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Let $\alpha \in P_R$ be a cut. Since there exists a cut that is not $\{q\in Q\mid q 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?2012-06-11
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    $\alpha \beta$ = {$p\in Q$|There exists $0 and $0 such that $p≦st$}2012-06-11
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    It's a definition for positive reals2012-06-11
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    Thanks; I’ll give it some thought.2012-06-11
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    It's equivalent to $0^* \cup$ {st | $0≦s \in \alpha$ and $0≦t \in \beta$}2012-06-11

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We have $\alpha\gamma=0^*\cup\{0\}\cup\{st:0t(1/r\notin\alpha)\}$. 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

We know that $q=a/b$ for some positive integers $a$ and $b$ such that $a$$q^{-n}=\left(\frac{b}a\right)^n=\left(1+\frac{b-a}a\right)^n\;.$$

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.