that $\alpha\subsetneqq(1/t)^*$, $q\ne st$. Equivalently, for every positive $s\in\alpha$, $\alpha$ is not a proper subset of $(s/q)^*$, i.e., $(s/q)^*\subseteq\alpha$. Fix a positive $s_0\in\alpha$. Given $s_n$ for some $n\in\omega$, let $s_{n+1}=s_n/q$; an easy induction shows that ${s_n}^*\subseteq\alpha$ for each $n\in\omega$. But $s_n=s_0q^{-n}$, so you’ll have the desired contradiction showing that $q\in\alpha\gamma$ once you show that the sequence $\langle q^{-n}:n\in\omega\rangle$ is unbounded in $\Bbb Q$.
We know that $q=a/b$ for some positive integers $a$ and $b$ such that $a<b$, so $$q^{-n}=\left(\frac{b}a\right)^n=\left(1+\frac{b-a}a\right)^n\;.$$
Let $p=\dfrac{b-a}a$; then $q^{-n}=(1+p)^n$, and it’s an easy induction to show that $(1+p)^n\ge 1+np$. Since $\langle 1+np:n\in\omega\rangle$ is clearly unbounded, we’re done.