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This question refers to the construction of $\mathbb{R}$ from $\mathbb{Q}$ using Dedekind cuts, as presented in Rudin's "Principles of Mathematical Analysis" pp. 17-21.

More specifically, in the last paragraph of step 4, Rudin says that for $\alpha$ a fixed cut, and given $v \in 0^*$, setting $w=- v / 2$, there exists an integer $n$ such that $nw \in \alpha$ but $(n+1)w$ is not inside $\alpha$. Rudin says that this depends on the Archimedean property of the rationals, however he has not proved it.

Could somebody prove the existence of the integer $n$?

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By definition there is some rational $q$ that is not in $\alpha$. Let $p=|q|+1$; then $p>q$, so $p\notin\alpha$, and $p>0$. The rational $w$ is also positive, so by the Archimedean property of the rationals there is a positive integer $m$ such that $mw>p$, and therefore $mw\notin\alpha$.

We also know that there is some rational $r\in\alpha$. Let $s=-|r|-1$; then $s, so $s\in\alpha$, and $-s>0$, so again by the Archimedean property there is a positive integer $k$ such that $kw>-s$. But then $(-k)w, so $(-k)w\in\alpha$. We now have positive integers $k$ and $m$ such that $-kw\in\alpha$ and $mw\notin\alpha$.

Let $S=\{i\in\Bbb N:(i-k)w\notin\alpha\}$; $m+k\in S$, so $S\ne\varnothing$. Since $\Bbb N$ is well-ordered, $S$ has a smallest element, $j$. Note that $0\notin S$, since $-kw\in\alpha$, so $j>0$. Let $n=(j-1)-k$. Then $nw\in\alpha$, and $(n+1)w\notin\alpha$.

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    @Addison: It’s rigorous as the term is normally used. To make it any more rigorous, one would have to go back to the construction of the rationals and the verification that they are an ordered field and recast the argument in that context, verifying that it still works. I’d say that you’re probably not missing anything.2016-10-02
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If $v\in 0^*$, then $v<0$, so $w>0$. Let $\gamma=\sup \alpha$. The Archimedean property says that for any $\gamma$ there is some integer $m$ such that $mw\geq\gamma$. Since $\mathbb{N}$ is well-ordered, there is a smallest $m$ such that $mw\geq\gamma$, call this $n+1$. This means that $nw<\gamma$, and hence $nw\in \alpha$. But also $(n+1)w\notin \alpha$ since it is larger than $\gamma$ (OK, so there might be some subtleties with endpoints you should think about, but essentially this is the idea).

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    The archimedean property says that there is a positive integer $m$ such that mw > \gamma, or m > -20, as you said. Just to be sure, by the second fact to "be used freely" (written just before step 2): if $r \notin \alpha$ and r < s then $s \notin \alpha$, it follows that $m \notin \alpha$, although we already knew this since $m$ is bigger than the supremum. Since $m \in \mathbb N$ and you are using its well-ordering, the smallest possible value for $m$ is $1$. If $m = n + 1$, then $n = 0$. It follows that $nw = 0 \notin \alpha$ and $(n+1)w \notin \alpha$.2014-12-01