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
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 such that $p≦rs$} This definition is equivalent to $0^* \cup$ {$rs$|$0≦r \in \alpha$ and $0≦s \in \beta$}