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I posted this problem yesterday and Brian gave me really nice answer using Bernoulli inequality, but I think this can be proved with the concept of Archimedean property of $ \mathbb{Q}$ and greatest or least element of a bounded subset of $ \mathbb{Z}$, not induction. I know that the 'Way' of the proof doesn't matter, but it's the first time I have learned how to construct reals, so I want to keep this proof in my way..

Let $\alpha \in P_R$ be a cut. Since there exists a cut that is not $\{q\in \mathbb{Q}\mid q<r\}=r^*$ for every $r\in \mathbb{Q}$, $\alpha$ 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

Multiplication of positive reals is defined as; $\alpha \beta$ = {$p\in \mathbb{Q}$| There exists $0<r \in \alpha$ and $0<s \in \beta$ such that $p≦rs$} This definition is equivalent to $0^* \cup$ {$rs$|$0≦r \in \alpha$ and $0≦s \in \beta$}

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With Brian's help, I have proved that when $p \in 1^*$, "If $p \notin \alpha\gamma$, then for every $0<s \in \alpha$ and $0<t \in \gamma$, $(s/p)^*\subsetneqq\alpha \subsetneqq(1/t)^*$" – Katlus Jun 12 '12 at 0:53
Now i proved it thanks – Katlus Jun 12 '12 at 3:04
Would you post your proof as an answer? – Brian M. Scott Jun 12 '12 at 3:23
I used induction which is not actually i was looking for..and it's almost same as yours, but i used induction on multiplication rather than exponentiation i.e. Let A={$n\in \omega$ | $ns/p \in \alpha$} and with the inequality above and induction to show that A=$\omega$. Then since $p,t,s$ are positive rational numbers and $Q$ is Archimedean $(pp)/(ts) < m$ for some $m\in \omega$. Since A=$\omega$, $m\in$ A. Thus $ms/p \in \alpha$. Since $(pp)/(ts) < m$, $p/t \in \alpha$. Then $t(p/t)$ = p, which is a contradiction. – Katlus Jun 12 '12 at 11:56

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