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A version of the proof that $\mathbb{Q}$ is everywhere dense in $\mathbb{R}$ (the one in "Stephen Abbott: understanding analysis") goes like this: Let $a,b\in\mathbb{R}$ and $a<b$. For simplicity assume $a\ge0$ ($a<0$ will be proved later). A rational number is a quotient of two integers so we must produce $m,n\in\mathbb{N}$ such that:$$a<m/n<b.....(1)$$ Then we choose $n$ so that $$1/n<b-a.....(2)$$ (existence of $n$ guaranteed by A.P. (Archimedean property)).Multiplying $(1)$ by $n$ gives $na<m<nb$. We now choose $m$ to be the smallest natural number greater than $na$ i.e$$m-1\le^{(3)}na<^{(4)}m$$$(4)$ directly gives $a<m/n$ so half the work is done. $(2)$ can be written as $a<b-1/n$. From $(3)$: $$m\le na+1$$ $$\quad\quad\quad\quad\quad<n(b-1/n)+1$$ $$=nb$$ $$\therefore m<nb\quad\implies m/n<b$$ The case in which $a<0$ can be proved without much work by converting it into the above case.QUESTION: I fail to understand why we have to take seperate cases for $a\ge 0$ & $a<0$. Assuming the problem is in $(3)$ & $(4)$, cant we just take $m\in \mathbb{Z}$ so that $(3)$ & $(4)$ are possible for -ve $a$ too ? I know $(4)$ requires the A.P. which is usually proved for $\mathbb{N}$ but isnt it obvious it holds for $\mathbb{Z}$ too ? Is there any other step which requires $a$ to be +ve or zero ?

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  • $\begingroup$ See this note for a slightly different proof. $\endgroup$ Jun 27, 2014 at 7:31
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    $\begingroup$ Shouldn't it be "smallest natural grater than.." (before (3) and (4))? Probably they have proved before that there exists a "smalles natural number in a nonempty set of natural numbers". $\endgroup$ Jun 27, 2014 at 7:32
  • $\begingroup$ Clearly $m=na+1$, right? So there is no need to be concerned with existence of such $m$ unless we are down to the level of Peano's Axioms or something similar ... $\endgroup$
    – String
    Jun 27, 2014 at 7:33
  • $\begingroup$ @PeterFranek: The OP writes "no" meaning "number" which is quite distracting, I must admit ... $\endgroup$
    – String
    Jun 27, 2014 at 7:34
  • $\begingroup$ Ok, I see :-) However, $m=na+1$ doesn't hold in general, only for rational $a$. So, you need to have some argument to show that $m$ exists. I haven't read the book but a standard argument is that every nonempty set of natural numbers has a minimum. $\endgroup$ Jun 27, 2014 at 7:37

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