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In Stephen Abbott's Understanding Analysis, when proving that the set of rational numbers is dense in the set of real numbers, Abbott picks an integer $m$ such that $$m - 1 \leqslant na < m,$$ where $n$ and $a$ are a natural number and a real number respectively.

Intuitively it makes sense that such an $m$ exists, but it seems to me that Abbott is taking liberties here. How do I know from the fact that the set of reals is a complete ordered field that such an $m$ exists? (Abbott doesn't discuss the real number axioms but I know them from elsewhere.)

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    $\begingroup$ I think that these things usually boil down to the Archimedean property of the reals. $\endgroup$ – MisterRiemann Sep 12 '18 at 8:25
  • $\begingroup$ Maybe it takes for granted the floor function $[na]$ . $\endgroup$ – dmtri Sep 12 '18 at 8:30
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    $\begingroup$ @Sobi Then construct a proof and post it as an answer. $\endgroup$ – PiKindOfGuy Sep 12 '18 at 9:35
  • $\begingroup$ I think you can find a proof of this (or at least a similar) claim in the first chapter of Rudin's Principles of Mathematical Analysis. $\endgroup$ – MisterRiemann Sep 12 '18 at 9:56
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How do I know from the fact that the set of reals is a complete ordered field that such an m exists?

From the Archimedean property you can recover the more general fact that for any real $r$ there is an integer $m$ such that $m-1 \le r < m$. Firstly note that it is obvious if $r$ is an integer. Next consider the case that $r > 0$. By the Archimedean property, there is a positive integer $k$ such that $1/k < 1/r$, since $1/r > 0$. But then there is a minimum $m$ of all such $k$. If $m = 1$, then $r < 1$ so you can finish trivially. If $m > 1$, then by minimality of $m$ we have $1/m < 1/r \le 1/(m-1)$, and hence $m-1 \le r < m$. The case of $r < 0$ is similar, but you need to be slightly more careful with the inequality.

From the (Dedekind-)completeness property you can also prove that same fact. Again consider the case that $r > 0$. Let $S = \{ n : n \in \mathbb{Z} \land n \le r \}$. Then $S$ is obviously non-empty, and hence has a supremum $c$ by the completeness property. Let $m \in S$ such that $m > c-1$, which exists by a basic property of the supremum. Then it cannot be that $m+1 \le r$, otherwise $m+1 \in S$ and so $m+1 \le c$. Therefore $m \le r < m+1$. The case of $r < 0$ is similar, but here too you need to be careful with the inequality.

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