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I was trying to show on my own that for every two reals $x,y$ (such that $x<y$) then there is a rational number $p$ in between them. i.e.:

$$\forall x \in R, \forall y \in R, \exists q \in Q : x<q<y $$

I was following Rudin's analysis book so technically I only know so far the Archimedean property of R and there exists an ordered field with the least upper bound property and the greatest lower bound property (more precisely I only know up to theorem 1.20).

I attempted to prove the theorem without looking at the answer (even when I looked at it it didn't provide much help really). I feel I have given it an honest attempt and it feels like I am really close to the answer (or at least applied some the facts I know non-trivially) so I was hoping I could get some hints on my current approach (if it has any chance of success) or what I am missing. I am fine if someone gives me the answer but I'd prefer it that part of the answer is hidden or if I mostly receive hints/ideas. So this is what I have tried:

We want $x < p < y$. Since $p$ is a rational we can have integers such that $x < \frac{n}{d} < y \iff dx < n <dy$. At this point is where I realized that maybe the Archimedean property might be useful. Lets focus on the RHS of the inequality first: $n <dy$. Recall the Archimedean property from Rudin's:

$$ \forall x \in R, \forall y \in R, x > 0, \exists n \in N : nx > y $$

i.e. there always exists a natural number to make a real number bigger than any other real number (or equivalently when x=1 there is always a natural number that bounds a real number). Now that we've recalled it we notice $n <dy$ is exactly of the form of the Archimedean property in the sense that for any real $n$ (and also any natural number $n$) and any real $y$, there exists a natural number $d$ that makes $n <dy$. The important thing to notice is that some $d$ exists and that this holds for our specific $y$ we are interested in. Notice we have not chosen $n$, just asserted the existence of such a $d$.

Then at this point I thought that maybe I needed to use the Archimedean property on the LHS. The LHS of what we want is $dx < n$. Unfortunately this seemed backwards from what I needed so I negated it and got $-d' x > n \iff d' (-x) > -n = \hat n$. This holds for some $d'$ (maybe different from the first $d$ I mentioned) and for all $-x \in R, -x > 0$ and any $-n \in N$ (in particular any natural number). I tried combining this with the first statement I got by requesting $\hat n = n$ (which is true since we the two statements held for any $\hat n$ and $n$) but that yielded $ d(-x) > n < dy $ which didn't seem useful.

Anyhow, I didn't give up and so I tried something else with the LHS $dx < n$. I noticed that when $d=1$ then $x < n$ is in fact the Archimedean property. In other words there is some real number $n$ that makes $x < n$. Since the RHS $n <dy$ holds for any $n$ I choose $n$ that made $x < n$ true. Thus I finally got something that connected $x$ and $y$ with only (seemingly) true statements:

$$ x < n < d y$$

then I divided by $d$ to get:

$$ \frac{x}{d} < \frac{n}{d} < y$$

Unfortunately, I can't call it a day since $\frac{n}{d}$ might be smaller than $x$.

At this point I noticed I could define the set $A$ as follow:

$$ A = \{ q \in Q: q < y \}$$

Notice $\frac{n}{d}$ is constructed from specific $n$ and $d$ that makes $x<n$ true and $n < dy$. Thus $\frac{n}{d}$ exists and its in $A$ (i.e. $\frac{n}{d} \in A$). Thus $A \not = \emptyset $.

At this point it was obvious that since $A \subset R$ (non empty) and $R$ has the least upper bound (LUB) and greatest lower bound (GLB) property, then it was clear that $A$ had a supremum. Let that supremum be $\alpha = sup A = LUB(A)$. At this point I noticed that if $y \in Q \implies y \in A$ and can take the average of $\frac{n}{d}$ and $y$ to construct the $q$ in between $x < q < y$ that we wanted. Not a very rigorous ending to this case but at least I had an idea that it could be done.

However, when $y \not \in Q$ seems to be the challenging case. Thus, I noticed that to complete the proof I just had to show that I can always construct a smaller fraction that is larger than $\frac{n}{d}$ but smaller than $y$ and eventually get it large enough that its larger than $x$. It seems thats the piece of the proof that I might be missing. What hints can people provide to show that indeed we con construct such a $q \in Q$? Or maybe this direction for the proof is not correct and I should go back? My intuition tells me right now that the Archimedean property has to be important because it the only statement that I know that generates relations between natural numbers and real numbers, so it feels I have to apply that property at least twice (to the RHS and LHS) to then generate two natural numbers that I can divide to generate the fraction I am looking for. Anyone has any ideas/hints alongs these lines?

As a side thought I remember that example 1.1 (in Rudin) had me show nearly an identical thing, that $A = \{ q \in Q_+ : p^2 < 2 \}$ and show there was no largest number in $A$. Unfortunately, that proof seemed to strongly rely on the fact of squares and square roots as shown here. Furthermore, it feels that I could just recycle that proof (and maybe with some additional logic show it is eventually is larger than $x$ but less than $y$) to compete this proof, however, it feels like cheating because I don't actually know right now that every real number has a square root (a fact introduced next page that seems to use the fact I am trying to prove right now).

So I've thought about it a bit right now but I seem a bit stuck. Ideally I want hints but the actual answer is fine too, though, it would be nice to hid that part or at least warn me its the answer. Ideally I'd like to think for myself as much as I can but right now I feel a bit disoriented.

