How to Prove Existence of Integer n in Dedekind Cut Using Archimedean Property?

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The discussion focuses on proving the existence of an integer n in a Dedekind Cut α using the Archimedean property of the rational numbers Q. The key conclusion is that if no such integer n exists, then nw would be a member of α for every n, which contradicts the properties of α not being equal to Q. The proof relies on the relationship between members of α and the Archimedean property, establishing that for any rational q not in α, there exists an n such that nw > q, ensuring nw cannot belong to α.

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Shaji D R
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Let α be a Dedekind Cut. w a positive rational.How to prove that there exists a integer n such that nw is a member of α and (n+1)w is not a member of α, using Archemedian propoerty of Q.

Suppose p is a member of α. we can find n such that nw < p < (n+1)w. So nw is
a member of α. Further I am not able to proceed.

Please help me.
 
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OK. Looks like I got the answer. It was a trivial case. If there is no such n then nw is a member of alpha for every n by induction. Since aplpha is not equal to Q and Q is archemedian it is not possible.If q is not a member
of alpha then q > any member of alpha and nw > q for some n.And nw cannot be a member of alpha.

Sporry for asking such a simple question.
 

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