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I was reading Calculus I by Tom Apostol and it has a question as following: Let $ x $ be any real number, prove that there exists exactly $ 1 $ integer $ n $ such that $ n \leq x < n + 1. $ In this exercise I am allowed to use the fact that the sets $ \mathbb R $ and $ \mathbb Z $ are both unbounded.

Here is my approach, I use proof by contradiction. Suppose, on the contrary that there is either no such $ n $ exists or there are more than $ 1 $ integers $ n $ that satisfy the condition.

If no such $ n $ exists, it means that for every $ x \in \mathbb R $ and for every $ n \in \mathbb Z $, either $ n > x $ or $ x \geq n + 1. $ It therefore means that $ \mathbb R $ is bounded above or bounded below, which is a contradiction. Hence there must be at least $ 1 $ integer $ n $ such that $ n \leq x < n + 1 \; \; \; \forall x \in \mathbb R. $

Now suppose that there are $ 2 $ distinct integers $ n \neq n' $ such that $ n \leq x < n + 1 $ and $ n' \leq x < n' + 1 \; \; \; \forall x \in \mathbb R. $ It follows that $ n - n' - 1 < 0 < n -n' + 1, $ or equivalently $ -1 < n - n' < 1. $ Since $ n - n' \in \mathbb Z, $ it must be the case that $ n - n' = 0, $ but this is a contradiction since $ n \neq n'. $ Hence there cannot be $ 2 $ distinct integers both satisfy the condition.

Can some please proofread my proof to see if there are any subtle errors? Thanks.

Jon Warneke
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  • Your proof of unicity seems correct to me (except that you get $-1<n'-n<1$ and not $-1<n-n'<1$). As for the existence proof, you seem to have forgotten that $x$ was fixed? – Guest Dec 13 '15 at 03:51
  • I see what you mean for the existence part, so if instead I say "For every $ x \in \mathbb R $, if no such integer $ n $ exists, then either $ n > x $ or $ x - 1 \geq n $ for every integer $ n $, meaning $ \mathbb Z $ is bounded above or below, which is a contradiction." then it would be correct right? –  Dec 13 '15 at 07:05

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For a totally ordered field $K$, the existence of the "largest previous integer" or "floor function" is one of the characterizations of $K$ being "archimedean". One definition of $K$ being archimedean is: $$ \forall k\in K,\exists n\in\mathbb{N}\text{ s.t. }k\leq n\tag{1} $$ Note that for this definition to make sense, $K$ must contain a copy of $\mathbb{N}$; in fact, one can show that every totally ordered field indeed contains a subfield isomorphic to $\mathbb{Q}$.

Presumably you know $(1)$ for $\mathbb{R}$ at this stage. If so:

Let $x\in\mathbb{R}$. Consider the set $$ E:=\{p\in\mathbb{Z}:x<p\} $$ By $(1)$, $E\neq\emptyset$. Also, $E$ is bounded below. Indeed, by $(1)$, you can find $s\in\mathbb{N}$ such that $-(x-1)\leq s$, that is, $s\leq x-1<x$ and so $s$ is a lower bound. Hence, $E$ is a subset of $\mathbb{Z}$ which is bounded below and so $E$ has a minimum, say $n+1$. Then $$ n\leq x<n+1 $$ where the left inequality comes from the minimality of $n+1$ in $E$.

Guest
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