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Charlie Parker
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    Sorry, I'm a bit too woolly in the head to read and understand the whole thing now, but as a variation of your approach, think about how you could finish when you use the Archimedian property to get a $d$ such that $d\cdot (y-x) > 1$. – Daniel Fischer Dec 24 '16 at 21:20
  • How do you define $<$ for real $x,y$? – copper.hat Dec 24 '16 at 21:20
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    @copper.hat: Following Rudin, we are assuming here that the reals are a complete ordered field. So we don't have to define $<$. – Nate Eldredge Dec 24 '16 at 21:22
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    Yeah, I think you are going down a rabbit hole here, trying to use the axioms to push symbols until you get what you want. Remember that you actually know how the real numbers work, even though you can't use all those facts in your proof. If you knew everything about the real numbers, how would you actually go about finding $n,d$ for a particular $x,y$? Once you know how you would do that, then try to use the axioms you have to prove that it works. – Nate Eldredge Dec 24 '16 at 21:24
  • @NateEldredge: Thanks Nate, but there still needs to be a connection between that order and the order on the rationals. – copper.hat Dec 24 '16 at 21:25
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    @copper.hat: From the field axioms you can define "integer" as any real number of the form $\pm (1 + 1 + \dots + 1)$. Then you can define "rational" as any real number of the form $n m^{-1}$ where $n,m$ are integers and $m \ne 0$. Then it's an easy exercise to show that the given order on the reals induces the usual order on "rationals". I think at this stage of development, we get to assume this has been done. – Nate Eldredge Dec 24 '16 at 21:28
  • @NateEldredge: Thanks, I fell into the rabbit hole :-). – copper.hat Dec 24 '16 at 21:29
  • @NateEldredge I'm assuming your second comment is addressed to me. This is probably not what you are referring to but I can't think of anything else about the real numbers I would to to find a $n$,$d$ for particular $x$,$y$, except appeal that $(x+y)2$ is in between $x$ and $y$ and since $(x+y)2$ is a dedekind cut (i.e. all rationals bellow it), then by definition there must be a rational between $x$ and $y$. Is that what you were referring with your second comment or what did you have in mind? (maybe another hint? or making it a bit more explicit? though that was definitively useful!) – Charlie Parker Dec 24 '16 at 22:20
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    I think Eric's answer says it better, but think about a specific example. Say $x = 5.300001$ and $y = 5.300002$. How are you going to choose $d$? $d=1$ will definitely not work, so that line of reasoning is going nowhere. Indeed, this shows that your choice of $d$ cannot only depend on how large $x,y$ are; it has to depend on how close they are to each other. – Nate Eldredge Dec 24 '16 at 22:25
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    Remember that "Dedekind cut" isn't something you can use here. All you know about $\mathbb{R}$ is that it's a complete ordered field. Maybe you know that it is possible to construct a complete ordered field from Dedekind cuts, but it would take a lot more work to show that every complete ordered field looks like that. – Nate Eldredge Dec 24 '16 at 22:28
  • @DanielFischer sorry for the long question. It was (very) hard for me to know what to leave out. Anyway, why should I consider $d(y-x) > 1$ and not say, greater than 0.5 or 2? – Charlie Parker Dec 24 '16 at 23:06
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    @Pinocchio: If you only had $d(y-x) > 0.5$, you could end up with, say, $dy = 506.9$ and $dx = 506.3$. (This would happen with $x=10.126$, $y=10.138$, $d=50$.) And you can't find an integer $n$ between $dy$ and $dx$, can you? Put another way, how far apart do two real numbers have to be in order to guarantee that there is an integer between them? – Nate Eldredge Dec 24 '16 at 23:42

1 Answers1

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The key thing you're missing is that you need to somehow take into account how much greater $y$ is than $x$. To provide a bit of intuition, imagine you were in a non-Archimedean world and $y$ was "infinitesimally" greater than $x$. For instance, maybe $x=0$ and $y$ is some positive but infinitesimal number (that is, $y<1/n$ for every positive integer $n$). Then you wouldn't be able to find any rational number between $x$ and $y$, since there are no (nonzero) infinitesimal rational numbers.

So this suggests that you should apply the Archimedean property not to $x$ or $y$ themselves, but to the difference $y-x$. That is, you want to choose $d\in\mathbb{N}$ such that $d(y-x)$ is large. Just thinking intuitively, how large would you need $d(y-x)=dy-dx$ to be to ensure there is an integer $n$ between $dx$ and $dy$? Choosing $d$ such that $d(y-x)$ is that large, you can now try to prove that such an $n$ really exists.

The details of one way to carry this out are hidden below.

Intuitively, if $dy-dx>1$, then there should be an integer between $dx$ and $dy$. So choose $d$ such that $d(y-x)>1$, by the Archimedean property. Now by the Archimedean property, there exist integers $a$ and $b$ such that $a<dx<b$. The set $S=\{c\in\mathbb{Z}:c>dx\}$ is thus nonempty and bounded below. Any nonempty bounded-below set of integers contains a least element; let $n$ be the least element of $S$. Then $n>dx$. If $n\geq dy$, then $n-1\geq dy-1>dx$ so $n-1\in S$, which contradicts minimality of $n$. Thus $n<dy$ and so $\frac{n}{d}$ is between $x$ and $y$.

Eric Wofsey
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  • Wow Eric, thanks so much. That was really insightful and useful for me. :)

    I have a follow up though. Why is it that the proof doesn't finish once you assert $dy - dx > 1$? Since that is definitively true by the Archimedean property, then it forces some integer to exist between $dx$ and $dy$. Since there exists one natural number such that $dx < n < dy$ then the proof is complete (since there difference is greater than 1). Thus, my question is, why do you start talking about $a,b$, isn't the proof done after your second sentence? Or am I missing or asserting something without proof?

     – Charlie Parker Dec 24 '16 at 23:25
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    Well, that depends what facts you already know. It is not immediate from the definition of a "complete ordered field" that if $t-s>1$, then there exists an integer between $s$ and $t$. So this statement requires proof, though it is possible you have already seen this statement as a theorem. – Eric Wofsey Dec 24 '16 at 23:29