# Existence of a smallest integer greater than some real number

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.)

• I think that these things usually boil down to the Archimedean property of the reals. – MisterRiemann Sep 12 '18 at 8:25
• Maybe it takes for granted the floor function $[na]$ . – dmtri Sep 12 '18 at 8:30
• @Sobi Then construct a proof and post it as an answer. – PiKindOfGuy Sep 12 '18 at 9:35
• 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. – MisterRiemann Sep 12 '18 at 9:56

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.