# How does this proof of density of $\mathbb{Q}$ in $\mathbb{R}$ require $a \geq 0$?

My Real Analysis text offers the following proof of "given two real numbers a and b, with $a < b$, there exists a rational number r strictly in between of those two", conditional on $0 \leq a < b$. It also says that the case of $a < 0$ follows easily from this proof after small rearrangement. My problem is that I don't see where $a \geq 0$ comes into play in this proof, as I can't find where any of the proof's arguments are dependent on a's sign. Here's the proof:

We need to produce m, n $\in \mathbb{N}$ such that $a < \frac{m}{n} < b$.

By Archimedean Property we know that there exists $n$ such that $\frac{1}{n}<b-a$.

Proceed to $na < m < nb$. Now we need to pick such an m so that it's the smallest natural number greater than na; in other words $m-1 \leq na < m$. From here we get $m > na$, which is half way. Going back to our Archimedean Property equation, we can rewrite it as $a < b-\frac{1}{n}$, and thus write

$$m \leq na+1$$

$$< n(b-\frac{1}{n})+1$$

$$=nb.$$

Thus, conditions for $a < \frac{m}{n} < b$ are satisfied. Once again, I don't see where a's positivity plays a role here. Thanks!!

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If $a\lt 0$, then there may be no natural number $m$ such that $m-1\leq na\lt m$ (e.g., if $na\lt -1$). But this is not really a problem: if you pick $m$ an integer with the desired property, the argument follows through. In short: technically you need $a\gt 0$ to ensure you can pick $m\gt 0$, but there's actually no need to do so in the first place. So you're right that it is not really needed, if you tweak the argument slightly. Note that the author is trying to produce a positive rational, though I don't know why. – Arturo Magidin Jun 29 '11 at 21:22
If $a<0<b$ then $0$ is the rational number you are looking for. If $0\leq a<b$ then your proof finds two POSITIVE integers $m,n$ with $a<\frac{m}{n}<b$. For $b<a<0$ it is just a symmetry argument. – Beni Bogosel Jun 29 '11 at 21:24
@Arturo: Oh so it's integer vs. natural number, I somehow missed that when reading... Thanks! You can post the answer so that I can accept it and remove the question from the unanswered list. – confused Jun 29 '11 at 21:27
Done. – Arturo Magidin Jun 29 '11 at 21:34

(As requested, posted as an answer).

For some reason, the author wants to find a positive rational between $a$ and $b$ (note he is looking for natural numbers $n$ and $m$); of course, that will not happen in general. Technically, because he is requiring $m$ to be a natural number (hence nonnegative [or possibly positive, depending on whether the author's version of $\mathbb{N}$ includes $0$ or not]), it may be impossible to find such a number that satisfies $m-1 \leq na \lt m$ (e.g., if $na\lt -1$). So he is asking that $a$ be nonnegative to ensure that one can pick $m$ nonnegative that satisfies the condition.

That said, there is absolutely no need to divide the argument like this. The exact same argument works if we pick $m$ to be the unique integer such that $m-1\leq na\lt m$. The argument carries through then to produce a rational between $a$ and $b$, regardless of the signs of $a$ and of $b$. So you are right that the essence of the argument does not require $a$ to be nonnegative, and a minor tweak to the argument would eliminate the need for that assumption.

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Dear Arturo and @confused, Although user11867's answer below is slighty wrong in its interpretation of the argument, it may be getting at something correct: the author may feel the need to justify the existence of the minimal $m$, and may be appealing to well-ordering of the natural numbers to do so. (Of course, there are lots of ways to rephrase things so that, even with this level of detail, the argument can proceed without making the negative/positive distinction. But the author has possibly stated the well-ordering of natural numbers, and hence may find it easier just to appeal to it ... – Matt E Jun 29 '11 at 22:33
... in the manner written, rather than to try to ad lib at this point to justify the existence of $m$ in the context when it is negative.) Regards, – Matt E Jun 29 '11 at 22:34
@Matt E: I thought briefly, as I was writing the answer, that there might be an issue involving that, or that it could be what was behind the restriction. Good call. – Arturo Magidin Jun 30 '11 at 2:59
@Matt: I checked the book, and there's no prior mention of any well-ordering. Would invoking a well-order on integers as opposed to natural numbers work just as fine? – confused Jun 30 '11 at 6:01
@confused: If the author is invoking (rather than "implying") well-ordering to justify the existence of $m$, then to deal with $a\lt 0$, we can do as Beni Bogosel suggests: if $a\lt 0\lt b$, then take $0$ to be your rational. If $a\lt b\lt 0$, then use the author's construction to find a rational $r$ with $0\lt -b\lt r\lt -a$, and then take $-r$. – Arturo Magidin Jun 30 '11 at 18:09

Let us begin from the point where we have established that $na < m < nb$. Let $$S = \{k\in\mathbb{Z}: na < k < nb\}.$$ Since $m\in S$, we know that $S$ is nonempty. Since $a\ge 0$, we know that $S\subset \mathbb{N}$. Hence, by the well-ordering principle (every nonempty set of positive integers has a smallest element), we know that $S$ has a smallest element. Call it $m_0$.

Since $m_0\in S$, we know that $na < m_0 < nb$. But since $m_0$ is the smallest element in $S$, we know that $m_0-1\notin S$, and this implies $m_0-1\le na$. Thus, $m_0-1\le na<m_0$, and we may proceed with the rest of the proof.

I would guess that this is what the author(s) meant.

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